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The following game is being played :

Player $\mathrm{B}$ pays to Player $\mathrm{A}$ an amount $\mathrm{X}$ and throws a coin at most $20$ times. If at the toss $k\space (k \leq 20)$ tail is thrown, the game stops and player $\mathrm{A}$ has to pay to player $\mathrm{B}$ $2^k$ cents. Else A gets to keep the stake $\mathrm{X}$.

  • Now the questions are: When is the game fair and what is a corresponding fair stake $\mathrm{X}$? I assume that since it is a binomial distribution, the expected value is $np$, hence here the expected value is $10$ and so $X = 10$ for a fair game. Does anything else has to be considered such that the game is fair?

  • Now the game is changed as follows: $\mathrm{A}$ has only a capital of $50$ dollars, but he lets player $\mathrm{B}$ throw the coin $20$ times anyway. The rules are now changed such that $\mathrm{A}$ pays to player $\mathrm{B}$ $ (5000+X)$ cents if $\mathrm{B}$ throws tail for the first time at the $k$-th throw and $2^k > (5000+X)$ holds. What would be here a fair amount $\mathrm{X}$?

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The distribution is geometric, not binomial. – Eckhard Apr 6 '13 at 10:03
up vote 0 down vote accepted

(i) The probability that the first tail appears at the $k$'th throw is $p_k=2^{-k}$. It follows that the expected payout is $$E=\sum_{k=1}^{20} 2^k p_k=20\ .$$ For a fair game the stake then should be $20$ cents.

(ii) Given the stake $x$ put $r:=\lfloor\log_2(5000 +x)\rfloor$. The probability of a win is then $$p=\sum_{k=r+1}^{20} p_k=2^{-r}-2^{-20}\ .$$ For a fair game we should then have $$x=(5000+x) p=(5000+x)\left(2^{-\lfloor\log_2(5000 +x)\rfloor}-2^{-20}\right)\ .$$ Plotting the function $$f(x):=(5000+x)\left(2^{-\lfloor\log_2(5000 +x)\rfloor}-2^{-20}\right)-x$$ with the help of Mathematica one finds a unique zero at $x\doteq 1.22$. Therefore a fair stake would be $1.22$ cents.

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