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Lavanya and Ketak each flip a fair coin (i.e., both heads and tails have equal probability of appearing) $n$ times. What is the probability that Lavanya sees more heads than ketak?

In the following, the binomial coefficient $\binom{n}{k}$ counts the number of $k$-element subsets of an $n$-element set.

- $\frac{1}{2}$

- $\frac{1}{2}\left ( 1-\sum_{i=0}^{n}\frac{\binom{n}{i}^{2}}{2^{2n}} \right )$

- $\frac{1}{2}\left ( 1-\sum_{i=0}^{n}\frac{\binom{n}{i}}{2^{2n}} \right )$

- $\frac{1}{2}\left ( 1-\frac{1}{2^{2n}} \right )$

- $\sum_{i=0}^{n}\frac{\binom{n}{i}}{2^{n}}$

2 votes

Option $(B)$ is the correct answer.

$P(H(L) > H(K)) + P(H(K) > P(L)) + P(H(K) == H(K)) = 1$

Now since, $P(H(L) > H(K))$ and $P(H(K) > P(L))$ are symmetrical, they will be equal (lets say = $Y$).

Assume, $P(H(K) == H(K)) = X$

So, $X + 2*Y = 1$

$\Rightarrow Y = (1-X)/2$

And, $X = P(H(L)==0$ AND $H(K)==0)$ + $P(H(L)==1$ AND $H(K)==1)$ + .... + $P(H(L)==n$ AND $H(K)==n)$

$X = \binom{n}{0}^2/2^{2n} + \binom{n}{1}^2/2^{2n} + .... + \binom{n}{n}^2/2^{2n}$

$X = \sum_{i=0}^{n} \binom{n}{i}^2/2^{2n}$

$Y = \frac{1}{2}(1 - \sum_{i=0}^{n} \binom{n}{i}^2/2^{2n})$

$P(H(L) > H(K)) + P(H(K) > P(L)) + P(H(K) == H(K)) = 1$

Now since, $P(H(L) > H(K))$ and $P(H(K) > P(L))$ are symmetrical, they will be equal (lets say = $Y$).

Assume, $P(H(K) == H(K)) = X$

So, $X + 2*Y = 1$

$\Rightarrow Y = (1-X)/2$

And, $X = P(H(L)==0$ AND $H(K)==0)$ + $P(H(L)==1$ AND $H(K)==1)$ + .... + $P(H(L)==n$ AND $H(K)==n)$

$X = \binom{n}{0}^2/2^{2n} + \binom{n}{1}^2/2^{2n} + .... + \binom{n}{n}^2/2^{2n}$

$X = \sum_{i=0}^{n} \binom{n}{i}^2/2^{2n}$

$Y = \frac{1}{2}(1 - \sum_{i=0}^{n} \binom{n}{i}^2/2^{2n})$