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A fair coin is flipped 5 times. What is the probability of getting more heads than tails?

Note: Since this is a GRE practice question, I want a method that can help me solve this problem in the fastest way possible. Of course, I can compute this probability directly by considering the cases of getting 5, 4, or 3 heads separately. However, this approach would be very time-consuming.

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    $\begingroup$ The GRE will almost certainly have at least one symmetry-themed problem, so one should be on the lookout for that. However, with small enough enumbers, and if you are familiar with the Pascal Triangle to modest depth, calculation is time-competitive. $\endgroup$ – André Nicolas Jul 31 '13 at 15:26
  • $\begingroup$ in this special case, calculation is not time-competitive :-) $\endgroup$ – mau Jul 31 '13 at 16:41
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Fast: there is a bijection between sequences with more heads than tails, and sequences with more tails than heads, by simply reversing each flip. There would be a problem if 5 were even, but it's odd, so you're golden.

Hence, the answer is 50%.

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    $\begingroup$ If $5$ were even, we'd have a lot of additional concerns ;) $\endgroup$ – Hagen von Eitzen Jul 31 '13 at 15:42
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One more way to look at it: $5$ is an odd number, so the probability of getting an equal number of $H$ and $T$ in this experiment is $0$. Since your coin is fair, the sets 'more H than T' and 'more T than H' are of equal size ($16$), and the answer follows.

So both the number $5$ and the fair coin are important.

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  • $\begingroup$ Thanks! This makes it easier to solve the problem. $\endgroup$ – user87274 Jul 31 '13 at 23:30
  • $\begingroup$ You are welcome $\endgroup$ – Alex Aug 1 '13 at 1:03

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