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You repeatedly flip a coin until you either get a heads followed by a tails, or two heads in a row. Which is more likely to happen first?

Solution 1: Both are equally likely, because each flip, there's a $\frac{1}{2} \times \frac{1}{2} = \frac{1}{4}$ of both the previous flip and this flip being heads, and a $\frac{1}{2} \times \frac{1}{2} = \frac{1}{4}$ of the previous flip being heads and this flip being tails.

Solution 2: Heads followed by tails is more likely, because if you want heads heads, flipping a heads followed by a tails makes you start over, but if you want heads tails, flipping a heads followed by another head gives you another chance of succeeding next turn.

Which reasoning is correct, and intuitively, why is the other one wrong?

  • The second is correct. The first assumes that if two flips fail to satisfy the desired conditions, they don't have an impact on the next flips (which is untrue). – Rushabh Mehta May 14 '20 at 00:57
  • Thank you! I get it now –  May 14 '20 at 00:58
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    No. If you are flipping a single coin and the question is "which is more likely to appear first, $HT$ or $HH$?" then the answer is that they are equally probable. You flip as many tails as you like until an $H$ appears, and then the next toss is $H$ or $T$ with equal probability. If you meant something else, please clarify. – lulu May 14 '20 at 01:09
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    Your reasoning in (2) is an informal argument as to why the average (or expected) waiting time for HH is longer than for HT ($6$ rolls vs. $4$ rolls). But in terms of "who wins a head to head race", it's 50/50 as lulu explained. – Ned May 14 '20 at 02:12
  • https://math.stackexchange.com/questions/238313/why-does-phh-differ-from-pth your question is answered here – SagarM May 14 '20 at 15:28

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