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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

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 $Pr[b > a] = \frac{1}{2}$

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