# Expected payout of throwing darts

I am doing a problem from Mark Joshi's Quant Job Interview book, just to have some fun with probability.

The (paraphrased) question is as follows:

Suppose I throw a dart at a circular board of radius $$R$$. For the sake of the problem, we will assume that we always hit the board, but we are equally likely to hit anywhere on the board. Suppose you win a dollar if you hit 10 times in a row inside a radius of $$R/2$$, but each time you throw you have to pay 10 cents. If you throw 100 hundred times, how much money will you have won/lost in expectation?

My work and my question:

First, I am assuming that the count of dart throws inside a radius $$R/2$$ does not reset after being paid 1 dollar. In other words, if I threw 11 darts in a row inside a radius of $$R/2$$, I would make 2 dollars.

With our assumptions, the probability of hitting the board inside a radius of $$R/2$$ is $$1/4$$.
Let $$X_i$$ be the random variable that gives the payout of the game on the $$i$$th throw, where $$10 \leq i \leq 101$$ ($$i$$ ends at 101 since we are 1-indexing). So $$X_i = 1-0.1 = 0.9$$ with probability $$(1/4)^{10}$$ and $$X_i = -0.1$$ with probability $$1-(1/4)^{10}$$. Let $$\tilde{X}$$ be the random variable that describes the expected payout of the game. Then $$E[\tilde{X}] = E[X_{10} + \dots + X_{101}] = 91E[X_i].$$ Furthermore, we have $$E[X_i] = (0.9)((1/4)^{10}) + (-0.1)(1-(1/4)^{10})$$ so $$E[\tilde{X}] \approx -9.099$$. Now, we just have to add the cost of the first 9 throws where we deterministically lost 0.1 dollars, i.e. the total expectation approximately $$(-9.099 + 9(-.1)) \approx -10$$.

Is this analysis correct?

• Seems plausible Commented Jul 12, 2023 at 22:51
• When you doubt your analysis, my advice (besides carefully justifying all your steps) is to try to simulate it. It's often a useful check, and with a good environment, can be done in a few moments. If you use R, replicate is a really handy tool for this Commented Jul 13, 2023 at 4:13