Probability of playing tennis (Explanation of answer given in book)

Question

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

Answer 1

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:

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

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

Finally

$\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:

If

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

then

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

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.

Answer 2

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