<p>Im reading a paper by Karl Menger [1] about the St. Peterbourg paradox :</p>
<blockquote>
<p>In the theory of probability, the &quot;Petersburg Game&quot; designates the
follow- ing gamei between two persons, A and B. Initially, A requires
that B place a certain bet with A. Then they flip a coin. If it shows
heads, then B receives 1 dollar from A, and the game is over. If it
shows tails, it is flipped again. If the second throw turns up heads,
then B receives 2 dollars from A, and the game ends. The coin is
flipped until heads appears for the first time. If this happens at the
nth throw (that is, if the first n - 1 throws aII come up tails, but
the nth throw is heads), then B receives 2n - 1 dollars from A and the
game ends. Depending on whether n = 1, 2, 3, 4, ... , B wins 2° = 1,21
= 2,22 = 4, 23 = 8, ... dollars. What bet should B be willing to make? What is B's mathematical expectation?</p>
</blockquote>
<p>his point is that the paradox does not rely on a mathematical expectation of infinite value, but on the discrepancy between the behavior of the player B and the expected result: in real life, normal people (Menger says) would not commit an infinite amount of money, in spite of expectation being mathematically infinite.</p>
<p>The lottery is always presented in terms that A will pay to B.
But the lottery never takes into consideration the expected value for A in committing the game.</p>
<ul>
<li>Which is the expected value of player A, in playing the Petersbourg game and risking to pay a sum to B ? Will A have a chance to win something ? Or is player A completely disregarded ?</li>
</ul>
<p>[1] Karl Menger, The Role Of Uncertainty In Economics, 1979.</p>

CC BY-SA 4.0
2021-12-23T16:13:19.037Z
Please clarify your specific problem or provide additional details to highlight exactly what you need. As it's currently written, it's hard to tell exactly what you're asking.
Community
CC BY-SA 4.0
2021-12-24T13:35:01.340Z
Will A have a chance to win something ? Yes, if the number of throws n is such that $2^{n -1}<K$ where K is the bet.  But this is obvious.  Is that what you are asking ?
dm63
CC BY-SA 4.0
2021-12-24T17:35:42.857Z
It is written that B places a certain bet with A: the game ends when B wins, and the coin keeps on being flipped until B wins. So I did not understand what for you it is obvious: is the Bet the prize at each throw, or is it the maximum number of throws that can be done by betting 1 dollar at each time ?
E.g. 
Say B wins with heads, and bet is 3 dollars. After 3 throws, the expected value should it be: `1/2*1^0 + 1/2*2^1 + 1/2 * 2^2`  Or `1/2* 3^0 + 1/2* 3^1 + 1/2 * 3^2`? If the bet is 3 dollar, A wins after 2 throws if both had tails, assuming each 1 dollar is bet at each throw ?
user305883
CC BY-SA 4.0
2021-12-26T04:28:12.717Z
No I think B bets an amount only once, before any coins are flipped.  The ‘paradox’ is that B’s expected value is theoretically infinite, but in real life B would not pay unlimited amounts to play this game.
dm63
CC BY-SA 4.0
2021-12-26T08:50:05.660Z
So, if B bets 1 dollars and heads does not show in the first 1 tosses, A would never win, right ? 

(1 -  1/2)^0 + (1- 1/2) ^1 + (1 - 1/2) ^2 + ...] => 1 + 0.5 + 0.25 + .. ] => which is never 2−1<. 

Instead, we can calculate how many tosses A may win if the bet > 1, correct ?

Can you unpack your suggestion, I thought A's expected value would be (1 - the probability B has to win) * the value for A,  but I am still in doubt how to calculated the value for A at each toss.
user305883
<utility><statistics><expected-utility><behavioral-economics>
Utility of both players in St. Petersbourg paradox - behavioural economics
user305883
# Utility of both players in St. Petersbourg paradox - behavioural economics

(Asked by: username_0 on 2021-12-23)

Im reading a paper by Karl Menger [1] about the St. Peterbourg paradox :

> In the theory of probability, the "Petersburg Game" designates the
> follow- ing gamei between two persons, A and B. Initially, A requires
> that B place a certain bet with A. Then they flip a coin. If it shows
> heads, then B receives 1 dollar from A, and the game is over. If it
> shows tails, it is flipped again. If the second throw turns up heads,
> then B receives 2 dollars from A, and the game ends. The coin is
> flipped until heads appears for the first time. If this happens at the
> nth throw (that is, if the first n - 1 throws aII come up tails, but
> the nth throw is heads), then B receives 2n - 1 dollars from A and the
> game ends. Depending on whether n = 1, 2, 3, 4, ... , B wins 2° = 1,21
> = 2,22 = 4, 23 = 8, ... dollars. What bet should B be willing to make? What is B's mathematical expectation?

his point is that the paradox does not rely on a mathematical expectation of infinite value, but on the discrepancy between the behavior of the player B and the expected result: in real life, normal people (Menger says) would not commit an infinite amount of money, in spite of expectation being mathematically infinite.

The lottery is always presented in terms that A will pay to B.
But the lottery never takes into consideration the expected value for A in committing the game.

* Which is the expected value of player A, in playing the Petersbourg game and risking to pay a sum to B ? Will A have a chance to win something ? Or is player A completely disregarded ?

[1] Karl Menger, The Role Of Uncertainty In Economics, 1979.

**Tags:** utility, statistics, expected-utility, behavioral-economics
> **Comments:**
> **username_1**: Please clarify your specific problem or provide additional details to highlight exactly what you need. As it's currently written, it's hard to tell exactly what you're asking.  
> **username_2**: Will A have a chance to win something ? Yes, if the number of throws n is such that $2^{n -1}<K$ where K is the bet.  But this is obvious.  Is that what you are asking ?  
> **username_0**: It is written that B places a certain bet with A: the game ends when B wins, and the coin keeps on being flipped until B wins. So I did not understand what for you it is obvious: is the Bet the prize at each throw, or is it the maximum number of throws that can be done by betting 1 dollar at each time ? E.g.  Say B wins with heads, and bet is 3 dollars. After 3 throws, the expected value should it be: `1/2*1^0 + 1/2*2^1 + 1/2 * 2^2`  Or `1/2* 3^0 + 1/2* 3^1 + 1/2 * 3^2`? If the bet is 3 dollar, A wins after 2 throws if both had tails, assuming each 1 dollar is bet at each throw ?  
> **username_2**: No I think B bets an amount only once, before any coins are flipped.  The ‘paradox’ is that B’s expected value is theoretically infinite, but in real life B would not pay unlimited amounts to play this game.  
> **username_0**: So, if B bets 1 dollars and heads does not show in the first 1 tosses, A would never win, right ?   (1 -  1/2)^0 + (1- 1/2) ^1 + (1 - 1/2) ^2 + ...] => 1 + 0.5 + 0.25 + .. ] => which is never 2−1<.   Instead, we can calculate how many tosses A may win if the bet > 1, correct ?  Can you unpack your suggestion, I thought A's expected value would be (1 - the probability B has to win) * the value for A,  but I am still in doubt how to calculated the value for A at each toss.  

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## 0 answers

