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Suppose you toss 6 fair coins. What is the probability of getting an even number of heads?

The answer is

$$p = \frac{1}{2^6}\left({6\choose 0} + {6\choose 2} + {6\choose 4} + {6\choose 6} \right) = \frac{1}{2}$$

Is there a more natural way to see this?

user1936752
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    See this question – lulu Oct 06 '23 at 10:29
  • That is very cool! – user1936752 Oct 06 '23 at 10:40
  • If the number of elements in the finite set is odd then it's obvious that the number of even ordered subsets is half the number of subsets (just pair each subset to its complement). Even if the number of elements in the finite set is even, a variant on the same idea works (just pair every subset containing some fixed element to the same subset without that fixed element). – lulu Oct 06 '23 at 10:46
  • Each configuration (eg. HTHTTH ) is equally probable. If you flip the first coin of an odd-numbered configuration eg. (HTHTTH goes to TTHTTH), then you get an even-numbered configuration. This bijection shows that the number of odd configurations is equal to the number of even ones. Thus the probability of getting even is equal to the probability of odd, giving $\frac{1}{2}$ – user3257842 Oct 06 '23 at 10:54

1 Answers1

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A single coin has an equal number of equally probable ways to get an even number of heads (1) as an odd number of heads (1). For each of these, a second coin will lead to an equal number of total evens (2) as total odds (2). And so on... up to six.

QED.

David G. Stork
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    You do not need induction or six fair coins, as just one of the coins being fair and independent of the others will do. See https://math.stackexchange.com/a/3872058/6460 – Henry Oct 06 '23 at 10:53