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Is the probability of tossing an odd/even number of heads is 0.5 for any $n$, or is it only valid for odd $n$?

I think it only holds for an odd $n$.

Consider the binomial theorem $$ (x+y)^n = \sum_{i=0}^n \binom{n}{i}x^{n-i}y^i $$

We see that for odd $n$, we have an even number of summands. And we see that $$ \sum_{i:odd} \binom{n}{i}x^{n-i}y^i = \sum_{i:even}^n \binom{n}{i}x^{n-i}y^i $$

But with an even $n$, we have an odd number of summands, so I don't think the above equality will hold.

user5965026
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2 Answers2

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The probability of getting an even number of heads in $n$ tosses is

$$\sum_{k\ge 0}\binom{n}{2k}2^{-n}=2^{-n}\sum_{k\ge 0}\binom{n}{2k}\;.$$

But $\sum_{k\ge 0}\binom{n}{2k}$ is just the number of even-sized subsets of a set of cardinality $n$, which is $\frac12\cdot2^n=2^{n-1}$ provided that $n\ge 1$. Thus, the probability of getting an even number of heads in $n$ tosses is

$$2^{-n}\sum_{k\ge 0}\binom{n}{2k}=2^{-n}\cdot 2^{n-1}=\frac12\;,$$

and of course the probability of getting an odd number of heads must also be $\frac12$.

Brian M. Scott
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  • Is there an easy way to see this from the symmetry of the binomial theorem? – user5965026 Jul 27 '20 at 05:11
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    @user5965026: I don’t know of one. Probably the easiest approach from the binomial theorem is to note that $$0=(-1+1)^n=\sum_k\binom{n}k(-1)^k\cdot 1^{n-k}=\sum_k\binom{n}k(-1)^k;,$$ where that last sum counts the even-sized subsets positively and the odd-sized subsets negatively: since this weighted sum is $0$, there must be exactly as many even-sized subsets as odd-sized subsets. – Brian M. Scott Jul 27 '20 at 05:16
  • I'd consider that an easy and intuitive proof :). Tbw's answer below actually made me realize a very nice visual by using a binary tree. We can label each node with the number of heads and each level as the number of tosses. Starting from level = 0 (0 tosses), we will have a single node labeled $0$. On level = 1, we will have 2 children nodes, 1 labeled "0" and 1 labeled "1". On each subsequent level, you will see that there will always be the same number of odd and even labeled nodes. – user5965026 Jul 27 '20 at 05:24
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Here's a simple proof that the probability is always $0.5$. After n tosses, there will either be an even or odd number of heads. On the next toss, there is $0.5$ probability that the parity will change and $0.5$ probability the parity will not change. Therefore, no matter what the probability of heads being even by the previous toss, on the n+1 toss, it will be $0.5$. This works for all n greater than or equal to $0$.

For example, after 0 tosses, the probability of even heads is $1$. Half the time, it will become even, half tails.

After 1 toss, the probability of even heads is $0.5$, as we just saw. If it's even, then the probability of staying even is $0.25$, and becoming odd is $0.25$. If it's odd, then the probability of staying odd is $0.25$, and becoming even is $0.25$.

Tbw
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  • Ahh, your approach is actually what inspired me to ask this question. When I read your first paragraph, I still had the confusion. But starting from 0 tosses really clears things up. – user5965026 Jul 27 '20 at 05:13
  • It also really helps if you draw a tree, where each node is labeled with the number of heads. Each level is the number of tosses. – user5965026 Jul 27 '20 at 05:15