# Dart expected number of throws problem

To win a game, you need to hit the bullseye $$n$$ times in a row (if easier, let $$n$$ be $$3$$). The probability of hitting the bullseye is $$p$$ for each throw, so it's independent (if easier, let $$p$$ be $$0.6$$). What is the expected number of throws it takes you to win?

The difficulty I couldnt get past was the fact that it has to be in a row, so you could for example hit, hit, miss and then still have to hit three times in a row again, so every miss kind of resets you. If it helps, I wrote a short program which estimates the number of throws by simulating the game a $$100000$$ times, and the expected score for $$n=3$$ and $$p=0.6$$ turned out to be about $$9.07$$, but I need to get the exact value for any probability $$p$$, not just an estimate.

• You have a Markov chain with four states corresponding to having the last $0,1,2,3$ shots being hits. The last state is absorbing. The transition probabilities are $p$ to advance one state and $1-p$ to go back to $0$. Define variables for the expected time to absorb from each state. You should get a set of three simultaneous equations. Jan 5 at 2:45

Let $$k_n$$ be the expected number of throws to get $$n$$ bullseyes in a row. Then

• With probability $$1-p$$ we miss, so waste the first throw.

• With probability $$p(1-p)$$ we score then miss, so waste the first two throws

$$\vdots$$

• With probability $$p^{n-1}(1-p)$$ we score $$n-1$$ then miss the $$n$$-th, so waste the first $$n$$ throws.

• With probability $$p^n$$ we score $$n$$ in a row.

So putting that all together, we get $$k_n=p^nn+\sum_{i=1}^np^{i-1}(1-p)(k_n+i)=\left(1-p^n\right)k_n+\sum_{i=1}^{n+1}p^{i-1}\implies\boxed{k_n=\frac{1-p^n}{p^n(1-p)}}.$$

Alternatively, note that it takes $$k_{n-1}$$ throws in expectation to get $$n-1$$ in a row. Once we get there, either we toss a bullseye and win (with probability $$p$$), or we miss and start over (with probability $$1-p$$). So $$k_n=k_{n-1}+p+(1-p)(k_n+1)\implies k_n=\frac{k_{n-1}+1}{p},$$ and we are done by induction, as $$k_1=\frac{1}{p}$$.

Just for the record, here's a fun martingale solution.

Suppose at time $$k$$, a new gambler $$g_k$$ enters the casino and bets £$$1$$ that throw $$k$$ will hit the bullseye. If you miss, then $$g_k$$ loses so leaves the game; else $$g_k$$ wins £$$p^{-1}$$. In that case, $$g_k$$ then bets all their money that throw $$k+1$$ will hit the bullseye, and so on. This continues until either you miss, or you hit $$n$$ bullseyes in a row and $$g_k$$ wins £$$p^{-n}$$.

If $$M_k$$ is the total winnings for the casino at time $$k$$, then it's clear that $$M_k$$ is martingale. Let $$\tau$$ be the time at which you hit $$n$$ consecutive bullseyes. Clearly $$\mathbb E\left[\lvert M_{n+1}-M_n\rvert\right]$$ is bounded, so we can apply the optional stopping theorem: $$0=\mathbb E[M_0]=\mathbb E[M_\tau]=\mathbb E\left[\tau-p^{-1}-p^{-2}-\dots-p^{-n}\right]\implies\mathbb E[\tau]=\frac{1-p^n}{p^n(1-p)}.$$