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In this game, I flip M fair coins and my opponent flips N coins. If I get more heads from my coins than my opponent, I win, otherwise I lose. I wish to know the probability that I win the game.

I came to this:

$$P(victory) = \sum\limits_{i=1}^M \left(\binom{M}{i}\left(\frac{1}{2}\right)^i\left(\frac{1}{2}\right)^{M-i} \times \sum\limits_{j=0}^{i-1} \left(\binom{N}{j}\left(\frac{1}{2}\right)^j\left(\frac{1}{2}\right)^{N-j} \right) \right)$$

My questions would be: Is the above formula correct? And can a closed form formula exist, and if not, is there a simple proof?

MJD
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Petr Hudeček
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3 Answers3

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You can simplify the problem by exploiting its symmetry. The probability of flipping heads is the same as the probability of flipping tails. See where this is going...?

Suppose that you flip $m$ heads out of $M$ coins, and your opponent flips $n$ heads out of $N$ coins. You win if $n<m$. Let $\bar m=M-m$ be the number of tails you flip. Then you win if $\bar m +n<M$.

Now relabel your coins so that heads become tails and tails become heads. Now you win if $m+n<M$. Only the total number of heads flipped out of all $M+N$ coins matters! Let's use the index $t$ to represent the total. Then your win probability is

$$P=\frac{1}{2^{M+N}}\sum_{t=0}^{M-1}\binom{M+N}{t}.$$

This is a cumulative binomial distribution; as far as I know, it doesn't have a simpler form.

Chris Culter
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If you toss $M$ coins, there are ${M \choose m}$ ways to get exactly $m$ heads.

Then, the number of ways you can win when your opponent tosses $N$ coins, given that you've tossed $m$ heads, is

$${M \choose m}\sum_{i=0}^{m-1}{N \choose i}.$$

Now, just sum over all possible values of $m$, and divide by the total number of outcomes:

$$P_{win}(M, N) = 2^{-(M+N)}\sum_{m=0}^{M}{M \choose m}\sum_{i=0}^{m-1}{N \choose i}.$$

This looks pretty close to what you have.

John
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Your formula is basically right, but you can simplify the $1/2$ factors. Letting $V\equiv M>N$ (victory), using conditional probabilities:

$$P(V)= \sum_M P(V , M) = \sum_M P(M) P(V | M) = \sum_{m=0}^M \frac{1}{2^M}{M \choose m} \sum_{j=0}^{m-1} \frac{1}{2^N}{N \choose j}=\\=\frac{1}{2^{M+N}}\sum_{m=0}^M {M \choose m} \sum_{j=0}^{m-1} {N \choose j}$$

and you must also impose the convention ${N \choose j}=0$ if $j>N$

I doubt that this has a simpler expression.

leonbloy
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