Every Sunday, Tom calls Alice to see if she will play tennis with him on that day. If Alice has not played tennis with Tom since i Sundays ago, the probability that she will say yes to him is $\frac{i}{k}, k ≥ 2, i = 1, 2, ..., k.$ Therefore, if, for example, Alice does not play tennis with Bob for $k − 1$ consecutive Sundays, then she will play with him next Sunday with probability 1. Let Z be the number of weeks it takes Alice to play again with Tom since they last played. Find the probability mass function of Z.

The textbook answer:

The event Z > i occurs if and only if Alice has not played with Tom since i Sundays ago, and the earliest she will play with him is next Sunday. Now the probability is $\frac{i}{k}$ that Alice will play with Tom if last time they played was i Sundays ago; hence $$P(Z>i)=1−\frac{i}{k}, i=1,2,...,k−1\tag{1}$$ Let p be the probability mass function of Z. Then, using this fact for 1 ≤ i ≤ k, we obtain $$p(i)=P(Z=i)=P(Z>i−1)−P(Z>i)= 1−\frac{i-1}{k} − \left(1−\frac{i}{k}\right) =\frac{1}{k}$$

My way of doing it for calculating $P(Z=2)$ is $$P(\text{she does not play with him the first Sunday, she plays with him the second Sunday)}=P(\text{she plays with him the second Sunday|she does not play with him the first Sunday})*\\P(\text{she does not play with him the first Sunday})\\=\frac{2}{k}\left(1-\frac{1}{k}\right)$$ Which is incorrect. Could someone explain to me why my answer is incorrect, and why equation (1) is correct? Many thanks


2 Answers 2


The main problem is how to interpret the given information.

Please note that

$P (Z \leq i-1| Z > i-1) + P (Z = i| Z > i-1) + P (Z > i| Z > i-1) = 1$

Since $P (Z \leq i-1| Z > i-1) = 0$

we must have $P (Z = i| Z > i-1) + P (Z > i| Z > i-1) = 1$

Also it is given that (Is this the intended meaning?)

$P (Z = i| Z > i-1) = \frac{i}{k}$

$ \equiv P (Z > i| Z > i-1) = 1 - \frac{i}{k}$ [The conditional probability]

Using this interpretation, we get very different expressions for $P(Z = 1), P(Z = 2), P(Z = 3)$ etc.

However, it looks like the book interpreted it differently and meant

$P (Z > i) = 1 - \frac{i}{k}$ [The unconditional probability]

I did some calculation to compute a closed form expression for $P(Z = i)$ without using the book's interpretation. It is as follows:

$P(Z = i) = \displaystyle \binom{k-1}{i-1} \frac{i!}{k^i}$

It looks like the above is correct and also $\displaystyle \sum_{i=1}^k P(Z = i) = 1$

How to prove/show the results?

Let me denote $P(Z = i)$ by $f(i)$ [only for notational convenience]

First, note that

Step (1): $f(i) = \displaystyle \frac{i}{k} \left(1 - \sum_{j=1}^{i-1} f(j) \right)$

or, $\displaystyle \sum_{j=1}^{i-1} f(j) = 1 - \frac{k}{i} f(i)$

Add $f(i)$ to both LHS and RHS to obtain

Step (2): $\displaystyle \sum_{j=1}^{i} f(j) = 1 + \frac{i-k}{i} f(i)$

In Step (1), replace $i$ by $(i+1)$ to obtain

$f(i+1) = \displaystyle \frac{i+1}{k} \left(1 - \sum_{j=1}^{i} f(j) \right)$

Now replace the quantity $\displaystyle \sum_{j=1}^{i} f(j)$ on the RHS of the above using step(2) to obtain

Step (3): $\displaystyle \frac{f(i+1)}{i+1} = \left(\frac{k-i}{k}\right) \frac{f(i)}{i}$

Next, put $\displaystyle \frac{f(i)}{i} = g(i)$ in the above step (3) to obtain

Step (4): $\displaystyle g(i+1) = \frac{k-i}{k} g(i)$

Also note that $g(1) = \frac{f(1)}{1} = \frac{1}{k}$

Now try repeated substitution:

$\displaystyle g(2) = \frac{k-1}{k} g(1) = \binom{k-1}{1} \frac{1!}{k} g(1)$

$\displaystyle g(3) = \frac{k-2}{k} g(2) = \left(\frac{k-2}{k}\right) \left(\frac{k-1}{k}\right) g(1) = \binom{k-1}{2} \frac{2!}{k^2} g(1)$


$\displaystyle g(i) = \binom{k-1}{i-1} \frac{(i-1)!}{k^{i-1}} g(1) = \binom{k-1}{i-1} \frac{(i-1)!}{k^i} $

or, $\displaystyle f(i) = \binom{k-1}{i-1} \frac{i!}{k^i}$

Next part: How to show $\displaystyle \sum_{i=1}^k f(i) = 1$?

If you try to show the above, you'll find that it is equivalent to prove the following:

$\displaystyle 1 + 2 \left(1 - \frac{1}{k} \right) + 3 \left(1 - \frac{1}{k} \right)\left(1 - \frac{2}{k} \right) + \ldots + k \left(1 - \frac{1}{k} \right) \left(1 - \frac{2}{k} \right) \ldots \left(1 - \frac{k-1}{k} \right) = k$

There should be a better way of proving the above but for now I have the following method only. We need the following result:


$\displaystyle S = 1 + 2(1-x) + 3(1-x)(1-2x) + \ldots + n(1-x)(1-2x) \ldots (1- [n-1]x)$


$\displaystyle S = 1 - \left( \frac{1-x}{x} \right) \left[(1-2x)(1-3x) \ldots (1-nx) - 1 \right]$

Now put

$\displaystyle n = k$ and $\displaystyle x = \frac{1}{k}$

in the above result to check that

$\displaystyle S = k$

Please let me know if you could follow all the steps. Else I can try to explain more.

  • $\begingroup$ I do not understand the following. Why is $P(Z\le i) =\frac{i}{k}$. Why is $P(Z=2)=\frac{1}{k}$ $\endgroup$
    – johnson
    Jul 28, 2019 at 7:13
  • $\begingroup$ I guess you could just explain why is $P(Z\le i)=\frac{i}{k}$. The complement is given in the textbook solution. My question is why is it that. I do not get it. $\endgroup$
    – johnson
    Jul 28, 2019 at 7:14
  • $\begingroup$ I edited the answer quite a bit to make the argument more direct. Please let me know if it is understandable now. $\endgroup$
    – PTDS
    Jul 28, 2019 at 18:18
  • $\begingroup$ I do not understand why $P(Z>i-1)=1-\frac{i-1}{k}$ from the question, which you assumed I did. $\endgroup$
    – johnson
    Jul 29, 2019 at 6:58
  • $\begingroup$ So the book's answer is incorrect? Can you also tell me how you obtained the close form expression and how it sums to 1. $\endgroup$
    – johnson
    Jul 30, 2019 at 15:18

Let us take $k=4$. I would read the question to mean she accepts the first week with probability $\frac 14$. If she doesn't play the first week, she accepts the second week with probability $\frac 12$, so the chance she plays the second week is $\frac 38$. If she hasn't played in the first two weeks she accepts the third with probability $\frac 34$, giving a chance of $\frac 9{32}$ for the third week. Finally she will accept the fourth week in any case, so plays the fourth week with probability $\frac 3{32}$ I agree with your answer for $Z=2$.

Then $P(Z=i)=\frac ik\prod_{j=1}^{i-1}\left(1-\frac jk\right)$

  • $\begingroup$ So the book's answer is incorrect? $\endgroup$
    – johnson
    Jul 30, 2019 at 15:18
  • $\begingroup$ Could you explain your answer for $P(Z=i)$ $\endgroup$
    – johnson
    Jul 30, 2019 at 15:18
  • $\begingroup$ It is just like your answer for $2$. The $\frac ik$ factor is the chance she accepts that week. The product is the chance she has not accepted any previous week. For $2$ there is only one previous week, so only one term in the product, which is the one you have. $\endgroup$ Jul 30, 2019 at 21:25

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .