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## 99 fair coins

Person \(A\) flips 99 fair coins and obtains \(a\) heads. Person \(B\) flips 100 fair coins and obtains \(b\) heads. What is the probability that \(a < b\) ?

**Source:**folklore

## Answers

Hint:

We have \(a\) = number of heads obtained by \(A\) after 99 flips.

We have \(b\) = number of heads obtained by \(B\) after 100 flips.

Let \(b'\) = number of heads obtained by \(B\) after 99 flips.

Note that \(Pr[b' > a] = Pr[b' < a]\).

Hence, \(Pr[b' > a] = \frac{1}{2}(1 - Pr[b' = a])\) ........ (1).

Using the law of total probability, we have

\(Pr[b > a] = Pr[b' < a]{\cdot}0 + Pr[b' = a]{\cdot}\frac{1}{2} + Pr[b' > a]{\cdot}1\) ........ (2)

From (1) and (2) we get \(Pr[b > a] = \frac{1}{2}\)