Probability that Alice Tossed a Coin Three Times If Alice and Bob Tossed Totally 7 Times

Problem 752

Alice tossed a fair coin until a head occurred. Then Bob tossed the coin until a head occurred. Suppose that the total number of tosses for Alice and Bob was $7$.

Assuming that each toss is independent of each other, what is the probability that Alice tossed the coin exactly three times?

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Hint.

Let $X$ be the number of Alice’s tosses until she got a head.

Then $X$ is a geometric random variable with parameter $p=1/2$.
Recall that a geometric random variable can be interpreted as the number of trials until getting the first success and each trial is successful with probability $p$ and fails with probability $q=1-p$.

Note that
\[P(X=k)= q^{k-1} p = (1-p)^{k-1} p.\]

Solution.

Let $X$ be the number of Alice’s tosses until she got a head. Also, let $Y$ be the number of Bob’s tosses until he got a head. As each toss is an independent event and the coin is fair, the random variables $X$ and $Y$ are geometric with parameter $p=1/2$ (the probability of success(head) is $p=1/2$ and the probability of failure(tail) is $1-p=1/2$).

This question asks to find the conditional probability
\[P(X=3 \mid X+Y=7) = \frac{P\left((X=3) \cap (X+Y=7) \right)}{P(X+Y=7)}.\]

Let us first compute the numerator as follows.
\begin{align*}
P\left((X=3) \cap (X+Y=7) \right) &= P\left((X=3) \cap (Y=7-X) \right) \\
&= P\left((X=3) \cap (Y=4) \right) \\
&= P(X=3) \cdot P(Y=4)
\end{align*}
since $X$ and $Y$ are independent.

Now, as $X$ and $Y$ follow the geometric distribution with parameter $p$, we have
\begin{align*}
P(X=3) &= \left(\frac{1}{2}\right)^2 \frac{1}{2} = \frac{1}{2^3}\\
P(Y=4) &= \left(\frac{1}{2}\right)^3 \frac{1}{2} = \frac{1}{2^4}.
\end{align*}
The product of these gives the numerator $\frac{1}{2^7}$.

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Next, let us calculate the denominator $P(X+Y=7)$.
The law of total probability yields
\begin{align*}
P(X+Y=7) = \sum_{k=1}^6 P(X=k) P(X+Y = 7 \mid X=k).
\end{align*}

As before, since $X$ is a geometric random variable with parameter $p=1/2$, we have
\[P(X=k) = \left(\frac{1}{2}\right)^{k-1} \frac{1}{2} = \frac{1}{2^k}.\]

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The second factor $P(X+Y \mid X= k)$ can be computed as follows.
\begin{align*}
&P(X+Y = 7 \mid X= k)\\
&= P(Y = 7 – X \mid X = k)\\
&= P(Y = 7 – k \mid X = k)\\
&= P(Y=7 – k)\\
&= \left(\frac{1}{2}\right)^{(7-k)-1} \frac{1}{2} = \frac{1}{2^{7-k}}
\end{align*}
Here, the third equality follows because $X$ and $Y$ are independent.

It follows that the denominator becomes
\begin{align*}
P(X+Y=7) &= \sum_{k=1}^6 \frac{1}{2^k} \cdot \frac{1}{2^{7-k}}\\
&= \sum_{k=1}^6 \frac{1}{2^7}\\
&= \frac{6}{2^7}.
\end{align*}

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Combining these computations, it follows that the desired conditional probability is
\begin{align*}
P(X=3 \mid X+Y=7) = \frac{1/2^7}{6/2^7}=\frac{1}{6}.
\end{align*}

Abstraction

If you solved this problem, then try the following problem as well. This is more abstract than the current problem but the idea is the same.

Its solution can be found in the post: Conditional Probability When the Sum of Two Geometric Random Variables Are Known

Click here if solved 2
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