Runs of Consecutive Successes

Each draw is a jack with probability $p=\frac{1}{13}$ , independently of the others, because the deck is replaced and reshuffled. Write $k=4$ and let $E$ be the expected number of draws.

Key idea. Condition on when the current streak of jacks first breaks.

Look at the first $k$ draws. Two things can happen.

The first non-jack is draw $j$ , for some $1\le j\le k$ . This has probability $p^{j-1}(1-p)$ . We have used $j$ draws and must start again from scratch.
All $k$ draws are jacks. This has probability $p^k$ , and we stop after $k$ draws.

Hence

E=\sum_{j=1}^{k} p^{j-1}(1-p)\,(j+E)+p^k k .

Collect the $E$ terms. Their total coefficient on the right is $\sum_{j=1}^k p^{j-1}(1-p) = 1-p^k$ , so

p^k E=\sum_{j=1}^{k} j\,p^{j-1}(1-p)+k p^k .

The right-hand side is $\mathbb{E}[\min(G,k)]$ , where $G\sim\mathrm{Geom}(1-p)$ is the position of the first non-jack. Since $\mathbb{P}(G>i)=p^i$ ,

\mathbb{E}[\min(G,k)]=\sum_{i=0}^{k-1}\mathbb{P}(G>i)=\sum_{i=0}^{k-1}p^i .

Therefore

E=\frac{1+p+\dots+p^{k-1}}{p^k}=\sum_{j=1}^{k}p^{-j}.

With $p=\frac1{13}$ and $k=4$ :

E = 13+13^2+13^3+13^4 = 13+169+2197+28561=30940.

\boxed{30940}

Generalisation. The expected wait for $k$ consecutive successes with success probability $p$ is

\sum_{j=1}^k p^{-j}=\frac{p^{-k}-1}{1-p}.

Variations

The same idea with a twist. Each solution gives only the changed step.

Variation 11.6a

Alice and Bob repeatedly toss a fair coin. Alice wins as soon as three heads appear in a row; Bob wins as soon as three tails appear in a row. Find the expected number of tosses until one of them wins.

Work it out, then compare with the answer.

Solution

Call a toss after the first a match if it equals the previous toss. Each such toss is a match with probability $\frac12$ , independently of all earlier tosses.

The game ends exactly when two consecutive matches occur, since that means three equal tosses in a row. By the main result with $p=\frac12$ and $k=2$ , the expected number of tosses after the first is

2+2^2=6.

Adding the first toss gives

\boxed{7}