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The game begins with $n$ coins.

  • Every round we pay 1 dollar, toss all the $n$ coins, and remove the $T$ coins which landed in Tail.
  • Then we start over with the remainder $n-T$ coins until there are no more coins.
  • You receive a prize $P$ when the Game ends (All coins landed Tails in any of the throws)

The excercise is to find the fair price of the final prize $P$ given $n$ (price in which in long term both players do not lose or earn any money from playing)

For n = 1 I thought on:

Coin toss Probability Total paid
T $1/2$ 1
HT $1/4$ 2
HHT $1/8$ 3
HHHT $1/16$ 4
HHHHT $1/32$ 5
HHHHHT $1/64$ 6
... ... ...

With $P(n=1) = \sum_{n=1}^{\infty}(\frac{n}{2^n}) = 2$

So the Fair Prize of this game for $n=1$ is 2 dollars.

How can it be calculated for a higher amount of initial coins? lets say $n=2$ or $n=16$.

My thoughts until now are, for 2 coins, in the first toss:

Coin results probability thoughts
(HH) 1/4 which I don't know how to proceed, because is recurrent, as with the 1 coin example
(HT) 1/4 the price of one coin
(TH) 1/4 same, price of one coin
(TT) 1/4 we finish the game, so you receive the prize.

Any thoughts?

Clues on how to push this to higher dimensions?

Oscar Flores
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  • 1
    The probability that a single coin comes up $T$ at least once in $m$ trials is $1-.5^m$. Therefore the probability that the game ends in $m$ rounds or fewer is $(1-.5^m)^n$ You can then get the probability that it ends in exactly $m$ rounds by subtraction. – lulu Sep 30 '23 at 15:23
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    The analysis in the comment of @lulu, which did not occur to me, is more elegant than what would have been my first try, which might or might not have succeeded: recursion. That is: $$E(n) = 1 + \left( ~\frac{1}{2^n} \times \sum_{i = 0}^n \left[ ~\binom{n}{i} \times E(n-i) ~\right] ~\right).$$ – user2661923 Sep 30 '23 at 16:28
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    You may find helpful: https://math.stackexchange.com/questions/3755804/interested-in-a-closed-form-for-this-recursive-sequence/3755824#3755824 https://math.stackexchange.com/questions/1361569/expected-flips-for-n-coins/1361582#1361582 https://math.stackexchange.com/questions/3944546/toss-100-fair-coins-and-take-away-the-tails-toss-the-remaining-coins-and-take-a/3944658#3944658 https://math.stackexchange.com/questions/26167/expectation-of-the-maximum-of-i-i-d-geometric-random-variables https://math.stackexchange.com/questions/2923416/asymptotic-growth-of-sum-k-1n-frac-left-1-rightk1-binomnk1-p – Hagen von Eitzen Sep 30 '23 at 19:27

1 Answers1

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Let $X$ be the amount of rounds at the end of the game.

Then $$P(X\le x) = \left(1- a^x\right)^n$$

for $a=\frac12$ (this is the same approach as @lulu's comment). Hence

$$P(X>x)= 1-P(X\le x) = \sum_{j=1}^{n} (-1)^{j+1} \binom{n}{j} a^{xj}$$

We are interested in $E[X]$, which is

$$\begin{align} E[X] &= \sum_{x=0}^\infty P(X>x)\\ &= \sum_{j=1}^{n} (-1)^{j+1} \binom{n}{j}\sum_{x=0}^\infty a^{xj} \\ &= \sum_{j=1}^{n} (-1)^{j+1} \binom{n}{j} \frac{1}{1-a^j} \\ \end{align} $$

I'm not sure if this can be further simplified.

PS: It seems not https://oeis.org/A158466 From the papers quoted there one can get some asymptotics.

A semiempirical approximation $E[X]\approx \log_2(n) +1.336 + {0.664}\frac{1}{n}$, quite precise for the full range $1\le n \le \infty$

leonbloy
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  • Amazing answer. I tried the two approaches, the one provided by @lulu (Calculating the difference between each of the values of the probability of the game ending in $m$ rounds or fewer), and the one provided by you (with the sum from j=1 to n) and both converge numerically to basically the same results.

    All of the steps are quite clear, but I don't get where the Sum is coming from. The one resulting from 1-P(X>x) Could you develop more on this one? Thanks a lot!

    (I also liked the semiempirical approximation ngl)

     – Oscar Flores Oct 01 '23 at 11:44
  • " I don't get where the Sum is coming from" The first one? It's just the Binomial theorem applied to the first equation. – leonbloy Oct 01 '23 at 14:31