Alice tosses a fair coin \(10\) times. Bob tosses it \(11\) times. Charlie tosses it \(12\) times. Who is the most likely to get more heads than tails?

[SOLVED]

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Prunthaban Kanthakumar solved this puzzle:

Bob. The probability is \(0.5\) for Bob using symmetry. For Alice and Charlie, there is an additional case in which the number of heads is equal to the number of tails, the probability of which gets subtracted, so they get less than \(0.5\).

Susam Pal from cotpi added:

Bob is the most likely to get more heads than tails.

Let the probability that a particular person gets more heads than tails, the probability that he or she gets more tails than heads and the probability that he or she gets an equal number of heads and tails be \(P_h(X)\), \(P_t(X)\) and \(P_{eq}(X)\), respectively, where \(X\) is one of Alice, Bob or Charlie. Thus, \(P_h(X) + P_t(X) + P_{eq}(X) = 1\).

For each person, the probability that he gets more heads than tails is equal to the probability that he gets more tails than heads due to symmetry. Thus, \(P_h(X) = P_t(X)\).

Since Alice and Charlie have an even number of coins each, the probability that each gets an equal number of heads and tails is positive. Thus, \(P_{eq}(\text{Alice}) \gt 0\) and \(P_{eq}(\text{Charlie}) \gt 0\). However, \(P_{eq}(\text{Bob}) = 0\) because it is impossible to get an equal number of heads and tails by tossing an odd number of coins.

Thus, \begin{align*} & P_h(\text{Alice}) + P_t(\text{Alice}) + P_{eq}(\text{Alice}) = 1 \\ & \implies P_h(\text{Alice}) + P_t(\text{Alice}) = 1 - P_{eq}(\text{Alice}) \\ & \implies P_h(\text{Alice}) + P_t(\text{Alice}) \lt 1 \\ & \implies P_h(\text{Alice}) + P_h(\text{Alice}) \lt 1 \\ & \implies P_h(\text{Alice}) \lt 0.5. \end{align*}

Similarly,

\[ P_h(\text{Charlie}) \lt 0.5 \]

However,

\begin{align*} & P_h(\text{Bob}) + P_t(\text{Bob}) + P_{eq}(\text{Bob}) = 1 \\ & \implies P_h(\text{Bob}) + P_t(\text{Bob}) + 0 = 1 \\ & \implies P_h(\text{Bob}) + P_h(\text{Bob}) = 1 \\ & \implies P_h(\text{Bob}) = 0.5. \end{align*}

Credit

This puzzle is taken from folklore.

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