This is a really simple problem, but it's driving me insane. I came up with two solutions, both with two completely different answers:

Solution 1: Since we are dealing with fair coins, there are 2^10 ways to flip the coins, since we have two choices for each toss (head or tails). If one coin is going to be heads, we can think of it as having 1 choice for that coin, but still 2 choices for the remaining 9 coins, so that gives us 2^9/2^10, which simplifies to 50%.

Solution 2: Finding the number times we can get at least one head is essentially the the same as finding 1 - P(getting no heads). P(getting no heads) would mean we get 10 straight tails, which has a (1/2)^10. Therefore, the chance is 1 -(1/2)^10 which is almost 100%

Which one is wrong, and why?

  • $\begingroup$ Solution 2 is correct. Solution 1 is a totally different scenario. Can you figure it out yourself? $\endgroup$ – Peter Oct 13 '15 at 17:41
  • $\begingroup$ In the first attempt you are just counting the cases where you get exactly one head in a particular toss instead of getting atleast one head $\endgroup$ – G-man Oct 13 '15 at 17:41

The second is correct.

The first solution describes the probability that the first flip results in head which is not needed in order to get at least one head.


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