text:
What are the odds in Scrabble of not being able to make a legal move on opening turn
I've been playing the game for 50 years and this happened to me recently:
I had the following rack
PBQZJDH

And so I could not  play a legal move.
I've been trying to work out the odds of this (no legal word on the first turn) happening.

I was going to calculate the odds of a consonant-only rack as the lower bound, but I was surprised to learn there are 40 valid Scrabble words with no vowel (Y included).

From a computer science perspective, this is a very interesting question.  My first guess would be to pare down the legal scrabble dictionary by removing all words that a subset of the letters forms another word (for example, AND would be removed because AN is a valid word), then for each word find the number of possible pulls from the bag that can form that word and no other word previously examined.  Problem is this this at least an NP problem, and we probably don't have enough time left in the universe to calculate it.

Brute-forcing all 9.33x10^157 possible combinations of letters is a non-starter, and the fact that it happened to you means that the number of possible combinations where there's no legal letter is probably in the ballpark of 10^150 (plus or minus a few digits).  I'm looking at the dictionary now and a cursory glance at the list shows that a good chunk of the 280K words can be removed as there is a subset of those words are also a word (first 900 words or so can be removed just by the use of AA and AB)

Did I just get nerd sniped?

I would say the odds must be astronomical.

I had the following rack. PBQZJDH.

@DenisS Where did you get  9.33x10^157?

Can you estimate:  (1) how many games you've played in 50 years.  (2) how many times you've seen this happen?  That will serve as a sanity check on any results that come out of the computations....

@John: Sufficient to closest order of magnitude for this purpose: so estimate  50 * 365 < N < 10 * 50 * 365; which simplifies to 18,000 < N < 180,000 and we use N = 18,000 * pi. (As an old slide rule master, pi is a ready and convenient approximation for sqrt 10 .)

(NOTE : Final revision of my original answer)
The odds of the first tileset not having a single valid word is exactly 91,595,416 / 16,007,560,800 or .5722%, with it occuring once every 174.76378 games.  This value is calculated by using the dictionary found in this answer, but can be adapted for any other dictionary.

This was brute-forced via python.  Code will be available at the end of the answer.

We start by noticing that any tileset that contains at least one blank tile can play a valid word.  The only letter that does not form a two letter word is the letter "V", and there are only 2 "V" tiles in a scrabble game.  Therefore, any tileset with one blank can form a valid word, because (?VV....) must contain 4 more letters, one of which will combine with the "?" to form a valid word.
We also discovered, after the fact, that all possible tilesets that contain the letter "O" are also, in fact, valid.  The only 6 tile tileset that contains "O" and cannot form a word is "OCCVVQ" and any other letter will form a word. "A" is the only other letter that doesn't form a two letter word with "O" but "AVO" and "OCA" are both valid words, and you can't include more than one O or one A or else you can form "AA" or "OO".
These observations remove 53.32% of all possible tilesets, and allows us to perform the rest of our calculations using a 90-tile scrabble set that has no blanks and no Os in it.
We then remove from the dictionary all words that satisfy one of the following criteria

Word is longer than 7 letters long (can't spell an 8 letter word on the first turn)
A subset of the tiles required to form that word can also make another word (we're just looking to see if we can form a valid word, therefore there is no reason to check to see if a tileset can form the word "AND" when "AN" is a perfectly valid word as well)
Word requires at least one blank tile to solve (any tileset with a blank is valid, therefore no reason to examine these words).  These are the words FUFF (3rd F), JUJU (2nd J), KECK and KUKU (2nd K), SYZYGY (3rd Y), ZIZ, ZUZ, and ZZZ (2nd/3rd Z).
Word requires an O to solve.  All O tilesets are valid and we've removed them from the bag.

By applying these three rules, we can reduce the scrabble dictionary down to 149 words out of the original 280k~.

AA
AB
AD
AE
AG
AH
AI
AL
AM
AN
AR
AS
AT
AW
AX
AY
BE
BI
BRR
BUB
BUD
BULK
BULL
BY
CH
CIRRI
CIVIC
CLY
CRY
CUB
CUD
CUE
CUFF
CULL
CUZ
CWM
DE
DI
DRY
DUD
EE
EF
EGG
EH
EL
EM
EN
ER
ES
ET
EUK
EW
EX
FA
FLU
FUB
FUCK
FUD
FY
GHYLL
GI
GRR
GU
GYP
HI
HM
HYP
ICE
ICY
IF
IN
IS
IT
IVY
IWI
JA
JEU
JIZ
JUD
JUKU
KA
KEG
KI
KUDU
KUZU
KY
LI
LUCK
LUD
LULL
LUV
LUZ
MI
MM
MU
MY
NTH
NU
NY
PA
PE
PFFT
PHT
PI
PLY
PRY
PWN
PYX
QI
QUA
RHY
RIZ
SH
SLY
SPY
ST
SWY
THY
TRY
TWP
TYG
TYPP
TYPY
UH
ULU
UP
UR
US
UT
UVA
VAC
VAV
VEG
VIE
VLY
WHY
WIZ
WRY
WUD
WULL
WUZ
XI
XU
XYLYL
YE
YIRR
YU
ZA
ZE
ZO

At this point, we are going to brute force the number of invalid tilesets to get the numerator.  We do this by examining tilesets that are in alphabetical order, but do not care if it's unique.  For example, we will examine (A1A2BCDEF) and (A1A3BCDEF) but not (A2A1BCDEF), where A1, A2, and A3 refer to distinct A tiles in the bag.
The first tileset examined is (A1A2A3A4A5A6A7), followed by (A1A2A3A4A5A6A8), then (A1A2A3A4A5A6A9), and then finally (A1A2A3A4A5A6B1).  We continue down that path in the way that only a computer can do to determine every single alphabetized combination.
(Note that in the final version of my code, we change the definition of "alphabetical order" in order to get a speed improvement, but the final solution is the same.  Also, for speed reasons, we would never actually examine (AA.....) because "AA" is a valid word, so we would skip all tilesets of the form (AA.....))
After running through all the tilesets, we end with a final count of 91,595,416 tilesets.  This is our numerator, and the denominator is very easily calculated

100! / ((100-7)! * 7!) = 16,007,560,800

This is the calculation for finding the number of combinations possible, without replacement, and not caring about order (which we don't because otherwise this problem would be an few orders of magnitude harder to brute force.

I'm going to be putting some interesting notes down here as I play around with the program.

There is exactly 1 distinct tileset that has an "A" in it and is still invalid.  This tileset is "ACCUUUU".  The proof is left as an exercise for the reader.
Of the 91,595,416 invalid tilesets found, there are only 22,308 that contain exclusively vowels.  This is a much smaller number than I originally thought, but from reading the valid word list it makes sense.  As previously discussed, A only has the invalid tileset "ACCUUUU" and all O tilesets are valid.  Combined with the fact that "EE" is also a valid word, all vowel only tilesets without a valid word have to be of the form "EI...U..." or "I...U...".
Tilesets with a mix of vowels and consanants takes up a slightly larger amount of invalid tilesets, at 106,201 possible combinations.
Shocking no one, the vast majority of invalid tilesets are consonant only tilesets.  Of the 91,595,416 invalid tilesets, 91,466,907 of them are consonant only sets, which makes up 99.86% of all invalid tilesets.
Once the dictionary is purged of words that contains the letters [AEIOUY] there are only 13 words that can still be formed.  These 13 words can be, in turn, checked by looking at the letters [HPMRS].  This means that any tileset that contains nothing but the letters [BCDFGJKLNQTVWXZ] is invalid.  This comprises 15 of the 26 letters and 38 of the tiles in the bag.  Tilesets that contain only these letters comprise roughly 12% of all invalid tilesets.

Code, if anyone wants to run it for themselves (WARNING: slow, running on a beefy computer it still about a half hour to spit out an answer)
words = []

words_in_dictionary = 0
words_short_enough = 0

def is_all_c(arr1):
   for c in arr1:
      if c in ['A','E','I','O','U']:
         return False
   return True

def is_all_v(arr1):
   for v in arr1:
      if v in ['B','C','D','F','G','H','J','K','L','M','N','P','Q','R','S','T','V','W','X','Y','Z']:
         return False
   return True

#basic function to check if an array, arr1, is fully within a second array, arr2
def i_in_j(arr1, arr2):
   for a in arr1:
      if a in arr2:
         arr2.remove(a)
      else:
         return False
   return True

#basic function to check to see if word can be made from the tileset passed in
def is_valid_tileset(tileset):
   for word in words:
      if i_in_j(word["sorted"][:], tileset[:]):
         return word["base"]
   return None

# we have a big dictionary, 270k-ish words long, we need to load it
print("loading dictionary")
with open("dictionary.txt", "r") as dictfile:
   for line in dictfile:
      words_in_dictionary = words_in_dictionary + 1
      base_word = line.strip()

      #we are going to parse out the words longer than 7 letters now because it's quicker
      #we are also going to get rid of words with "O" in it
      if len(base_word) <= 7:
         if not "O" in base_word:
            words_short_enough = words_short_enough + 1
            word = {"base": base_word, "sorted": sorted(base_word)}
            words.append(word)

print("total words in dictionary is " + str(words_in_dictionary))
print("words 7 letters or shorter is " + str(words_short_enough))

# now we need to build our dictionary of unique words
# any word where a subset of the letters in that word can be used to build another word will be discarded
# for example, "AND" will be discarded because we can make "AN" out of that word
i = 0
while i < len(words):
   temp_sorted_working_word = words[i]["sorted"]

   j = 0
   while j < len(words):
      if i == j:
         j = j + 1
         continue

      if i_in_j(words[i]["sorted"][:], words[j]["sorted"][:]):
         del words[j]
         if i > j:
            i = i - 1
            j = j - 1
         elif j > i:
            j = j - 1

      j = j + 1
   i = i + 1

# there are also 8 words from this list that cannot be built without blanks, and we know that any tileset
# with at least one blank is valid
i = 0
while i < len(words):
   if words[i]["base"] in ["FUFF", "JUJU", "KECK", "KUKU", "SYZYGY", "ZIZ", "ZUZ", "ZZZ"]:
      del words[i]
   else:
      i = i + 1

print("unique minimilized word combinations is " + str(len(words)))

# this is an array of all tiles in scrabble, minus the blanks (since we don't care about them)
# because we're pruning words from the dictionary, we've rearranged the letters to try and remove as many
# words as possible as quickly as possible, to make lookups run faster, this means that the vowels and
# R,S,T,L,N have been moved up the list, it doesn't really impact anything, as alphabetical order is arbitary
# and nothing from this point forward depends on us having a sorted list
letters = ['A', 'A', 'A', 'A', 'A', 'A', 'A', 'A', 'A', 'Y', 'Y', 'E', 'E', 'E', 'E', 'E', 'E', 'E', 'E', 'E', 'E', 'E', 'E', 'I', 'I', 'I', 'I', 'I', 'I', 'I', 'I', 'I', 'U', 'U', 'U', 'U', 'H', 'H', 'P', 'P', 'M', 'M', 'R', 'R', 'R', 'R', 'R', 'R', 'T', 'T', 'T', 'T', 'T', 'T', 'W', 'W', 'C', 'C', 'N', 'N', 'N', 'N', 'N', 'N', 'S', 'S', 'S', 'S', 'B', 'B', 'F', 'F', 'G', 'G', 'G', 'L', 'L', 'L', 'L', 'D', 'D', 'D', 'D', 'J', 'K', 'Q', 'V', 'V', 'X', 'Z']

invalid_tilesets = 0

for a in range(0, 84):
   # if we've finished using a letter in the tileset, we're gonna remove the words in the condensed dictionary
   # that utilize those letters, this is more of a speed thing than anything else because those dictionary lookups
   # are expensive
   if a > 0 and letters[a-1] != letters[a]:
      i = 0
      while i < len(words):
         if i_in_j([letters[a-1]], words[i]["sorted"]):
            del words[i]
         else:
            i = i + 1

   print("invalid_tilesets = " + str(invalid_tilesets) + " | new_first_letter = " + letters[a] + " | words in dictionary = " + str(len(words)))

   for b in range(a+1, 85):
      if not is_valid_tileset([letters[a], letters[b]]):
         for c in range(b+1, 86):
            if not is_valid_tileset([letters[a], letters[b], letters[c]]):
               for d in range(c+1, 87):
                  if not is_valid_tileset([letters[a], letters[b], letters[c], letters[d]]):
                     for e in range(d+1, 88):
                        if not is_valid_tileset([letters[a], letters[b], letters[c], letters[d], letters[e]]):
                           for f in range(e+1, 89):
                              if not is_valid_tileset([letters[a], letters[b], letters[c], letters[d], letters[e], letters[f]]):
                                 for g in range(f+1, 90):
                                    if not is_valid_tileset([letters[a], letters[b], letters[c], letters[d], letters[e], letters[f], letters[g]]):
                                       invalid_tilesets += 1

                                       if invalid_tilesets % 10000 == 0:
                                          print("invalid_tilesets = " + str(invalid_tilesets) + " | " + str([letters[a], letters[b], letters[c], letters[d], letters[e], letters[f], letters[g]]) + " | " + str([a,b,c,d,e,f,g]))

print("invalid_tilesets = " + str(invalid_tilesets))

If someone knows the answer could they put it in odds of like 1000 to 1.

CH is in the list you're using.

@studog fixed, i just did a cursory lookup and found an article but the dictionary says it's there

I used the beginning of your program to run a quick monte-carlo experiment and got 0.7% probability of not being able to make a word.

A few comments:  Your word list is slightly longer than required, e.g. you include ACE but that is unnecessary as AE is already in the list.  You also have some words in your list which have too many of some letters (e.g. ZZZ) as the list you used is allowing for words which require blanks.  I ended up with 173 words in my tests.

My attempt at a solution to the program gave a probability of 0.57% or approximately 1 in 175.

@borrible interesting, i'm not sure how I missed ACE, because I did check the entire array both times

I was going to correct for blanks by not considering them as part of the set of tiles in scrabble, as any 7 set with blanks has a valid word in it.  I was probably going to ignore ZZZ and ZUZ because I was going to probably go through the ~70 Trillion non-blank combinations and try and do a lookup on the list of letters to determine validity.  There's probably a better way to do it that doesn't involve that many computations.

@DenisS There are others in your list as well that are not necessary, e.g. HUFF, FIR, ...  Ultimately it's unimportant, as you've reduced your list to <200 words out of the 280k.

I'm going to make an estimate from the following assumption:
Any hand that contains at least one vowel, y, or a blank allows a valid move. Any hand that contains entirely consonants does not. Obviously there are exceptions, but they should be rare enough to have a negligible effect (and the false positives and false negatives work to cancel each other out).
There are 46 of these tiles and 54 that are not. The chance of consecutively drawing 7 consonants is therefore:
54/100 * 53/99 * 52/98 * 51/97 * 50/96 * 49/95 * 48/94
This works out at 1.11%, or about 1 in 90 games.

I like the approximation approach but your problem is that your false positives and false negatives are highly unbalanced.  In particular, there are a huge number of all consonant combinations that include valid words.  The top problematic short valid words being HS, HNT, CH and HM.

@borrible Fair point, and I was a bit off.

I love the simplicity of this approximation. Would be even better if it could also give an upper and lower bound on the true probability by estimating by how much it is off.

I used the beginning of the program in @DenisS's answer to build the Scrabble dictionary, then I used it to write a small monte-carlo program to estimate the probability that no word can be formed with seven random tiles.
The result is a 0.58% probability that no word can be formed.
Output
$ python3 get_proba.py 1000 50
loading dictionary
total words in dictionary is 279497
words 7 letters or shorter is 77459
Running for 50 experiments of 1000 draws...
Ran for 50 experiments of 1000 draws.
Successes: [996, 996, 996, 995, 992, 996, 998, 993, 994, 993, 992, 993, 998, 994, 994, 986, 994, 996, 990, 994, 997, 998, 994, 993, 993, 991, 999, 991, 997, 996, 993, 989, 995, 996, 998, 996, 995, 996, 992, 992, 998, 994, 993, 989, 993, 991, 991, 999, 995, 995]
Proba of failure = 0.00582000000000005

Code
def build_dict():
    words = []
    words_in_dictionary = 0
    words_short_enough = 0
    print("loading dictionary")
    with open("dictionary.txt", "r") as dictfile:
        for line in dictfile:
            base_word = line.strip()
            if len(base_word) > 0:
                words_in_dictionary = words_in_dictionary + 1
                if len(base_word) <= 7:
                    words_short_enough = words_short_enough + 1
                    word = {"base": base_word, "sorted": sorted(base_word)}
                    words.append(word)
    print("total words in dictionary is " + str(words_in_dictionary))
    print("words 7 letters or shorter is " + str(words_short_enough))
    ok_combinations = [''.join(word["sorted"]) for word in words]
    return(ok_combinations)

def flatten(ll):
    return [x for l in ll for x in l]

def build_letter_bag():
    return flatten([['A']*9, ['B']*2, ['C']*2, ['D']*4, ['E']*12, ['F']*2, ['G']*3, ['H']*2, ['I']*9, ['J']*1, ['K']*1, ['L']*4, ['M']*2, ['N']*6, ['O']*8, ['P']*2, ['Q']*1, ['R']*6, ['S']*4, ['T']*6, ['U']*4, ['V']*2, ['W']*2, ['X']*1, ['Y']*2, ['Z']*1, ['*']*2])

dico = build_dict()
letter_bag=build_letter_bag()

from itertools import chain, combinations

def powerset(iterable):
    "powerset([1,2,3]) --> () (1,) (2,) (3,) (1,2) (1,3) (2,3) (1,2,3)"
    s = list(iterable)
    return chain.from_iterable(combinations(s, r) for r in range(len(s)+1))

def can_make_word(letters):
    if '*' in letters:
        return True
    return any((''.join(subset) in dico) for subset in powerset(sorted(letters)))

import random

def montecarlo(n):
    nb_ok = 0
    for i in range(n):
        letters = random.sample(letter_bag, 7)
        nb_ok += (1 if can_make_word(letters) else 0)
    return nb_ok

import statistics

def run_experiments(nb_draws, nb_experiments):
    nb_ok_list = [montecarlo(nb_draws) for i in range(nb_experiments)]
    average = statistics.fmean(nb_ok_list)
    stdev = statistics.pstdev(nb_ok_list, mu=average)
    return average, stdev, nb_ok_list

def get_args(argv):
    nb_draws, nb_exp = 1000, 1
    if len(argv) > 1:
        nb_draws = int(argv[1])
        if len(argv) > 2:
            nb_exp = int(argv[2])
    return nb_draws, nb_exp

def main(argv):
    random.seed()
    nb_draws, nb_experiments = get_args(argv)
    print('Running for {} experiments of {} draws...'.format(nb_experiments, nb_draws))
    average, stdev, l = run_experiments(nb_draws, nb_experiments)
    print('Ran for {} experiments of {} draws.'.format(nb_experiments, nb_draws))
    print('Successes:', l)
    print('Proba of failure = {} +- {}'.format((nb_draws - average)/nb_draws, stdev/nb_draws))

import sys
if __name__=='__main__':
    main(sys.argv)

Rendering unto Caesar:

The code in build_dict() is from @DenisS's answer;
The rest of the code is from me;
The file dictionary.txt is the 2019 Collins Scrabble Words file linked in this answer to a related question;
The justification that a hand with a blank tile can always score is in @DenisS's answer (if '*' in letters: return True in my code);
The basic idea of the algorithm is to use a Monte-Carlo method, because browsing the dictionary is acceptable but trying out all possible hand combinations is unreasonable.

Considering that both myself and borrible calculated the odds as 0.572% you're pretty close with the Monte Carlo method.

91592097 in 16007560800 which is approximately 0.572% (or 1 in 175).

Some of what follows is already covered in @DenisS's answer and I have used the same dictionary of words (Collins Scrabble Words (2019))
for easy comparison.  Note in particular in that answer the argument for discounting blanks when looking for valid combinations without words (i.e. that the only letter not in a 2 letter word is a V and that there are not enough of those to fill up our 7 selected tiles) and the discussions on pruning.
The following approach is rather “quick and dirty” and relies on several tools that are available on multiple platforms.
Firstly, I took the dictionary and alphabetised the letters in each word.  (I removed duplicates, caused by words which were anagrams of each other, although this was not necessary.  This resulted in a dictionary containing 247491 words.)
The 2 letter words (93 unique alphabetised words) were then removed and the dictionary pruned so that it no longer contained any words which contained all the letters of one of those words.  (For example, the word AE removed words from the list including those where the letters were adjacent such as AESSSY and where they were not adjacent AABCELN).
This was done as a simple iteration over the 2 letter words in bash using grep with some shell parameter expansions.
for f in $(cat 2LetterWords) ; do grep -v ${f:0:1}".*"${f:1:1} temp > a; rm temp; mv a temp; done

The 3 letter words (61 unique alphabetised words) were then extracted and the new dictionary pruned in a similar manner.  4 letter words (15) and 5 letter words (4) were similarly extracted.  At each stage, the handful of dictionary words that could not be formed without using blanks were also removed.
As all other words in the dictionary contain the letters that enable us to make one of these 2-5 letter words, these are the only ones we
need to consider.  I.e. we only need to find the combinations of 7 tiles where we cannot make any of the following 173 words:

AA
AB
AD
AE
AF
AG
AH
AI
AJ
AK
AL
AM
AN
AP
AR
AS
AT
AW
AX
AY
AZ
BE
BI
BO
BY
CH
DE
DI
DO
EE
EF
EH
EL
EM
EN
EO
EP
ER
ES
ET
EW
EX
EY
EZ
FI
FO
FY
GI
GO
GU
HI
HM
HO
HS
HU
IK
IL
IM
IN
IO
IP
IQ
IS
IT
IX
JO
KO
KY
LO
MM
MO
MU
MY
NO
NU
NY
OO
OP
OR
OS
OT
OU
OW
OX
OY
OZ
PU
RU
ST
SU
TU
UX
UY
ACO
ACV
AOV
AQU
AUV
AVV
BBU
BCU
BDU
BFU
BRR
CDU
CEI
CEU
CIY
CLY
CMW
CRY
CUZ
DDU
DFU
DJU
DLU
DRY
DUW
EGG
EGK
EGV
EIV
EJU
EKU
FLU
GPY
GRR
GTY
HNT
HPT
HPY
HRY
HTY
HWY
IIW
IJZ
IRZ
IVY
IWZ
LPY
LSY
LUU
LUV
LUZ
LVY
NPW
PRY
PSY
PTW
PXY
RTY
RWY
SWY
UWZ
BKLU
BLLU
CFFU
CFKU
CKLU
CLLU
DKUU
FFPT
IRRY
JKUU
KUUZ
LLLU
LLUW
PPTY
PTYY
CCIIV
CIIRR
GHLLY
LLXYY

There are 16,007,560,800 (100 C 7) combinations of tiles we can pick, although some of these combinations will be indistinguishable from each other.  If we only consider the number of combinations that are distinguishable
we are reduced to 3,199,724 which is a far more tractable value and, from any given distinguishable combination we can easily calculate the number of different combinations of tiles which are indistinguishable.
That value can be calculated using some brute-force methods.  A bunch of nested loops in C such as
for (A=0;A<=anMax[0];A++)
for (B=0;B<=anMax[1];B++)
for (C=0;C<=anMax[2];C++)
for (D=0;D<=anMax[3];D++)
…

where the anMax array (offset from 0) is set to the number of available tiles for each letter struggles but a few short-circuit checks to ensure that we do not go over the required number of tiles
…
for (C=0;C<=anMax[2];C++) if (A+B+C<8)
…

is sufficient to run the calculation in a couple of seconds.  (My first attempt, adding checks spaced out on the C, E, G, L, O, S and W was good enough.)
A little more shell scripting in awk, such as:
awk '{print (substr($0,1,1)" && "substr($0,2,2)") ||"}' 2LetterWords

with a little bit of editing (to account for repeated letters), e.g. (for the two letter words)
if (
    (A>1) || (A && B) || (A && D) || (A && E) || (A && F) || (A && G) || (A && H) || (A && I) || (A && J) || (A && K) || (A && L) || (A && M) || (A && N) ||
    (A && P) || (A && R) || (A && S) || (A && T) || (A && W) || (A && X) || (A && Y) || (A && Z) || (B && E) || (B && I) || (B && O) || (B && Y) || (C && H) ||
    (D && E) || (D && I) || (D && O) || (E>1) || (E && F) || (E && H) || (E && L) || (E && M) || (E && N) || (E && O) || (E && P) || (E && R) || (E && S) ||
    (E && T) || (E && W) || (E && X) || (E && Y) || (E && Z) || (F && I) || (F && O) || (F && Y) || (G && I) || (G && O) || (G && U) || (H && I) || (H && M) ||
    (H && O) || (H && S) || (H && U) || (I && K) || (I && L) || (I && M) || (I && N) || (I && O) || (I && P) || (I && Q) || (I && S) || (I && T) || (I && X) ||
    (J && O) || (K && O) || (K && Y) || (L && O) || (M>1) || (M && O) || (M && U) || (M && Y) || (N && O) || (N && U) || (N && Y) || (O>1) || (O && P) ||
    (O && R) || (O && S) || (O && T) || (O && U) || (O && W) || (O && X) || (O && Y) || (O && Z) || (P && U) || (R && U) || (S && T) || (S && U) || (T && U) ||
    (U && X) || (U && Y)
   ) return 0;

gave some simple conditional checks to ensure the word list did not appear.
There are 309831 distinguishable combinations where none of the 2-letter words can be formed.  252242 if we ensure 2 and 3 letter words can not be formed.  251180 excluding 2,3 and 4 letter words and down to 251021 excluding the complete list.
We cannot just look at 251021 / 3199724 as our probability as different distinguishable combinations have different numbers of corresponding tile combinations.  The distinguishable combinations excluding the word list tend to use the rarer tiles which means they tend to have fewer corresponding tile combinations.
We can count the number of combinations that correspond to a given distinguishable combination by looking at the number of ways the corresponding letters might have been chosen, which was calculated as:
Cr(0,A)* nCr(1,B)* nCr(2,C)* nCr(3,D)* nCr(4,E)* nCr(5,F)* nCr(6,G)* nCr(7,H)* nCr(8,I)* nCr(9,J)*
nCr(10,K)* nCr(11,L)* nCr(12,M)* nCr(13,N)* nCr(14,O)* nCr(15,P)* nCr(16,Q)* nCr(17,R)* nCr(18,S)*
nCr(19,T)* nCr(20,U)* nCr(21,V)* nCr(22,W)* nCr(23,X)* nCr(24,Y)* nCr(25,Z)

This gives us 91,592,097 combinations (of which there are 251,021 distinguishable sets) out of 16,007,560,800.

Those odds seem very high, given that OP has played for 50 years and never had this happen before....

@John If the OP had played once a month for 50 years, the odds of never having personally drawn such a rack would be (1-0.00572)^600 which is about 3%.  If they've played half as often it's about 18%.  Neither of those values look unreasonable.  This answer is also in accordance with Stef's answer which was arrived at using a different methodology (Monte Carlo simulation rather than combinatoric enumeration).

Also this can only happen when the OP is the first player, which is not every game.

@borrible fair enough.  I was thinking of my aunt, who averaged 30 games a month for 50 years, not someone playing only once / month.

@borrible I finally got off my lazy ass and finished my program, which went about finding the combinations differently, but got an answer that agrees with yours


text:
Help me score my cribbage hand
In Cribbage, my hand was 2,2,2,5 and the cut card was a 3.  I wanted to use the cut card with each of my twos for a count of five and subsequently one 15 for two points.
I thought the hand should have been 6 points for the three 2's plus 2 points for one 15 (2h+3, 2s+3, 2c+3= 15) however my partner said thats wrong.
Who is right?

Your partner is correct.
The cut card is "just" another card for your hand for almost all aspects of scoring except flushes and knobs.  So, given a 3, 2, 2, 2, 5 you can score no 15's, and your hand score is 6 for the three 2's.

IF THE CUT CARD WAS A 5 AND I HAD 3 TENS IN MT HAND, WOULDNT EACH BE COUNTED AS 15?

Yes, the same way if you had 3 tens and a 5 in your hand. The difference is that you're not using the cut card multiple times for the each 15.

source: Stack Exchange

license: Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/

sort: votes


text:
Does the "unnatural" blackjack after splitting push with the dealer?
If I split aces, and get an "unnatural" (as they say in casinos), does the ante push if the dealer draws into a blackjack too, or does the dealer win?
I know, that it usually pays 1:1 at the casinos in this case. But what if it's a draw?
The word `unnatural' just confuses me.

It depends on the house rules, which may be either determined by law (or subordinate regulation) or by the casino by publication.
The dealer will often make the rule explicit by stating that splits aces don't become blackjacks. In this case they cannot push on the dealer's blackjack, because they only have a 21 at best.

This scenario will never happen when playing blackjack in a typical Vegas-style casino. The hand ends immediately if the Dealer has (natural) Blackjack. So there will never be an opportunity for you to split your aces if the dealer has blackjack.
In European-style casino rules, the specific rules about how to handle the additional bets of double-downs and splits can vary, with some casinos returning the additional bets and others taking all the bets. But no matter how the additional bets are handled, all split hands (even if they are 2-card 21s) lose to a dealer's blackjack.

This is patently false in many casinos, wherein the dealer has only one card until all players have stood or bust and cannot know at all whether they will have a blackjack. The player can split aces, receive tens on both, before the dealer gets a blackjack and takes the player's money (variously, all or it or only the original bet).

@Nij sounds like a bad casino making up their own rules. Are those "typical"? References?

Sounds like a lawful requirement that the casinos cannot change... New Zealand or Australia Capital Territory for example, as the initial deal.

@Nij huh, I was unaware that Blackjack variations were that different outside Las Vegas. I've updated my answer to clarify the Vegas-style rules.

all_licenses: Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/


text:
Rummikub, replacing a joker
When replacing a joker from the board the rules state I must replace the joker gained from the board with 2 tiles from my hand.

A joker that has been replaced must be used in the player's same turn with 2 or more tiles from his rack to make a new set

My opponent interprets this meaning that I can only use the tiles from your hand. For example J,4,5 or J,6,6. Though I agree that I must use at least 2 tiles from my hand I find nothing that prohibits me from using another tile from the board along with the joker and the 2 tiles from my hand.  For example I replaced a Joker from board, a 9+12 from my hand, and an 11 from another run to create the following. 9,J,11,12.  My opponent thinks this is an illegal move.
Is this move legal ?

Primarily it depends on which ruleset you're using. Pick your rule and stick with that. (My vote is "yes" you can use other tiles from the table with your two from hand and a joker.)


text:
In spades, is a deuce of clubs considered a spade?
In Spades, is a deuce of clubs/hearts considered a spade?  I played with some people that argued they consider their suit to be spade.

Why would it be considered a spade? As far as I am aware it can't take a trick unless everyone is out of both clubs and spades and depending on the rules the 2 of clubs should be the first card played so that will never happen.

@JoeW, the rule: "2 of clubs should be the first card played".  is a Hearts rule, it is not common in Spades.

@Cohensius Yes and I did say it was depending on the rules that it was played first and I mentioned that as a side note as in some cases it will be the first card played.

The deuce of club and hearts do not considered a spade in the standard game, however there are variants where they do.
From Pagat:

In some versions of Spades, some or all of the four twos are elevated to the top of the spade suit, are ranked in some specified order, and are considered to be spades. The rest of the cards rank as in normal.

For example, in the New York City rules the trump suit ranks is as follows: big joker, small joker, 2♦, 2♠, A, K, Q, J, 10, 9, 8, 7, 6, 5, 4, 3
In the Deuces high rules the trump suit is is as follows:2♠, 2♦, 2♣ 2♡, A, K, Q, J, 10, 9, 8, 7, 6, 5, 4, 3
Those variants are pretty common, and are implemented for example at AI-factory's Spades free app.

Like any game, Spades has a lot of variations. Long unnecessary story short, I've moved A LOT. 48 states, 6 provinces, dozens of cities in Mexico...
I like to play cards, it seems to dissolve borders. Anyway...
When I learned to play as a kid in Syracuse NY.  Rank of cards from highest to lowest was:
Big Joker, Little Joker, 2 H, 2 C, 2 D, 2 S, A S, --- 3 S.
3 H, and 3 C weren't used.
Total number of tricks remained the same, but total number of spades was 18 instead of 13.
Since then, I've played a bunch of variations from "no nil bid allowed" "max 5 bid/hand" "reneg" "forced blind when -200" "2 D, and 2 S were trump behind the jokers" "jokers high then A S --- 2 S" and any number of combinations of these rules.
The "original" rules without the jokers and only the 13 spades as trump is probably the most common across the US as a whole. Maybe not the way they'd usually play, but they're familiar with this way most often.
Apples and Oranges, though.  They're really different games.  The mechanics are different, the tracking becomes a little more challenging, and the game seems to be less predictable.  Just like with Hearts, spades gets to be a little mechanical and repetitious, but the additional 5 trump cards make it a little less so.
Best practice with any game is to make sure you all know the rules BEFORE the situation presents itself unless you're all very patient :-)


text:
Do you know the name of this Rummy variant?
[ Context: My Lebanese grandma taught me this card game from her grandparents. We call it "Pinaq" I don't know how to spell it in english but it's pronounced "pee-nak". I thought it was Pinnochle bc it sounds similar but it isn't. I think the name of the card game derived from a French card game starting with a B. Not sure. Also, I'm really bad at explaining things so sorry if this looks more complicated than it seems ]
You use a 52 card deck. Ace being the highest and 3 being the lowest. The 2s are special cards which we call "Pinaq(s)". Which I'll explain about later.
it's a 2 - ? player game (not sure about the limit).
How to Play
Dealer deals out 9 cards to each player,  in a clockwise direction (don't know if it's crucial for the direction). These cards become their hand. Then they put the rest of the cards in a pile in the middle. flipping one card over from the middle pile face up next to it.
The player left to the dealer starts, and the turns move in a counter-clockwise direction.
The player can either take a card from the face-down pile of cards in the middle, or the face-up card beside the pile.
Generally speaking, If a player chooses to take from the face-up card pile, they can take all of them, 1 of them, any amount of them basically. And If they take from the face-up card pile, they don't HAVE to discard a card from their hand to the face-up pile. But if they want to, they can only discard one card from their hand.
Since at the beginning of the game, there's only 1 card in the face-up card pile, if they choose to take that card, they must discard one card from their hand, if the pile is empty basically you have to discard one card.
and when discarding a card, it can't be the same one you picked from that face-up pile of cards. But let's say you picked 2 or more cards from the face-up pile of cards, you can discard a card from the selection of cards you picked, since you haven't picked up only one card.
Now let's say you chose the pile of face-down cards, you can only pick up one card. And when you do choose from the face-down pile, you must discard one card from your hand to the face-up pile, it can be the same card you just picked from the face-down pile. Unlike earlier as I explained, if you pick up one card from the face-up pile, you can't discard the card you just picked up.
Alright, the objective of the game is to try to reach your targeted goal of points. We usually set the game to 52 points. you can change it to whatever your goal of points you wish to be. Whoever reaches that amount of points first by the end of the rounds win. Also the objective is to place down as many sequences as you can and get rid of all the cards in your hand first!
To get points: you have to make, I guess you can call them sequences. 3-13 card sequences. So for example, you can lay down cards from your hand like: 3 of hearts, 4 of hearts, and 5 of hearts, they MUST be the same suit. now if you have a 2 in your hand, that's the special card called a "pinaq" which can be used as any card in any suit. So, you can use it here, for example: 6 of spades, 7 of spades, then a 2, and 9 of spades. The only rule is you can't use the 2s "pinaqs" beside eachother in a sequence, like: 10 of diamonds, a 2, then another 2. You can use as many 2s "pinaqs" as you can (there's only 4 of them in a deck lol) in a sequence but they have to be at least one card away from eachother. Like: 4 of clubs, a 2, 6 of clubs, another 2, 8 of clubs, 9 of clubs, another 2.
How a player ends it's turn is by either: discarding a card from their hand to the middle where the face-up pile of cards are intended to be. Or, if they started their turn by picking one or more cards from the face-up pile (not leaving it empty), and choose not to discard one from their hand. they can just announce that they've ended their turn, or they can end the turn with putting down sequence(s) if they choose to. If you want to put one or more sequences of cards down, do it before you discard a card. You cannot end your turn by placing down your sequence(s), you must do that before. Unless, like I explained earlier, it's an exception if you're in a situation where you do not choose to discard.
A player must start their turn by either: choosing one card from the face-down pile in the middle, choosing one or more cards from the face-up pile in the middle, or- I don't know if this is allowed, but if they think they can win, they can start by placing down their sequence(s) and ending their turn by discarding their last card. That is also how a player wins a round^^
How to count points at the end of Rounds
A round is finished when a player has no cards in their hand, which also means they have won the round. When a round has ended, the winner player from that round gets +10 additonal points. The rest of the players do not get additional points. To count your points, each card from 3-Ace is +1 point, the 2 (pinaq) is worth +2 points. By the time the round ends, if you have sequences placed down, those are your positive points, all the cards in your hand are negative points. Calculate each player's total of points at the end of each round and write them down on a piece of paper or keep count. First to reach 52 points (or your own targeted goal of points) wins the game!
The thing about this game, is it's about luck, but also a bit of strategy. You can put down sequences early on in the game, that will let other players know what you're looking for to add to it, you have to be careful not to hang onto your cards for too long, incase a player wins the round then you get many negative points from all the cards in your hands you were saving up for. even if you have sequences in your hand, by the time the round ends whatever's in your hand is negative. You can make the game fun with more people, team up with a player to destroy the leading player haha, or demolish the losing player hehe, make deals with  others and help eachother out. (I don't know if that's cheating lol) it's fun to play with more than 2 players, I say 3-4 is the best!
I'm sure I could've explained this in simpler terms so I'm sorry about how long and not very detailed this explanation is. Hope there's a name for it, thanks for your time !!

Sounds like one of the Rummy Games

You describe a variant of a Rummy game.  With the following special rules:

2's act as jokers.
Special pick-up-from-discard pile rule: can pick up any number of cards from discard or a single face-down card from the deck.
Target goal of points, similar but not the same as in Pagat:Kalooki
starting hand of 9 cards

Paget:Rummy have a long list of Rummy games, divided into groups, you might be able to find it there.  Your variant sound like group 5/6/7

Basic Rummy Games
Conquian group
Asian Rummy Games
Contract Rummy Games
Manipulation Rummy Games
Knock Rummy Games
Meld Scoring Games
Canasta Group


text:
Board games Online
The questions and answers I found are pretty old, hence, here I am:
I played a few games on Tabletop simulator on steam. I found it rather rubbish and not working very well, especially with a not-so-fast connection. On the other hand, I played Dominion on dominion.games, and the experience has been quite positive. Is there any other platform where games can be played in similar manner? (no download, played on browser and possibly lightweight).
I don't mind downloading a game if the playing experience is good.

I'm playing on Board game arena (BGA) and I find it excellent for several reasons (IMHO):

Great catalogue of games, esp. for euro + family games.
Free to play along in all games
Cheap premium membership where you can start all games yourself
Functional, relatively lightweight interface - works a treat on PC and on tablets/phones
Very active in these times (2020)
Sign up with Google(...) account, so barrier of entry is low.

A possible Con might be that it's not as pretty as the real thing. Nor as pretty as some dedicated app could be.
But, seriously, just google "board games online" and invest an hour to find the platform you like.

Yucata is a good one. It's designed for turn based games rather than live games and ones where everyone takes their turn simultaneously. It's almost exclusively card games.
You can play multiple games at once; if you're on turn in a number of different games, you can easily click through each of them and make your turns.
Games can take a number of days to finish. I tend to log on every few hours and make all my turns; if my opponent is also logged on we might get a few turns in each, otherwise just the one until a few hours later. Providing you're ok with playing like that, it works great (e.g. as a distraction from work...)
It is possible to play live games (either with friends, or with strangers by requesting a live opponent), but there is no way to ensure your opponent plays live i.e. there are no time limits for turns.
I didn't know most of the games on there, but it's been nice to learn some new ones.
Finally, there's a cool level/points based system; if you're the kind of person who likes getting Stack Exchange badges, you'll enjoy this feature.
I also back BGA too, as the other comment recommends.


text:
What are the odds for a “19” (zero pt) hand in Cribbage?
I am doing a math paper for school using Cribbage.  I know how to use factorials to calculate the odds of getting the perfect 29 hand.
My question is, how many 5 card hands (4 in hand and turn up card) add up to zero?

Maybe it’s worth showing what’s you’ve tried so far?   Whilst it seems a valid question for the site I wouldn’t feel comfortable doing someone’s school work for them when they should be making every effort themselves and the teacher being able to show how to improve it.

Please add details that will allow someone that do not know Cribbage but knows probability (or just combinatorics) could help you

I can use C(52,6) to get the total number of hands possible with a 6 card deal.  Certain sites say that the probability of getting a 0 pt hand is about 7%.  A 0 pt hand implies 5 cards with no pairs, no runs, no cards adding up to 15. I can calculate this manually but am curious about the mathematical equation.

Take any one card. Calculate the number of cards you should NOT draw next. Put the different numbers (depending on the card drawn) in a (wheighted) tree. Repeat for all cards.
You might see a patern at some point and do a mathematical induction, but I would not count on it.

This is a significantly more difficult computation than computing the probability of a 29.     There are a ton of edge cases to work out.  If you are not familiar with inclusion-exclusion in combinatorics, you may want to look that up.

I am discovering that!  I have been looking I to inclusion/exclusion combinatorics which I believe will be what I will use in the final calculation..so thank you for that lead.  I am continuing my struggle and am learning a lot%. Any other ideas are welcome!

Do you need to calculate the number of 0-point hands as part of your work, or is it just an input to some other calculation? If the latter, use some other means to work out the answer. My go-to plan in these scenarios (because I've forgotten almost all my maths but I can still program) is to write a software simulation.

I wrote a software simulation (20 million deals) to work this out, because the maths is far too complicated. The chances of getting a zero-point hand of five cards (assuming the starter card is chosen randomly among those five) is roughly 7.4%.
However, this assumes that your five cards are chosen at random. In reality a player chooses four cards from the six they are dealt, so you would actually need to calculate the odds of getting a six-card hand of zero points (about 1.36%), then getting a starter card that means you still end up on zero points. This comes out at approximately 0.43%, assuming the four cards you keep from the initial six are chosen randomly.


text:
In Monopoly, what happens if we run out of houses and/or hotels?
How the game proceeds after all the properties, houses and hotels already bought?

It is unclear if you are asking a question about the rules, or about what will practically happen in such a situation.. can you clarify?

You proceed as normal, rolling the dice, moving, paying rent and, importantly, selling off houses to pay rent as needed. If the  properties are evenly owned, it may take quite a while for luck of the dice to kick one of the players into having to sell, but it should eventually happen.

Building shortages are part of the strategy of the game and it is intended that players might not be able to buy houses because of a lack of them in the bank. It is also intended that when a player has to sell a hotel they also have to sell all all houses as well if there are not enough in the bank.

BUILDING SHORTAGES
  When the Bank has no houses to sell,
  players wishing to build must wait for some player to return or sell
  his/her houses to the Bank before building. If there are a limited
  number of houses and hotels available and two or more players wish
  to buy more than the Bank has, the houses or hotels must be sold at
  auction to the highest bidder.


text:
Card game with animated and holographic cards
I recently remembered a card game that I saw couple years ago, here are the things I (vaguely) remember:

most of the cards being holographic and animated.
It came in big plastic container that opened up into a little area to put your cards.
The little area the container opens to has a little card holder that makes the cards both animated and holographic, it had a see through film.

We are going to need some more details such as what the game involved.

As a note we like to identify games so if you can provide more information we can try and help.

The see through part reminds me of Gloom. But that does not match with the rest of the description.

Other details that will help.  What shape/colour was the container.  Do you know what was on the cards? People, words, monsters etc?  Most importantly can you remember what you did in the game?  Were the cards fighting each other for example?  The more details you can give the better chance of an answer

I found the game after searching for it myself its called Redakai
It's on sale on eBay and for clearance in some stores since it wasn't popular

The nostalgia hit me like a freight train, this is it! Thanks!


text:
Checkers - with customized starting setups and AI agents
Does anyone know a Checkers program that allows customized starting setups when playing vs an AI?
My current app - Dammen (in Dutch) - has the possibility to customize the setup, but then I have to play both sides myself. If I could play against the computer, I could learn a great deal from it.

I'm not including this as an answer, because I'm unsure if it's entirely correct/what you are looking for, but I believe the Checkers application on the kindle store (for the kindle) has this option.

I found a Windows program on https://dammen.startpagina.nl/ (checkers is called 'dammen'in Dutch). On this page I installed the program "dam 2.2". This is amazing. I can customize the setup (positions of the pieces on the board) and save the game. Then I can choose for two players, one player against the computer, or only the computer. This last option is awesome, because now I can learn from the computer what is best and see why it is so.

"learn from the computer what is best" - you cant learn what is best, just what is probably very good.  The AI do not play optimally from any position.  Checkers has been weakly solved: From the starting position, we have a computational proof that checkers is a draw.
https://science.sciencemag.org/content/sci/317/5844/1518.full.pdf?casa_token=fP9QReDlUZgAAAAA:tu_6yU_rT3ntCLLXTBUzRsqEUSRc_OP00aJwHd5LQ808csPHsGpn1Aw2AKGmGrkpPLsj0jLGrj_Wc18

Thanks for the comment and interesting article. I am a beginner and did not know this, thanks!


text:
2 player 2 decks card game
This game is called Touch in Egypt (my country) but I can't find it online.  Can anyone help me with the real name of the game?  Also I would love to hear any competitive strategy I can use to win!
Setup:

Two players
Two shuffled decks of cards
Players create 8 columns that can be completed like K-Q-J-10-etc but in different colors.
Aces are drawn out so the both players can complete the Aces' columns according to their suite and order.

Rules:
Each player has two piles to draw from, one is faced up that other player can put his cards on if they are the same suite and according to their order like if you have 3 of hearts, I can put my 4 of hearts on it.
Game play:
Players take turns...
TOUCH:
if any player made unnecessary move, like arranging the columns when he/she had the chance to complete an Ace column, the other player can say "TOUCH" ending the first's turn.
Win condition: you need to finish both of your piles.

This sounds like Russian Bank, which seems to have a wide variety of different names (including Touch) in different countries. (I knew it as Cravat when I grew up playing it in the UK).


text:
Probability of being dealt a 7-card flush in Rummy
Playing 7-card rummy with 4 players, I was dealt a 7 card Spade flush (not straight) and on the first card picked from the deck pulled an 8th card spade and immediately won (it was 23456 8910).
What are the odds of being dealt all 7 cards in the same suit?

What are the odds of being dealt all 7 cards in the same suit?

The odds of pulling 7 spades in 7 cards the number of ways to pull 7 spades (from among the 13 spades) divided by the number of ways to pull 7 cards (from among the 52 cards).
13c7 = 1716
52c7 = 133784560
1716 / 133784560 = 0.0000128 = 0.00128 %
For general suit (not just spades), multiply that by four:
= 0.0000513 = 0.00513 %
If you want to continue to the eighth card:
The odds of pulling 8 spades in 8 cards the number of ways to pull 8 spades (from 13) divided by the number of ways to pull 8 cards (from 52).
13c8 = 1287
52c8 = 752538150
1287 / 752538150 = 0.00000171 = 0.000171 %
And again multiply by four for general suit:
= 0.00000684 = 0.000684 %
(subtract the number of ways to get a straight if you want to exclude straights from "all cards in the same suit".)
Notation explanation:
13c8 is read as "13 choose 8" and means "the number of ways to choose 8 things from a set of 13 things". It is computed as 13!/((13-8)! * 8!).
In general, NcM = N! / ( (N-M)! * M! )

Adjusted for the 7-card version ( I got excited about the 8th card matching when I read the description and somehow focuesed on that ).

+1 argument is correct, but to someone who doesn't know maths, it must surely not look very intuitive. Your first sentence is good. Then you just state three things. It's not at all clear what these mean without knowledge of how to do the problem in the first place. For non-mathematicians, I think it would be really helpful if you add a few words -- eg, "13 of the cards are spades, so we wish to know 13c7, namely the number of distinct (unordered) ways of choosing 7 objects from an original collection of 13", and then similarly for 52c7. I think this would be illuminating for many people :)

Also, a little bugbear of mine: one should not think of NcM as N! ( (N-M)! * M! ). (I wanted to write an exclamation mark there for emphasis, but thought it would be confusing with the factorial! Also, you have typos in yours.) Instead, it is N (N - 1) ... (N - M + 1) / M!. Indeed, how many ways are there to choose M from N when you care about ordering? Precisely the numerators. Now divide by the number of reorderings, which is precisely the denominator. Simple! :) -- your formula (after being corrected) is just one way of writing this. Not that helpful for working out by hand either!

@SamT thanks for the catch. I've corrected the typos in the formulation.

@BenjaminCosman Exactly so, as I said. "multiply by four if you don't care that it's spades in particular"

I'm not sure where your mistake is, but your math doesn't match up with the very logical way of calculating this given in Kevin's answer (and the spreadsheet in my comment on his answer.) Are you able to explain the discrepancy?

@corsiKa Not sure what discrepancy you're seeing. My 0.00513% matches Kevin's 0.00513%

Some slightly watered down math (that still gives the right answer).
You have a 100% chance of your first card being a "good" card.
There are then 51 cards left, of which 12 will be a card that matches suit, so you have a 12/51 chance that the second card will be "good".
11/50 chance that the third card will be "good".
....
So, you have 1x(12/51)x(11/50)x(10/49)x(9/48)x(8/47)x(7/46)=0.0000513
So you have 0.00513% chance of drawing a 7 card flush

You have 7/56 but I think that should be 7/46 - but also your math doesn't add up - I threw it into Excel and got .000051 - https://i.imgur.com/3Bs6Bgd.png

However despite the answer being off a bit, this is definitely the way I would solve the problem!

@corsiKa good catch. Had a typo in my formula in my first cell that I copied and pasted into the other cells.

L. Scott Johnson is mostly correct except for 2 things:

You specified it wasn't a straight flush (there are 7 8-card straight flushes per suit he should have removed.)
You asked about the 7 card flush, which you were dealt before the draw, not the 8 card flush you drew into.

This gives us:
13C7 as the nCr we care about, giving us 1716 possible flushes of 7 cards.
Then we remove the 8 7-card straight flushes in each suit to give us 1708 possible flushes of 7 cards per suit.
We multiply that result by 4, since we want a flush of 7 cards, not specifically caring about suit giving us 6832 possible non-straight 7 card flushes.
Now we need 52C7, the number of combinations of 7 cards in 52 cards, which is 133,784,560.
This gives us 6832 out of 133,784,560 different 7 card hands that gives us odds of 1194444:61 against, or a 0.005107% chance of such a 7 card flush on the deal (or 0.001277% if you want a specific suit, or odds of 4777959:61 against)
If we do include the 7-card straight flushes, we have 6864 combinations in a deck of 7-card flushes, which gives us odds of 643162:33 against or 0.005131% of possible combinations.

In his particular scenario he didn't have a straight flush, but he asked "What are the odds of being dealt all 7 cards in the same suit?" Excluding a straight flush would lead to an incorrect answer to the question.

@GregSchmit Absolutely correct

@GregSchmit then take a look at the last paragraph where I put the 32 straight flushes back in and give the odds on those.

@Andrew No worries, I only mentioned that because you said that Scott was "mostly correct except...", indicating he was mistaken to include those.


text:
Bridge simple open bid for 7♠, 2♥, 1♦, 3♣?
With the following hand should I open with 3 spades to let my partner know I have a 7 spades?
This is my hand

♠ AK87642
♥ T3
♦ 4
♣ JT8

I am a beginner.

I can understand that you may not have enough information to answer, but I find it odd that everyone would just vote to close without telling OP exactly what is needed to answer it, especially given that they're new here.

Playing any natural system, or any system where the opening of 3 Spades or 4 Spades is played as natural, I would open 3 Spades Vulnerable and 4 Spades Not Vulnerable.
Some may prefer to open just 3 Spades in both vulnerabilities, but that is likely more a matter of partnership style and agreement. This should be discussed with partner, as some partnerships are stricter than others on the need for an 8-card suit to open 4S.
The key qualities of a hand considering an opening of 3 or 4 Spades are:

A 7-card or longer suit of good quality - a standard requirement for opening bids of either 3S or 4S for most partnerships playing traditional preempts. This Spade suit clearly qualifies
No significant defensive values outside the suit opened. Your holding of:
♥: T3
♦: 4
♣: JT8
meets that eminently. The Club holding even has the potential of a slight additional offensive value, with little defensive prospect.
Roughly 8 -10 High Card Points. At just 8, all but one in the opened suit. this hand is perfect. The point here is that as your hand will take just one call, this one, the hand should be such as to have no other values that might result in an erroneous decision by partner.
The reasonable prospect of taking within 3 tricks of the bid Not Vulnerable and 2 tricks of the bid Vulnerable (to limit loss if the opponents opt to double and defend). With 6.5 Spade tricks and 1/4 offensive trick in Clubs in hand, again the expectation is met.

In summary: you could do far worse than to hold this hand in your mind as being relatively close to a model opening preempt in a Major suit - of 4 Not Vulnerable and 3 Vulnerable. (The minors have more complex issues even for early intermediates, but at your stage just ignore that with the proviso of a pending amendment.)

Alexander Woo makes the point below that opposing vulnerability is also relevant to a decision between opening 3♠ or 4♠. Less aggressive partnerships might choose to open 4♠ only at favourable Vulnerability (i.e. Not Vulnerable vs Vulnerable opponents). This is probably the more general practice when considering a 4♠ opening, as here, on a 7-card suit. However this should be discussed with partner an an agreement reached. An accumulation of even small differences in choice of call can, over time, undermine the partnership confidence required to play good bridge.
A consideration that may seem irrelevant when one is learning bridge, but becomes of ever increasing significance as one's own, and one's opponents', skill increases, is that preempts gain in the bidding but lose in the play. If opponents win the hand, the opening preempt has frequently provided a roadmap to avoiding the consequences of a bad break.
This is where the significance of playing sound preempts shows up. When partner can make sound and successful raise/pass/double decisions, boldly and with confidence, your preempt has truly achieved the goal of interfering with the opponents' successful bidding. If partner must instead hesitate, your preempt is probably just helping the opponents overall by keying their Declarer (and even Defense) play.

I think I'm opening 4S not vul vs vul and 3S otherwise.  But I agree it's a matter of taste, partnership agreement, seat, and how eager to bid LHO looks.

@AlexanderWoo: Excellent point - I have added that to the answer.


text:
Is it ok to show my cards when I know I I’m the only one with spades left
Is there any formal rule, or perhaps etiquette, that prevents me from showing my cards once I know I am the only one with spades left? I was recently playing a game and I got down to 5 spades left and knew the other players did not have any (I led with a spade and nobody else played one).
I considered just notifying the other players to save us a couple of minutes but wasn’t sure if this was allowed.
I also couldn’t think of any strategic advantage or disadvantage there might be to not letting the hand play out normally.

Just saw this statement above: "(I led with a spade and nobody else played one)". If this is the only way you knew the opponents were out of Spades - instead of by counting them - you really should not be attempting any claims or TRAMs.

@ForgetIwaseverhere, your comment might sound a bit patronize.  Even if you don't count the cards, it is easy to notice that you are the only player left with spades after you lead spades and no one follows.  I mean that one can be a very amateur player and still use TRAM.

The rules (in most spades rulesets) prohibit showing your cards, yes.
That said, what you describe is a pretty common practice in friendly games, as (as you note) the rest of the hand's play is foregone.
Note the Wiki entry's inclusion of the TRAM section (added in 2012, with no cited sources) also includes a paragraph on penalties for doing so prematurely -- penalties ultimately derived from the fact that showing the cards in your hand before playing them is not allowed.
None of the rules for spades that I've found (including all of the ones cited in the wiki page for Spades) allow it. Indeed I've only found one that mentions it at all, and it does so only to disallow the practice explicitly. From The Spades Connection:

Each hand must be played to completion rather than announcing the rest
  are mine to avoid potential scoring errors.
Cards dropped from any player's hand, (revealed to any other player)
  should remain face up on the table and played at the first
  opportunity.

This procedure, called Claiming or Conceding as the case may be, is common in Contract Bridge (both Rubber and Tournament). However it is also error prone for beginners and can be used unethically by those morally challenged. In consequence there are several Laws governing the process to protect both sides.
When playing casual Bridge all players typically want to play as many hands as possible, so claims and concessions are frequent. I imagine your Spades group feels likewise. However, if your group is unfamiliar with making Claims and concessions I would suggest laying out a few ground rules in advance to avoid misunderstandings and hurt feelings. The following excerpts from the Laws of Rubber Bridge may assist with that. You may need to make minor adjustments to compensate for the absence of a Dummy in Spades.

Law 68 – Declarer’s Claim or Concession of Tricks
....
Declarer should not make a claim or concession if there is
any doubt as to the number of tricks to be won or lost.

Law 69 – Procedure Following Declarer’s Claim or Concession
... declarer must ... make a comprehensive statement as to his proposed plan of play, including the order in which he will play the remaining cards.

Declarer’s claim or concession is allowed, and the deal is scored accordingly if both defenders agree to it. ....

When his claim or concession is not allowed, declarer must play on, leaving his hand face up on the table. ....

The objective of subsequent play is to achieve a result as equitable as possible to both sides, but any doubtful point must be resolved in favor of the defenders.

Note that the above is from the Laws of Rubber Bridge - the non-tournament variety of the game. Stricter constraints apply to Duplicate (Tournament) Bridge, where a Director is available both to adjudicate and to ensure that fairness is maintained for the players at other tables.

This is a common play, called: The rest are mine (TRAM).
Wikipedia:

A common play among more experienced or skilled players is for a
  player who realizes that he cannot help but win all remaining tricks
  to simply lay down his hand and declare "the rest are mine" or
  similar. This is known as "TRAMing", and can help speed up the
  game.


text:
How to solve Free Cell games fast? (in under 2 min)
Are there any known tricks to play faster at FreeCell?  I make the same number of moves as the best players but it takes me about 5.5 minutes.

I've never done it in 2 minutes

The OP is surely talking about some electronic version of the game, which could be considered off-topic. But since speed is based on strategy, it could be considered on-topic.

And “winner of my game” is confusing if your talking about the solitaire game Free Cell. Assuming you are talking about a competitive computer version of the game; the question could be improved by simply asking for strategies on how to solve Free Cell games faster.

My best time in Expert in FreeCell in Microsoft's Solitaire Collection is 1:04.
I think I obtained that time in one of the "Daily Challenges", so it may not have been a "real game". That said, I have reached close to that time on other occasions. In fact, while trying to obtain a screenshot of the score screen to include in this post, I believe I have gotten my second best time ever.

(I play all games to completion now, but I abandoned some games early on. My goal is to get the win rate to 99%, which requires playing another 2692 games.)
Rather than spending time pondering my next move, I prefer to experiment and use Undo with impunity. Undo can be performed rapidly by pressing Ctrl-Z. (Holding allows to quickly undo multiple moves, and even reset the board.)
Additionally, the discovery of "Single Tap To Move" option has greatly sped up my game significantly (although my best time was made without it this setting off).

With this option, I regularly get times around 2:00, which I would have considered exceptionally good before.
As for my actual strategy? I really wouldn't know how to describe it.

My fastest game thus far has been 1:08 on a regular game. I play all games at medium difficulty and I currently have an average time of 3:08, with a 99% win rate. I am over 2,500 games right now. Since my time has gotten better, most games now are played in under 3 minutes.
My strategy is, before starting the game, identify the location of all the aces. Of course, those cards have to be uncovered first. Make getting the aces out a priority. Then work on the 2s, etc.
Hint: It helps to clear a row, if possible; that allows you to move more cards, and it gives you a chance to move a bunch of cards off another row.
On the more difficult games, I have to think out some moves, but I often let the cards play themselves. That is, I just see what cards are available to play and use those, and go from there.
Be mindful of playing patterns. There are certain types of moves of the cards you will use over and over. Be aware of how you are playing. I've never developed names for these moves, though I started to do so once, but just be aware of the type of moves you are making. It helps your awareness, and it helps you better understand your options.
I hope that helps.

One thing to improve speed is turn on "single tap to move" in the settings.  With that (and some luck of the draw), I've done 0:48

please note that i have "single tap to move" turned on in the settings


text:
In Pinochle, what is the probability of a thousand aces and a double run in one hand?
What are the odds of getting a hand containing:
A♣ A♦ A♥ A♠ A♣ A♦ A♥ A♠ Tx Kx Qx Jx Tx Kx Qx Jx for some suit x, and another four unspecified cards.
Rules Background
Double-deck Pinochle is a four player game using a deck of 80 cards (four lots of ATKQJ in each of ♣♦♥♠) (cf. Wikipedia and this rules page).
A double run is two sets ATKQJ in whichever suit is declared trumps. Trumps is decided by a bidding mechanism, but for this question we'll assume that the bidding was won by the player in question, and trumps were chosen to match the hand. Consequently we're interested in the chance of being dealt two sets of ATKQJ both in the same suit.
"A thousand aces" is a term from standard (single-deck) Pinochle, and refers to two "rounds" of aces (i.e. A♣ A♦ A♥ A♠ x 2). In standard Pinochle this is all of the aces in the deck and is worth 1000 points. In double-deck Pinochle this is only half the aces, so is much more common and is only worth 100 points.
Melds in Pinochle are allowed to reuse cards (as long as they are different types of melds), and so the aces for the runs can be reused for the double run. Consequently the required cards for the hand are A♣ A♦ A♥ A♠ A♣ A♦ A♥ A♠ Tx Kx Qx Jx Tx Kx Qx Jx for some suit x, which leaves four cards unspecified.

What do you mean by getting a thousand aces?

In pinochle "1000 aces" or "aces abound" is a meld containing all 8 aces.

It would help you could edit in the exact hand or if someone with more knowledge could edit in exactly what a thousand aces and a double run means. With proper information someone with no knowledge of the game could provide the answer as long as they know what is in the hand and deck

The problem here is double deck pinochle is dealt exhaustively, and commonly played with 2-8 players, with some games including passing cards to a partner, so there's a lot of variation on the probability.

As KMR and JoeW indicate: please fill in the details of the set-up you wish to have analyzed: deck composition, number of cards dealt, composition of a hand that you're counting, and explicit disallowance for human decision making in the process (e.g., passing cards).

TLDR: This will happen roughly once every 1.7 million hands.
Rules Background
Double-deck Pinochle is a four player game using a deck of 80 cards (four lots of ATKQJ in each of ♣♦♥♠) (cf. Wikipedia and this rules page).
A double run is two sets ATKQJ in whichever suit is declared trumps.  Trumps is decided by a bidding mechanism, but for this answer we'll assume that the bidding was won by the player in question, and trumps were chosen to match the hand.  Consequently we're interested in the chance of being dealt two sets of ATKQJ both in the same suit.
"A thousand aces" is a term from standard (single-deck) Pinochle, and refers to two "arounds" of aces (i.e. A♣ A♦ A♥ A♠ x 2).  In standard Pinochle this is all of the aces in the deck and is worth 1000 points.  In double-deck Pinochle this is only half the aces, so is much more common and is only worth 100 points.
Melds in Pinochle are allowed to reuse cards (as long as they are different types of melds), and so the aces for the runs can be reused in the arounds.  Consequently the required cards for the hand are A♣ A♦ A♥ A♠ A♣ A♦ A♥ A♠ Tx Kx Qx Jx Tx Kx Qx Jx for some suit x, which leaves four cards unspecified.
Simulation
The calculation of this seemed quite involved (although see L. Scott Johnson's full calculation), so I wrote a Python script to estimate the probability:
import random
from multiprocessing.dummy import Pool as ThreadPool

suits='CDHS'
ranks='ATKQJ'
cards = [r+s for s in suits for r in ranks] * 4

def simulate(_):
    matches = 0
    for i in range(innerLoops):
        hand = random.sample(cards, 20)
        if hand.count('AC') >= 2 and hand.count('AD') >= 2 and hand.count('AH') >= 2 and hand.count('AS') >= 2:
            for s in suits:
                if hand.count('T'+s) >= 2 and hand.count('K'+s) >= 2 and hand.count('Q'+s) >= 2 and hand.count('J'+s) >= 2:
                    matches += 1
                    break
    return matches

poolSize = 8
pool = ThreadPool(poolSize)
loops = 100000000
innerLoops = loops / poolSize
for j in range(5):
    results = pool.map(simulate, range(poolSize))
    print('{} / {}'.format(sum(results), loops))

When I ran this the output was:
58 / 100000000
65 / 100000000
61 / 100000000
57 / 100000000
54 / 100000000

Combining these results gives an average of about 295 times in five hundred million hands, or 1/1694915.  This result supports the full calculation by L. Scott Johnson.

Typo: you verify hand.count('AD') twice and don't check for AH. The actual odds are much smaller, therefore. When I run your code with the typo corrected, I get around 51-69 / 100,000,000. I'll post the the true calculation shortly....

@L.ScottJohnson Thank you for spotting this!  I have corrected the typo and results. This also lead me to learn a little about multithreading in python - thank you!

TL;DR: The odds are 1 in 1,733,866
To compute:
First calculate how many ways you can select a "winning hand":
Winning configuration count for your choice of suit x:
Each digit in the configuration represents either

the number of aces of a suit, without regard to a specific suit

or

the number of TJQK in suit X, without regard to a specific rank:

CASE 1
You might do it exactly (with no extra aces or extra cards of suit x):
8 aces: 2222 and 8 suit-X-non-aces (SXNA): 2222 with four other cards (neither Aces nor of suit X -- I'll call these "Off" cards): 4 I'll write this as case 1:
1) 2222 2222 4

There are C(4,2) ways of choosing each Ace and each SXNA. With four choices of X. And there are C(48,4) ways of choosing the other four cards
C(4,2)^8 * 4 * C(48,4) = 1,307,278,725,120 ways of drawing such a hand.
There are C(80,20) different hands, so dividing shows us a 1 in 2,704,332 chance of getting that kind of hand.
Other cases
The tricky part comes in factoring in the other ways to get it: drawing some of the extra aces and/or the extra SXNAs
Let's say you only draw 3 "Off" cards: getting an extra ace or an extra SXNA. There are two configurations to do this (cases 2 and 3):
2) 2223 2222 3
3) 2222 2223 3

And these configurations have a count (you have to choose which suit the extra Ace is in, or which rank the extra SXNA is)
The 2223 part has a count of C(4,1) = 4. This 4 representz the ways to select the order (if the columns corresponded to distinct suits/ranks): 2223, 2232, 2322, or 3222
The 2222 part still has a count of just one, but we can write that as C(4,4), meaning we're choosing four of the suits to be 2 (or four of the ranks in the case of SXNA).
So to make the counts explicit:
1) C(4,4)*C(4,4) -- choosing Aces 2s then SXNAs 2s
2) C(4,3)*C(1,1) * C(4,4) -- choosing Aces 2s then Aces 3s then SXNAs 2s
3) C(4,4) * C(4,3)*C(3,3) -- choosing Aces 2s then SXNAs 2s then SXNAs 3s

Continuing with having 2 "off" cards (and two extra Aces and/or SXNAs:
4) 2222 2224 2
5) 2222 2233 2
6) 2223 2223 2
7) 2224 2222 2
8) 2233 2222 2

And so on, making the whole table of configurations:
 1) 2222 2222 4

 2) 2222 2223 3
 3) 2223 2222 3

 4) 2222 2224 2
 5) 2222 2233 2
 6) 2223 2223 2
 7) 2224 2222 2
 8) 2233 2222 2

 9) 2222 2234 1
10) 2222 2333 1
11) 2223 2224 1
12) 2223 2233 1
13) 2224 2223 1
14) 2233 2223 1
15) 2234 2222 1
16) 2333 2222 1

17) 2222 2244 0
18) 2222 2334 0
19) 2222 3333 0
20) 2223 2234 0
21) 2223 2333 0
22) 2224 2224 0
23) 2224 2233 0
24) 2233 2224 0
25) 2233 2233 0
26) 2234 2223 0
27) 2333 2223 0
28) 2244 2222 0
29) 2334 2222 0
30) 3333 2222 0

The count of each configuration is:
 1) 2222 2222 4 C(4,4) * C(4,4)
 2) 2222 2223 3 C(4,4) * C(4,3)*C(1,1)
 3) 2223 2222 3 C(4,3)*C(1,1) * C(4,4)
 4) 2222 2224 2 C(4,4) * C(4,3)*C(1,1)
 5) 2222 2233 2 C(4,4) * C(4,2)*C(2,2)
 6) 2223 2223 2 C(4,3)*C(1,1) * C(4,3)*C(1,1)
 7) 2224 2222 2 C(4,3)*C(1,1) * C(4,4)
 8) 2233 2222 2 C(4,2)*C(2,2) * C(4,4)
 9) 2222 2234 1 C(4,4) * C(4,2)*C(2,1)*C(1,1)
10) 2222 2333 1 C(4,4) * C(4,1)*C(3,3)
11) 2223 2224 1 C(4,3)*C(1,1) * C(4,3)*C(1,1)
12) 2223 2233 1 C(4,3)*C(1,1) * C(4,2)*C(2,2)
13) 2224 2223 1 C(4,3)*C(1,1) * C(4,3)*C(1,1)
14) 2233 2223 1 C(4,2)*C(2,2) * C(4,3)*C(1,1)
15) 2234 2222 1 C(4,2)*C(2,1)*C(1,1) * C(4,4)
16) 2333 2222 1 C(4,1)*C(3,3) * C(4,4)
17) 2222 2244 0 C(4,4) * C(4,2)*C(2,2)
18) 2222 2334 0 C(4,4) * C(4,1)*C(3,2)*C(1,1)
19) 2222 3333 0 C(4,4) * C(4,4)
20) 2223 2234 0 C(4,3)*C(1,1) * C(4,2)*C(2,1)*C(1,1)
21) 2223 2333 0 C(4,3)*C(1,1) * C(4,1)*C(3,3)
22) 2224 2224 0 C(4,3)*C(1,1) * C(4,3)*C(1,1)
23) 2224 2233 0 C(4,3)*C(1,1) * C(4,2)*C(2,2)
24) 2233 2224 0 C(4,2)*C(2,2) * C(4,3)*C(1,1)
25) 2233 2233 0 C(4,2)*C(2,2) * C(4,2)*C(2,2)
26) 2234 2223 0 C(4,2)*C(2,1)*C(1,1) * C(4,3)*C(1,1)
27) 2333 2223 0 C(4,1)*C(3,3) * C(4,3)*C(1,1)
28) 2244 2222 0 C(4,2)*C(2,2) * C(4,4)
29) 2334 2222 0 C(4,1)*C(3,2)*C(1,1) * C(4,4)
30) 3333 2222 0 C(4,4) * C(4,4)

And finally we have to factor in the group count: how many ways can we select the N aces or a given suit from the 4 available? For 2222, the count is C(4,2)^4 : choose 2 of the 4 available aces for each suit. for the Aces and the SXNAs, there are four cards available for each column (four suits or four ranks). For the Off column, there are 48 available cards (four copies of four non-ace ranks in three suits).
1) 2222 2222 4 C(4,2)^8*C(48,4)
2) 2222 2223 3 C(4,2)^7*C(4,3)*C(48,3)
3) 2223 2222 3 C(4,2)^7*C(4,3)*C(48,3)
4) 2222 2224 2 C(4,2)^7*C(4,4)*C(48,2)
5) 2222 2233 2 C(4,2)^6*C(4,3)^2*C(48,2)
6) 2223 2223 2 C(4,2)^6*C(4,3)^2*C(48,2)
7) 2224 2222 2 C(4,2)^7*C(4,4)*C(48,2)
8) 2233 2222 2 C(4,2)^6*C(4,3)^2*C(48,2)
9) 2222 2234 1 C(4,2)^6*C(4,3)*C(4,4)*C(48,1)
10) 2222 2333 1 C(4,2)^5*C(4,3)^3*C(48,1)
11) 2223 2224 1 C(4,2)^6*C(4,3)*C(4,4)*C(48,1)
12) 2223 2233 1 C(4,2)^5*C(4,3)^3*C(48,1)
13) 2224 2223 1 C(4,2)^6*C(4,3)*C(4,4)*C(48,1)
14) 2233 2223 1 C(4,2)^5*C(4,3)^3*C(48,1)
15) 2234 2222 1 C(4,2)^6*C(4,3)*C(4,4)*C(48,1)
16) 2333 2222 1 C(4,2)^5*C(4,3)^3*C(48,1)
17) 2222 2244 0 C(4,2)^6*C(4,4)^2
18) 2222 2334 0 C(4,2)^5*C(4,3)^2*C(4,4)
19) 2222 3333 0 C(4,2)^4*C(4,3)^4
20) 2223 2234 0 C(4,2)^5*C(4,3)^2*C(4,4)
21) 2223 2333 0 C(4,2)^4*C(4,3)^4
22) 2224 2224 0 C(4,2)^6*C(4,4)^2
23) 2224 2233 0 C(4,2)^5*C(4,3)^2*C(4,4)
24) 2233 2224 0 C(4,2)^5*C(4,3)^2*C(4,4)
25) 2233 2233 0 C(4,2)^4*C(4,3)^4
26) 2234 2223 0 C(4,2)^5*C(4,3)^2*C(4,4)
27) 2333 2223 0 C(4,2)^4*C(4,3)^4
28) 2244 2222 0 C(4,2)^6*C(4,4)^2
29) 2334 2222 0 C(4,2)^5*C(4,3)^2*C(4,4)
30) 3333 2222 0 C(4,2)^4*C(4,3)^4

To find the number of each case, just multiply 4 (the number of choices of suit X) by the configuration count and the group count:
1) 2222 2222 4 -- 1,307,278,725,120
2) 2222 2223 3 -- 309,873,475,584
3) 2223 2222 3 -- 309,873,475,584
4) 2222 2224 2 -- 5,052,284,928
5) 2222 2233 2 -- 20,209,139,712
6) 2223 2223 2 -- 53,891,039,232
7) 2224 2222 2 -- 5,052,284,928
8) 2233 2222 2 -- 20,209,139,712
9) 2222 2234 1 -- 429,981,696
10) 2222 2333 1 -- 382,205,952
11) 2223 2224 1 -- 573,308,928
12) 2223 2233 1 -- 2,293,235,712
13) 2224 2223 1 -- 573,308,928
14) 2233 2223 1 -- 2,293,235,712
15) 2234 2222 1 -- 429,981,696
16) 2333 2222 1 -- 382,205,952
17) 2222 2244 0 -- 1,119,744
18) 2222 2334 0 -- 5,971,968
19) 2222 3333 0 -- 1,327,104
20) 2223 2234 0 -- 23,887,872
21) 2223 2333 0 -- 21,233,664
22) 2224 2224 0 -- 2,985,984
23) 2224 2233 0 -- 11,943,936
24) 2233 2224 0 -- 11,943,936
25) 2233 2233 0 -- 47,775,744
26) 2234 2223 0 -- 23,887,872
27) 2333 2223 0 -- 21,233,664
28) 2244 2222 0 -- 1,119,744
29) 2334 2222 0 -- 5,971,968
30) 3333 2222 0 -- 1,327,104

For a total of 2,038,978,759,680
Dividing by the number of hands, C(80,20) = 3,535,316,142,212,180,000
We find the odds to be about 1 in 1,733,866
(nb: there's no reason you have to handle the aces distinctly from the SXNAs: there are four of each. So you could reduce the number of cases. They're left distinct here for approachability and transparency.)


text:
Looking to identify Solitaire game
My dad (now 93 years old) used to play a two deck solitaire game with the cards laid out in 10 columns as shown in the picture.  In brief:

Two Decks
10 columns
1st column has one card, second column has two cards . . . tenth column has ten cards.
Cards alternate face up, face down
Remaining cards are used one at time and only the most recent card that is top of the pile can be used.

I am looking to identify this variant and see if I can find an online version for him.

Please add more rules, such as the goal of the game.  What do you mean by the top card can be "used"?

ryanyuyu - Not very good at the terminology, but the remaining cards that are not originally placed in the 10 columns are revealed one at a time.  If the revealed card can be used then it is placed in the play area, if not another card is dealt on top it, now that card must be placed on the board, before the card below it can be played.

How do you play the cards on top of each other? Red on red? Red on black? Clubs on clubs? 5's on 4's or maybe 4's on 5's? Do you use a foundation, and do you build from ace up?

The closest I’ve found is Batsford.
Batsford uses 2 decks, and has 10 tableau stacks going from 1 to 10. You didn’t mention how many foundation piles; but given 2 deck I would assume 8 foundation piles.
The problem is that Batsford only has the top card facing up in the tableau stacks. Perhaps this is a Batsford variant that alternates the cards face up and face down in the tableau stacks...
Batsford is available online at https://www.classicsolitaire.com/batsford.html.

Christopher, I tried Batsford and has all of the same rules that my Dad uses.  This is a very close approximation, the only difference is the alternating pattern.  Thank you for your help.

Christopher, on second look, looks like Batsford allows you to go though the Reset deck more than once.  That is not allowed in the variation I am seeking.  It makes solving it difficult.  Using shuffled two decks of cards, you end up solving it 1 out 20 times (estimated).    Also the fact that there are face-up cards that you are seeking makes the game more fun and adds some strategic elements I found missing in Batsford.

With only two rule differences it seems very likely to be a variant of Batsford. Hopefully someone else will recognize and know it’s name. Good luck!

Maybe your dad made this variant. So you can call it Dadsford.

You can use this like to play Batsford since the other link doesn't seam to work anymore: https://www.isolitairegames.com/batsford.html

I know this question was asked a while ago, but my Dad taught me this game - the setup is as is shown in the picture.  You play cards on the tableau on each other ie alternating red and black descending in value.  in this example the nine of clubs can be played on either of the red tens.  The goal is like other solitaire games.  Get the piles of Ace through King in each suit - you'll have 8 piles as there are two decks.
RULES:
As you play, when a column empties out you can place a King at the top of the column (promotion) and then play descending cards in alternating colors.
An Ace starts the pile and you can place same suit cards on the Ace piles in ascending order - ie 2, 3, 4 etc - these piles are placed on the side of the tableau (or on top) like regular solitaire
You can take a card from the Ace pile and bring it back to the tableau as needed as long as it's put on a card of immediate higher value of opposite color - ie bring the six of clubs back and play it on the seven of hearts. IF you don't play with this rule, it is very hard to win at this solitaire.
GAMEPLAY:
After dealing the tableau, you will have 49 cards remaining in the deck - this is not a 'reset' deck - the cards in this deck will be played on the tableau - here's how:
After you've dealt the tableau, play as many cards on each other as you can, flipping over any face-down cards as they get revealed. In the picture above, you can play the four of hearts on the five of clubs and flip over the face-down card that is revealed when the four is moved.  The Ace and Two of clubs can be removed to start a pile of clubs (placed to the side or top  of the tableau the two on top of the Ace).
When you can't play on any other cards on the tableau or play cards on the Ace piles, you need to start placing the remaining deck cards on the tableau.  This is why it's not a reset deck.
To place the cards from the deck on the tableau:
Never put a card on the first column - it's like a free column
Never put a card on a column that is empty
Never put a card on a column that is headed by a King - that is, a column that is headed by a King that you promoted. (in the example picture you would put a card on the third column, because even though it is headed by a King, it hasn't been promoted
Do place one card at the bottom of every column (from column 2-10 and in keeping with the above rules) - this is considered one turn.
Then do the gameplay as in the beginning until you're stuck again.  Once you can't move, place another card on each column.  You will alternate between the gameplay and placing cards from the deck until you have no more cards on the deck and can't make any legal moves.
The game can get very messy and there are a lot of moving parts to consider.  It is a difficult game to win - the estimate of 1 in 20 sounds about right, but it is satisfying.
I think this version may be popular in the Middle East - hope this helps


text:
After my Nil bid was set, should I try to take or duck?
In Spades, when someone is forced to takes a trick on a nil bid, some-times, players keep giving away their high cards.
In which situations it is better to try to take tricks and in which it is better to avoid taking more tricks?

I'm voting to close this question as off-topic because the question inquires as to the motivations of unspecified unknown players of the game.

I think that after my edit the question is on-topic

It's perfectly possible that having failed in the nil, your hand is still pretty weak in trick-taking value and plays better by trying to sandbag your opponents rather than trying to make them fail their bid (particularly if the hand was grossly underbid). Your partner presumably needs little assistance to make his.

The question you should ask yourself is: can we set the opponents?
If the nil bidder have a chance of setting her opponents then it is good idea to try and take tricks, however when she realize that the chances of setting the opponents are slim then there is no point in collecting bags*.  Sum of bids equal to 10 is my cutoff point, when the sum of bids is lower than 10, usually I don't go for the set.  When sum is 11 or more I will try.

*Note that takes of the nil-bidder does not count toward her partner's contract.


text:
Which version of solitaire this is?
Setup: single deck, all cards are laid face up in 4 rows of 13.
Goal: to get all cards into one pile.
Moves: only cards that are next to each other or 4 spaces away.

Maybe you can find your game in this huge list of Solitaire games https://www.bvssolitaire.com/rules/

I believe you are talking about Gaps (or some variant of it).

The cards are dealt into four rows of thirteen. The aces are removed and discarded from further play. The gaps that they leave behind are filled by cards that are the same suit and a rank higher than the card on the left of the gap. (For example, 4♣ can be placed beside 3♣.) However, any gap at the right of a King is considered dead and no card can fill it.

Quote from: https://en.m.wikipedia.org/wiki/Gaps

I don't think Gaps/Montana variants meet the "into one pile" nor the "4 spaces away" movement criteria of the OP.

Possibly related to Accordion Patience, where a pile of cards can be moved either on to the previous adjacent pile or the pile that is three before.  The aim is to reduce the cards to a single pile.
It differs from your description in that the cards are treated as being in a single line rather than in 4 rows of 13 cards, although for space reasons you could lay out cards in that form. It's not stated in your description how adjacency works for cards in different rows - and given your "next to each other or 4 spaces away" - you might be considering edges of the rows to connected in a manner that's equivalent.  Accordian Patience typically uses 3 spaces away rather than 4 but that seems a small change.


text:
In two-player games with perfect information, if both players play optimally, will the game always end in a draw?
I've been running simulations of two strong agents playing Chess which always ended  in a draw.  Followed by this simulations, can we state that two-player perfect-information games like Go and Chess will always end in a tie when both players play optimally?

I find your  premise that homo economicus is considered the "perfect human" rather questionable. The homo economicus is driven by their rationality, but usually their goal is a self-centered one. This behavior is optimal for the individual, but can be harmful for many others. For as long as the homo economicus is not harmed by the harm inflicted to others, it's rational for them to continue this behavior. Yet, this is only a "perfect human" in the same sense that the robot Ash in the movie Alien describes the predatory alien as "the perfect organism".

Right. To prevent confusion I guess I'll change it to super-intelligence.

In general, this assertion clearly cannot be true. Imagine a game called "Player 1 wins". Here are the complete rules: Player 1 wins. When played by two super-intelligences, this will not end in a tie.

@PhilipKendall Well in the question I clearly meant board games like chess or go - so we’re talking about that kind of board games

A super intelligence doesn't care if it's a board game, a card game or an abstract game. It's just a framework of rules that leads to a conclusion.

Imagine a turn-based two-player game where on each turn you choose to either say "win" or "pass". If you choose "win", you win. If you choose "pass" the other player gets a turn. It should be trivial from this example that this game will not end in a draw with perfect play.

@GendoIkari right. Any idea how to rewrite the question to be complete clear?

@Schmuddi only true if there is only one such intelligence in the universe (if even then).

Accurding to Zermelo's theorem, in finite two-player games of perfect information in which the players move alternately and no affect of chance, one of those three possibilities is true:

First player can always win.
Second player can always win.
Both players can force a draw.

For example, Tic-Tac-Toe is known to have a strategy by both players that will force a draw.  In 4-in-a-row it is known that the first player has a winning strategy.  Chess haven't been solved yet, I think that the conjecture is that a draw can be forced.

A list of solved games: https://en.wikipedia.org/wiki/Solved_game
The field of this question is called Combinatorial game theory (CGT)
see: Solving Chess

On a bit of a tangent, for some games it's possible to prove which of these possibilities is true without knowing the optimal strategy needed to achieve it. For example, if the first player can (effectively) skip their first turn (and if the rules of the game are symmetric for both players) then the second player cannot possibly have a guaranteed winning strategy — if they did, the first player could skip and then use the same strategy to win instead. This is an example of a strategy-stealing argument.

@IlmariKaronen, right, this non-constructive proof of the winning player is called Ultra-weak solution.  It makes very elegant solutions.

No - there need not always be a tie in the general case. Even in games of perfect complete information there may still be a bias towards one player.  For example the game of nim cannot end in a tie, and depending on the starting position gives an advantage to either the first or second player - e.g.
Size of heaps | Result with
A     | B     | perfect play
----------------------------------
1     | 1     | First player wins
2     | 2     | Second player wins

For the specific games you mentioned (go and chess) then it is not known what the result is with perfect play.  For more details see the corresponding Wikipedia articles:

First move advantage in Chess - Solving chess
First move advantage in Go - Perfect Komi

Thank you for your answer! But isn't the Chess article based on statistics of played games (by not that super intelligent humans)?

Another well known counterexample: Hex. Perfect information, yet we can trivially prove that the first player has a winning strategy.

@AnatolyWein The chess article covers perfect play, and includes statistics from games played by computers (significantly better than humans) as well as theoretical studies.  For example in the opening section it says "White's winning percentage is about the same for tournament games between humans and games between computers" and "Chess is not a solved game, and it is considered unlikely that the game will be solved in the foreseeable future."

You are missing a couple of important factors in your question that make it impossible to answer as it stands. Mainly that a game with perfect information available to both players doesn't mean that it is not balanced in one players favor.
While it may be true that there are games out there where it is balanced between both players it is also true that there games where it is not the case it can be balanced in favor of either player (such as the second player is more likely to win which is why they go second).
A secondary point is that there are games where a tie or draw is not possible and if it gets to a point where nothing more can be done a player is declared the victor by default.


text:
Is Chess with more than three players exists or would be playable?
Chess board as a function of the number of players:

2 players - 8x8 square board
3 players - a special hexagonal board
4+ players - theoretically a (players * 2) gonal board.

First Question: Does this gonal board exist? Is it playable (talking about practical needs, e.g. size of a field)? How would the desk look like?
Second question: Would a 2+ players 3D-chess room be possible to create?

Comment and body do not match: three-player chess may exist (see answer), but four-player is just a thought-experiment.

what is GONAL?  google only finds Follicle-stimulating hormone

Please try to narrow your question.  Your two questions aren't really that related, and there are too many aspects of your questions to address.

There was a middle-ages chess variant for 4 players, known as 4-seasons chess. It was played on an 8x8 board. Each player had 4 pawns, 1K, 1B, 1Kt, & 1R formed as an angle in a corner of the board. Sets were often red, green, white, and black. I have played this a few times, years ago. There may be other 4-player varsions

Also, a '(players+2)gonal board' is a mathematical nonsense. A board can have triangles, squares or hexagons (equivalent to squares with alternate rows offset). Nothing else tesselates properly.

@TimLymington pretty sure OP isn't asking about tesselation - the spaces would still be squares in all scenarios (see the hexagonal 3-player board for example). It's possible to imagine octagonal and decagonal boards populated by squares too (well, quadrilaterals at least). The point at which they meet in the middle is where it gets interesting: the larger the board, the more distorted that point becomes, but it would still be theoretically playable at any size.

To answer the first question as posed: no, the n-gonal chessboard doesn't exist for n>6 (assuming Euclidean geometry).
You can't tile the plane with octagons -- you end up with small empty square spaces at every other side, so the symmetry is lost. Regular tilings (using regular polygons) are possible only with 3-gons (triangles), 4-gons (squares) and hexagons.

Yes, chessboards for 3 players exists. I have a similar to the following at home:
https://www.chesshouse.com/products/3-player-medium-wood-chess-set
It is playable and it is fun, however don't expect a really deep chess game. The main tactic is to be not the one who gets ganged upon. The better chess player won't win. The player who can manipulate the best will win (by making the other one vulnerable to attacks or just by talking).
It is almost impossible to defend against two opponents. For example one captures a piece and the second player captures another piece that was defending the first piece. Another situation is as one attacks your king ('check') and the other one attacks your queen. There are also some funny checkmates, like Scholar's mate where one person uses the bishop and the other uses the queen.
Be prepared for these situations, otherwise it feels very unfair. By times very funny, but also very unfair :)

I've got the same board and reached the same conclusions - in our games we actually ended up playing in silence or at least disallowing any conversation related to the game, because it was just too easy otherwise for two players to gang up, even inadvertently. It's an ingenious board though.

The question is a bit vague. But there already is a 4 player chess.
It would be possible to expand that to more players, but it's not sure if that is playable.

4 player chess does, in fact, exist. I've played it before, at my school. It's set up as an 8x8 board, with 3 extra rows on each side, skipping the corners, like such:
https://www.google.com/imgres?imgurl=https%3A%2F%2Fupload.wikimedia.org%2Fwikipedia%2Fcommons%2Fthumb%2F9%2F95%2FFour-handed_chess.png%2F1200px-Four-handed_chess.png&imgrefurl=https%3A%2F%2Fen.wikipedia.org%2Fwiki%2FFour-player_chess&docid=HAixsx0fsJsBPM&tbnid=aICeu9NGaThNRM%3A&vet=10ahUKEwiJw-SClsvkAhVhm-AKHQ_vBKkQMwhqKAMwAw..i&w=1200&h=1200&safe=active&bih=609&biw=1366&q=4%20player%20chess&ved=0ahUKEwiJw-SClsvkAhVhm-AKHQ_vBKkQMwhqKAMwAw&iact=mrc&uact=8