All throws are independent. Both players start with zero points. When Emily throws a six she scores two points. When Greg throws a five or a six he scores one point. The winner is the first player to score two points.

(a) For each integer $k\geq 2$, what is the probability that Greg wins on his $k$-th throw?

(b) What is the overall probability that Greg wins?

I've already attempted part (a) and using the fact that Greg must have 1 success in k-1 throws and a success on k-th throw and also that Emily must fail to throw a 6 k times. With the formula: $P(X=k) = \binom{k-1}{r-1} * (1-p)^{k-r} * p^r$ and got

$P(X=k) = ((2/3)^{k-2} * (5/6)^k * (k-1))/9$ and this simplifies to:

$P(X=k) = 1/4 * (5/9)^k * (k-1)$

Is my answer to part (a) correct? If yes how do I tackle part (b) and if not where did I go wrong?

Many thanks.

  • $\begingroup$ Without formatting this is very hard to read. Do you mean $P(X=k)=\frac 14 \times \left(\frac {625}{6561} \right)^{k(k-1)}$ or $P(X=k)=\frac 14 \times \left(\frac {625}{6561} \right)^{k}\times (k-1)$ ? Both seem wrong...at least, neither sums to $1$. $\endgroup$ – lulu Dec 2 '16 at 12:44
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    $\begingroup$ I see the $(2/3)^{k-2} \cdot (k-2) \cdot (1/9)$ from Greg, and from Emily the $(5/6)^k$ for her not getting a six at all, so your first version for part (a) looks right. [I didn't check the simplification.] For part (b) you're summing a series related to a geometric one, but with $n$ factor before, can be done via derivative of usual geometric series. $\endgroup$ – coffeemath Dec 2 '16 at 12:45
  • $\begingroup$ @lulu Why should the $P(X=k)$ sum to $1,$ since that would be probability of Greg winning at some point, which excludes Emily's chances to wil? $\endgroup$ – coffeemath Dec 2 '16 at 12:48
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    $\begingroup$ If $P(X=k)=(2/3)^{k-2}\cdot (5/6)^k \cdot (k-1)/9$ then you are right. $\endgroup$ – callculus Dec 2 '16 at 12:56
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    $\begingroup$ @coffeemath I see the overall probability of a Greg win as $\frac 49$. I'll post the calculation below. Might be wrong, been that kind of morning. $\endgroup$ – lulu Dec 2 '16 at 12:59

To determine the overall probability that Greg wins:

Each round begins with some state, depending on whether Greg has scored a point or not. That is, we have two states: $S_0,S_1$. Accordingly, we let $p_0,p_1$ denote the probability that Greg wins assuming we are starting from the associated state. As we start in $S_0$ the answer we want is $p_0$.

Starting in $S_1$ we see that the possible outcomes for the round are: Emily wins (probability $\frac 16$), Greg wins (prob $\frac 56 \times \frac 13=\frac 5{18}$), we end up back in $S_1$ (prob $1-\frac 16 - \frac 5{18}=\frac 59$). Thus $$p_1=\frac 16\times 0 +\frac 5{18} \times 1 +\frac 59 \times p_1\implies p_1=\frac 58$$

Starting in $S_0$ we see that the possible outcomes for the round are: Emily wins (probability $\frac 16$), Greg scores and we move to $S_1$ (prob $\frac 56 \times \frac 13=\frac 5{18}$), we end up back in $S_0$ (prob $\frac 59$ as before). Thus $$p_0=\frac 16\times 0 +\frac 5{18}\times p_1 + \frac 59 \times p_0\implies p_0=\frac {25}{64}$$

Note: Using the formula $p(X=k)=\left( \frac 23 \right)^{k-2}\times \left( \frac 56 \right)^k \times \frac {k-1}9$ and summing over $k$ we confirm this probability.

  • $\begingroup$ This is a good approach based on a Markov idea and avoids the need to sum a series. +1 $\endgroup$ – coffeemath Dec 2 '16 at 13:04
  • $\begingroup$ @coffeemath Doesn't make it right though. I think I have the probabilities off...in the first, say, Greg needs Emily to lose, so the probability of his victory is actually $\frac 56 \times \frac 13$. I'll correct. $\endgroup$ – lulu Dec 2 '16 at 13:06
  • $\begingroup$ @coffeemath I believe it's right now. Even better, it matches the result given by the infinite sum. $\endgroup$ – lulu Dec 2 '16 at 13:14
  • $\begingroup$ @lulu I agree with you. $\endgroup$ – callculus Dec 2 '16 at 13:16
  • $\begingroup$ @coffeemath could you tell me how I could go about finding the infinite sum as I haven't yet learned about the "Markov" method. $\endgroup$ – user337254 Dec 2 '16 at 13:47

As OP asks (since Markov method is unfamiliar) the method for summing the series in this case applies when it is of the form $A=\sum_{n=0}^\infty n x^{n-1}.$ ) Of course here one doesn't need the first term which is zero, and the series we have converges provided $|x|<1.$

One starts with the known geometric series $G=\sum_{n=0}^\infty x^n=1/(1-x)$ and differentiates each side. On the left side this derivative is our series $A,$ while on the right it is $\frac{1}{(1-x)^2}.$

To finish using this in your example requires [using the series as in lulu's answer] manipulating the latter so one can apply the above to it. One factors out $(1/9)(5/6)^2$ so the rest is $(k-1)[(2/3)(5/6)]^{k-2}$ and then since this $k$ goes from $1$ to $\infty$ a shift of index shows our above formula should give $(1/9)(5/6)^2[(9/4)]^2=25/64$ as expected. Note the inside $x=(2/3)(5/6)=5/9,$ so $1-x=4/9,$ which gets "reciprocalled" and squared in the formula.


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