# 2 player throw darts alternately. Find the probability generating function and expected number of throws

Max and Mark alternately throw darts.

The first one who scores a bull's-eye wins the game. the probabilities that Max and Mark score a bull's-eye in every shot are $\frac{3}{4}$ and $\frac{5}{8}$ respectively.

Max throws first.

Their successive throws are independent and the random variable $X$ denotes the number of throws before the game is over.

a) Find the probability generating function of the variable $X$ and verify your result.

b) What is the expected number of throws before the game is over?

Thanks!

• I suggest splitting this into two cases: Max wins and Mark wins. The number of rounds for each is basically geometrically distributed.
– amd
May 3 '18 at 20:17

Call the two people A and B, and denote by $a$ and $b$ their hitting probabilities in a single throw. Denote by $p_k$ $(k\geq1)$ the probability that the game ends after exactly $k$ throws. If A begins we have $$p_1=a,\quad p_2=(1-a) b,\qquad p_k=(1-a)(1-b)p_{k-2}\quad(k\geq3)\ .$$ Let $$p(x):=\sum_{k\geq1} p_k\,x^k$$ be the probability generating function. Then $$p(x)=p_1 x+p_2 x^2+\sum_{k\geq3}(1-a)(1-b)p_{k-2}\,x^k=p_1x+p_2x^2+(1-a)(1-b)x^2p(x)\ .$$ It follows that $$p(x)={ax+(1-a)b x^2\over 1-(1-a)(1-b)x^2}\ .$$ The expected number of throws is given by $$E=\sum_{k\geq1}k p_k=p'(1)={2-a\over a+b-a b}\ .\tag{1}$$ When $a={3\over4}$, $b={5\over8}$ one obtains $$E={40\over29}\ .$$ Formula $(1)$ can also be obtained without recurse to the function $p(x)$: Denote by $E_{\rm A}$ the expected number of additional throws when it is A's turn, and by $E_{\rm B}$ the expected number of additional throws when it is B's turn. Then we obviously have $$E_{\rm A}=1+(1-a)E_{\rm B},\qquad E_{\rm B}=1+(1-b)E_{\rm A}\ ,$$ which immediately leads to $(1)$.

• Wow, thanks for your solution, it's way cleaner than mine, using induction relationship instead of a definition at each $k$ allows for a nice cascading. I'm glad we have the same results, but gosh, your solution is awesome. May 7 '18 at 16:57

As Amd said, it's a matter of splitting between who wins first. But since they're playing in a defined order, it's an easy distinction: Max can only win on odd numbers of throws (1,3,...) while Mark can only on even ones (2,4,...).

Max loses until he wins means there are a number of $(1-3/4)(1-5/8)$ factors followed (he fails, then his buddy fails, and repeating) by a $3/4$ (he eventually lands the shot). Assign $1-p$ and $1-q$ to Max and Mark winning respectively, so it's easier to write your calcs.

Law governing $X$ is then: $$\forall k \geq 1, P(X=k)= \begin{cases}{p^{k/2}q^{k/2-1}(1-q)} \mbox{ if } k \equiv 1 \pmod 2\\ p^{(k-1)/2}q^{(k-1)/2}(1-p) \mbox{ if } k \equiv 0 \pmod{2}\end{cases}$$ To justify this calc, you have to indicate that events are independent (each shot is a new event, with no relation to the precedent allowing you to say that the probability of this multi-event is the product of the probability of each event)

You then sum these terms (note $A$ the random variable at which Max wins, and $B$ the others, which splits your variable $X$), alternating k and you'll find that: $$\sum_{k=0}^{\infty}{P(A=2k+1)} + \sum_{k=1}^{\infty}{P(B=2k)} = 1$$ if you don't make a mistake (notice that for the odd $k$, the sum with $2k$ as variable starts at $k=1$` as you can't win without throwing), which I guess is the intent of verifying your probability.

For the second question, you need to calc the expectation, which is $$\sum_{k=1}^{\infty}kP(X=k)$$, using the law above (again splitting on odd and even numbers). It's all calcs, with the main point being the ability to calc $$\sum_{k=1}^{\infty}{kx^{k}}$$ where you need to be clean (this infinite sum converges only under certain conditions).