Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

$5$ 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.

share|improve this question
    
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
add comment

1 Answer

up vote 0 down vote accepted

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: $P(X=4)=\frac{24}{120}=\frac15$.

share|improve this answer
    
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
    
Five arbitrary numbers might as well be 1,2,3,4,5 for the purposes of this problem, @CAF –  Thomas Andrews Dec 4 '12 at 13:23
    
I see, so we are essentially simplifying the problem without loss of generality? –  CAF Dec 4 '12 at 13:25
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.