Problem Statement
Suppose you flip a fair coin 10 times. We say a certain outcome (heads or tails) is leading after \(n\) flips if there are strictly more than \(\frac{n}{2}\) flips of that outcome within the first \(n\) flips.
Find the probability that the outcome of the first flip of the coin is leading after 10 flips.
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Solution
Let’s say the outcome of the first flip is \(k\) (could be either heads or tails — doesn’t matter, due to symmetry).
Now, for \(k\) to be leading after 10 flips, the total number of \(k\)’s in the entire 10-flip sequence must be strictly greater than 5, since we’re considering \(n = 10\).
But since the first flip is already \(k\), we just need to count how many additional \(k\)’s show up in the remaining 9 flips so that the total count of \(k\) becomes greater than 5.
So we’re looking for how many ways we can get:
- \(5\) more \(k\)’s out of 9 → total \(6\)
- \(6\) more \(k\)’s → total \(7\)
- \(7\) more \(k\)’s → total \(8\)
- \(8\) more \(k\)’s → total \(9\)
- \(9\) more \(k\)’s → total \(10\)
So the total number of favorable outcomes is: \(\binom{9}{5} + \binom{9}{6} + \binom{9}{7} + \binom{9}{8} + \binom{9}{9} = 126 + 84 + 36 + 9 + 1 = 256\)
Each of these outcomes has equal probability (since each coin flip is independent and fair), and there are \(2^9 = 512\) total possible outcomes for the remaining 9 flips.
So the final probability is: \(\frac{256}{2^9} = \frac{256}{512} = \frac{1}{2}\)
So the required probability is \(\boxed{\frac{1}{2}}\)
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