# Find pmf for $i=0,1,2,3,4$

I have a problem that I'm having trouble with. Here is the problem:

"Five distinct numbers are randomly distributed to players numbered 1 through 5. Whenever two players compare their numbers, the one with the higher one is declared the winner. Initially, players 1 and 2 compare their number; the winner the compares his number with that of player 3, and so on. Let X denote the number of times player 1 is the winner. Find $P(X=i), i=0,1,2,3,4.$ "

Of course $P(X=0) = \frac{1}{2}$, but I went on to $P(X=1)$, which I thought equaled $\frac{1}{2} \times \frac{1}{2}$. When I looked up the solution it showed $P(X=1)=\frac{1}{2} \times (1- \frac{2}{3})$.

My thinking was $P(X=1)=P$(player 1 wins the first and loses the second)=P(player 1 wins the first)P(player 1 loses the second)=P(player 1 wins the first)*(1-P(player 1 wins the second).

Why is P(player 1 loses the second) $= \frac{1}{3}$?

• How are the comparisons done exactly? That's not clear from your description. Jun 10, 2015 at 21:27
• Initially, players 1 and 2 compare their number; the winner the compares his number with that of player 3, and so on. Jun 10, 2015 at 21:31
• Note that by your computation, the sum of your probabilities is not 1..... Jun 11, 2015 at 1:10

We assume that Player 1 compares with Player 2. If Player 1 wins,she next compares with Player 3, and so on.

We use a little trick, first finding the probability that $X\ge k$. By your calculation, $\Pr(X\ge 1)=\frac{1}{2}$.

Next we find $\Pr(X\ge 2)$. The event $X\ge 2$ happens if Player 1 wins at least twice in a row. Whatever numbers Players 1,2,3 are holding, the probability that Player 1 has the largest is, by symmetry, $\frac{1}{3}$. It follows that $\Pr(X\ge 2)=\frac{1}{3}$.

Finally, $$\Pr(X=1)=\Pr(X\ge 1)-\Pr(X\ge 2)=\frac{1}{2}-\frac{1}{3}.$$

The same method can be used to calculate $\Pr(X=2)$. We have $\Pr(X\ge 3)=\frac{1}{4}$. It follows that $\Pr(X=2)=\frac{1}{3}-\frac{1}{4}$.

The rest can be done the same way, though we can take a shortcut since $\Pr(X=4)$ is very easy.

The same idea works with $n$ players and $n$ distinct numbers.

You can't multiply those probabilities because winning the first and losing the second are not independent. He's more likely to win the second given that he's won the first.

Initially there are $3! = 6$ equally likely orderings for the first 3 players. Beating the first and losing to the second is 1 of those orderings, so its probability is $1/6$. Once he beats the first, that leaves just 2 out of the 6 orderings, so the probability that he then goes on to beat the second becomes $1/3$.

P(win first and lose second) = P(win first)*P(win second | win first)

$= \left(\frac{1}{2}\right)\left(\frac{1}{3}\right) = \frac{1}{6}$