# Combinatorial coin game

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?

• It is a case of independent events Nov 19, 2019 at 9:08

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