# $5$ numbers distrubuted to $5$ players game

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

Attempt: $P(X=0) = P(\text{player 1 loses}) = P(\text{player 2 has higher number than player 1}) = 1/2$ I reasoned a half because we know know nothing about the distribution of numbers, so either he has a number lower or higher (not the same, because it is given that they are distinct), and both are equally likely.

$P(X=1) = P(\text{player 1 has higher than player 2 and player 1 has lower than player 3})$ To find this I am pretty sure I need to condition on player 1 beating player 2 first. Also, is it correct that these events are not independent, since the event that player 1 is lower than player 3 is only valid provided that player 1 beats player 2, that is requires knowledge about whether player 1 beat player 2?

So I said $P(\text{p.1 > p.2 and p.1 < p.3}) = P(\text{p1 < p3|p1 > p2})P(p1 > p2)$ I am struggling to compute the first term, but I think the second term is just 1/2 (this is the same 1/2 I computed for the first part) I tried working with a reduced sample space and listing possible outcomes but I didn't manage to attain the right answer this way.

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In the first case you mean $P(\text{player }2\text{ has higher number than player }1)$. –  Brian M. Scott Dec 4 '12 at 13:02
Yes, sorry typing error. Will edit. –  CAF Dec 4 '12 at 13:04

In your second case, if Player $1$ has $2$, Player $2$ must have $1$, and the remaining three players can have any permutation of $3,4$, and $5$, so there are $3!=6$ possibilities. If Player $1$ has $3$, there are $2$ possibilities for Player $2$ ($1$ and $2$), $2$ for Player $3$, and $2$ orders for the remaining two numbers, so there are $2\cdot2\cdot2=8$ possibilities. If Player $1$ has $4$, there are $3$ possibilities for Player $2$ and just $1$ for Player $3$, while the last two numbers can be assigned in either order, so there are $3\cdot2=6$ possibilities. The grand total for this case is therefore $6+8+6=20$, and $P(X=1)=\frac{20}{120}=\frac16$.
If $X=2$, Player $1$ must have $3$ or $4$, so a similar calculation isn’t too painful; I’ll leave it to you.
If $X=3$, Player $1$ must have $4$, Player $5$ must have $5$, and it’s not hard to see that there are just $3!=6$ possible arrangements: $P(X=3)=\frac6{120}=\frac1{20}$.
If $X=4$, Player $1$ must have $5$, and there are $4!=24$ possibilities: $P(X=4)=\frac{24}{120}=\frac15$.
What you have above is what I was trying to do with possible outcomes etc.. But does the above not assume we are dealing with the numbers 1,2,3,4,5. Are we not to deal with 5 arbritary numbers, $a_1, a_2, a_3, a_4, a_5, a_i \neq a_j?$ –  CAF Dec 4 '12 at 13:22