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I found this in my math book. I have the answer but no explanation for it. I don't have any ideas on how to explain the formula.

$N$ persons are playing for $2m$ coins the following way: First, person 1 throws all the coins. If exactly half of them are heads, they win the game and all coins. If not the turn goes to person 2 who does exactly the same thing. If none of the $N$ persons win the first round, the turn goes back to person 1, and the game continues. The probability that person $k$ wins is $\frac{r(1-r)^{k-1}}{1-(1-r)^N}$ where $r=\frac{\binom{2m}{m}}{2^{2m}}$. Can you explain the formula?

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  • $\begingroup$ It is a case of independent events $\endgroup$
    – mathsdiscussion.com
    Nov 19, 2019 at 9:08

2 Answers 2

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It's called Geometric probability. This means that for person $k$ to win, $k-1$ must lose, I.e. $(1-p)^{k-1}p$. The probability of winning is sampling $m$ coins out of $2m$. This solution is for arbitrary $k$. Can you extend it to the rotating case, with a total of N players?

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The probability that player $1$ wins on the first toss is $r$. Let's call the probability that player $1$ wins (at all, not just on the first toss) $p_1$. If player $1$ doesn't win on the first toss, then player $2$ is in the same position as player $1$ was just in, so the overall probability that player $2$ wins is $p_2=(1-r)p_1$. It's an easy induction to see that the probability that player $k$ wins is $p_k=(1-r)^{k-1}p_1$.

One of the players eventually has to win. Thus:

$$1= \sum_{k=1}^N p_k = \sum_{k=1}^N (1-r)^{k-1}p_1 = p_1\sum_{k=1}^N (1-r)^{k-1} = p_1\frac{1-(1-r)^N}{1-(1-r)} =p_1 \frac{1-(1-r)^N}{r}.$$

Thus, $$p_1 =\frac{r}{1-(1-r)^N} \text{ and }p_k=\frac{r(1-r)^{k-1}}{1-(1-r)^N}.$$

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  • $\begingroup$ How do you get $$ p_1\sum_{k=1}^N (1-r)^{k-1} = p_1\frac{1-(1-r)^N}{1-(1-r)} $$ ? $\endgroup$
    – Benjamin Arvola
    Nov 20, 2019 at 0:01
  • $\begingroup$ That's the sum of a (finite) geometric series with ratio $1-r$: $$\sum_{k=0}^{N-1} t^k = \frac{1-t^N}{1-t}$$. $\endgroup$
    – Robert Shore
    Nov 20, 2019 at 0:57

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