question: if I flipped a coin and got heads twice the odds are 1/4 or 40%? correct? What's the math behind working it out beyond this? getting it right three time in a row, then four, then five?
comment: 1/4 is 25%, not 40%. Every time you add another flip, the chance is cut in half. So the odds of flipping heads X times in a row is 1/2^(x).
passages: BULLET::::- The probability of 20 heads, then 1 head is 0.5 × 0.5 = 0.5
   The first statistical assumption is this: for each of the dice, the probability of each face (1, 2, 3, 4, 5, and 6) coming up is . From that assumption, we can calculate the probability of both dice coming up 5:    More generally, we can calculate the probability of any event: e.g. (1 and 2) or (3 and 3) or (5 and 6). 
   A simple example of the occasion for this concept may be seen by observing that Pearson's chi-squared test is an approximate test. Suppose Pearson's chi-squared test is used to ascertain whether a six-sided die is "fair", i.e. gives each of the six outcomes equally often. If the die is thrown "n" times, then one "expects" to see each outcome "n"/6 times. The test statistic is
   BULLET::::- Odds of 9/4 ("top of the head") – both hands touching the top of the head.
  
  BULLET::::- Odds of 10/1 ("cockle" or "net") – fists together with the right-hand thumb protruding upwards, to resemble the number 10.
  
  BULLET::::- Odds of 11/10 ("tips") – hands together and touching all fingers on both hands together.
  
  BULLET::::- Odds of 5/4 ("wrist") – the right hand is moved to touch the left wrist.
  
  BULLET::::- Odds of 33/1 ("double carpet") – arms crossed, hands flat against the chest.
   The expected return of a $1 wager on a particular symbol is calculated as follows:
  
  BULLET::::- The chance of the symbol coming up on none of the dice is 5/6 × 5/6 × 5/6, which is 125/216. The player loses $1 in this case.
  
  BULLET::::- The chance of the symbol coming up on all three dice is 1/6 × 1/6 × 1/6, which is 1/216. The player wins back the $1 plus $3 more.
   In this case, the probability of getting 51 or more 6s on a fair die is 0.02654. If we were looking for significance at the 5% level, this result indicates that the die is loaded to give many 6s (one-tailed test).
   Bernoulli came across expected utility by playing the St Petersburg paradox. This paradox involves you flipping a coin until you get to heads. The number of times it took you to get to heads is what you put as an exponent to 2 and receive that in dollar amounts. This game helped to understand what people were willing to pay versus what people were expected to gain from this game.
   Analogously, given odds as a ratio, the probability of success or failure can be computed by dividing, and the probability of success and probability of failure sum to unity (one), as they are the only possible outcomes. In case of a finite number of equally likely outcomes, this can be interpreted as the number of outcomes where the event occurs divided by the total number of events:
  
  Given a probability "p," the odds as a ratio is formula_19 (probability of success to probability of failure), and the odds as numbers can be computed by dividing:
   If you bet on rolling one of the six sides of a fair die, with a probability of one out of six, the odds are five to one against you (5 to 1), and you would win five times as much as your wager. If you bet six times and win once, you win five times your wager while also losing your wager five times, thus the odds offered here by the bookmaker reflect the probabilities of the die.
   Suppose we have a board game that depends on the roll of one die and attaches special importance to rolling a 6. In a particular game, the die is rolled 235 times, and 6 comes up 51 times. If the die is fair, we would expect 6 to come up 235/6 = 39.17 times. Is the proportion of 6s significantly higher than would be expected by chance, on the null hypothesis of a fair die?
   BULLET::::- Winning positions in the mathematical game of Mogul: a position in Mogul is a row of 24 coins. Each turn consists of flipping from one to seven coins such that the leftmost of the flipped coins goes from head to tail. The losing positions are those with no legal move. If heads are interpreted as 1 and tails as 0 then moving to a codeword from the extended binary Golay code guarantees it will be possible to force a win.
   If the first flip was heads, we must calculate the conditional probability that the chosen coin was heads-only as well as the conditional probability that the coin was fair, after which we can calculate the conditional probability of heads on the next flip. The probability that it came from a heads-only coin given that the first flip was heads is the probability of selecting a heads-only coin times the probability of heads for that coin divided by the initial probability of heads on the first flip, or .5 * 1 / .75 = 2/3. The probability that it came from a fair coin given that the first flip was heads is the probability of selecting a fair coin times the probability of heads for that coin divided by the initial probability of heads on the first flip, or .5 * .5 / .75 = 1/3. Finally, The conditional probability of heads on the next flip given that the first flip was heads is the conditional probability of a heads-only coin times the probability of heads for a heads-only coin plus the conditional probability of a fair coin times the probability of heads for a fair coin, or 2/3 * 1 + 1/3 * .5 = 5/6 ≈ .8333
   If after tossing four heads in a row, the next coin toss also came up heads, it would complete a run of five successive heads. Since the probability of a run of five successive heads is (one in thirty-two), a person might believe that the next flip would be more likely to come up tails rather than heads again. This is incorrect and is an example of the gambler's fallacy. The event "5 heads in a row" and the event "first 4 heads, then a tails" are equally likely, each having probability . Since the first four tosses turn up heads, the probability that the next toss is a head is:
   For example, when drawing a single card at random from a regular deck of cards, the chance of getting a heart or a face card (J,Q,K) (or one that is both) is formula_19, because of the 52 cards of a deck 13 are hearts, 12 are face cards, and 3 are both: here the possibilities included in the "3 that are both" are included in each of the "13 hearts" and the "12 face cards" but should only be counted once.
  
  Section::::Mathematical treatment.:Conditional probability.
   BULLET::::- The chance of the symbol coming up on two of three dice is C(3,1) x 1/6 × 1/6 × 5/6, which is 15/216. The player wins back the $1 plus $2 more.
  
  BULLET::::- The chance of the symbol coming up on one of three dice is C(3,1) × 1/6 × 5/6 × 5/6, which is 75/216. The player wins back the $1 plus $1 more.
   Now the same coin is reused and you are asked to bet on the outcome again.
  
  If the first flip was tails, there is a 100% chance you are dealing with a fair coin, so the next flip has a 50% chance of heads and 50% chance of tails.
   During a coin toss, the coin is thrown into the air such that it rotates edge-over-edge several times. Either beforehand or when the coin is in the air, an interested party calls "heads" or "tails", indicating which side of the coin that party is choosing. The other party is assigned the opposite side. Depending on custom, the coin may be caught; caught and inverted; or allowed to land on the ground. When the coin comes to rest, the toss is complete and the party who called correctly or was assigned the upper side is declared the winner.
   All pieces start at the village. The players move their mothers according to the roll of a die (traditionally pieces of stick were used). A player must throw a six to get their mother from the village onto the first circle. A player must throw the exact number to reach the well; if the number thrown is higher than the number of days to the well they must try again on their next turn. Once they reach the well the mother washes her clothes until the player rolls a six again. Then the mother starts the return journey to the village.
   BULLET::::- There are 7 possible outcomes where only one die will match the number
  
  BULLET::::- There are 4 possible outcomes where only two dice will match
  
  BULLET::::- There are 4 possible outcomes where all three dice will match
  
  At odds of 1 to 1, 2 to 1 and 3 to 1 respectively for each of these types of outcome, the expected loss as a percentage of the stake wagered is:
  
  1 - [(7/54) * 2 + (4/54) * 3 + (4/54) * 4] = 22.2%
   Section::::Examples.:Why the probability is 1/2 for a fair coin.
  
  If a fair coin is flipped 21 times, the probability of 21 heads is 1 in 2,097,152. The probability of flipping a head after having already flipped 20 heads in a row is . This is an application of Bayes' theorem.
  
  This can also be shown without knowing that 20 heads have occurred, and without applying Bayes' theorem. Assuming a fair coin:
  
  BULLET::::- The probability of 20 heads, then 1 tail is 0.5 × 0.5 = 0.5
   The probability of observing k heads in 10 tosses of a fair coin, with p(heads) = 0.5, is given by the binomial formula:
  
  The probability for each value of "k" is given in the table below.
  
  The probability of 0, 1, 2, 3, 7, 8, 9, or 10 heads in 10 tosses is the sum of their individual probabilities:
   As for the slim prospect of getting ten out of ten heads from a fair coin—the outcome that made the coin appear biased—many may be surprised to learn that the chance of any sequence of heads or tails is equally unlikely (e.g. H-H-T-T-H-T-H-H-H-T) and yet it occurs in "every" trial of ten tosses. That means "all" results for ten tosses have the same probability as getting ten out of ten heads, which is 0.000976. If one records the heads-tails sequences, for whatever result, that exact sequence had a chance of 0.000976.
   So the player has an 87.2% chance of scoring even when they only have 5 dice left and have not kept a 1 or a 4 on the first throw. They have a 93.5% chance of scoring at this stage if they kept a 1 or 4 on the first throw. When a player hasn't kept either a 1 or a 4 and has only 2 dice left, the chance of getting a 1 and 4 with these 2 dice is 13.9%.
   If there are no other limitations on the number of flips, the probability that the game will eventually end this way is 1. (One way to see this is as follows. Any given finite string of heads and tails will eventually be flipped with certainty: the probability of not seeing this string, while high at first, decays exponentially. In particular, the players would eventually flip a string of heads as long as the total number of pennies in play, by which time the game must have already ended.)
   While a run of five heads has a probability of = 0.03125 (a little over 3%), the misunderstanding lies in not realizing that this is the case only before the first coin is tossed. After the first four tosses, the results are no longer unknown, so their probabilities are at that point equal to 1 (100%). The reasoning that it is more likely that a fifth toss is more likely to be tails because the previous four tosses were heads, with a run of luck in the past influencing the odds in the future, forms the basis of the fallacy.
  
presuppositions: One of four is equal to forty percent. One of four is equal to forty percent.
corrections: One of four does not equal forty percent, one of four equals twenty-five percent. One out of four is twenty-five percent.
raw_presuppositions: One of four is equal to forty percent. One of four is equal to forty percent. One of four is equal to forty percent.
raw_corrections: One of four does not equal forty percent, one of four equals twenty-five percent. One out of four is twenty-five percent. One out of four is twenty-five percent.