Question

Five distinct numbers are randomly distributed to players numbers 1 through 5. Whenever two playerscompare their numbers, the one with the higher one is declared the winner. Initially, players 1 and 2compare their numbers; the winner then compares her number with that of player 3, and so on. LetXdenote the number of times player 1 is a winner.

Answer 1

Answer:

Observe that the first and the second player have equal probability to get any of number. Using the principle of that symmetry, we have that  

$P(X=0)=\frac{1}{2}$

Event X = 1 means that the first player has got greater number than the second player, but not than the third player. So, choose any three numbers out of five of them and say that the minimal number out of these three goes to the second player, mean number to the first one and the largest to the third one. Permute remaining two numbers on remaining two people. Hence

$P(X=1)=\frac{1}{6}$

Event X = 2 means that the first player has got greater number than the second and the third player, but not than the fourth player. So, choose any four numbers out of five of them and say that the minimal number and the next minimal out of these four go to the second and the third player (and permute them), third number to the first one and the largest to the fourth player. Give remaining number to the last person. Hence

$P(X=2)=\frac{1}{12}$

Event X = 3 means that the first player has got greater number than the second, the third, and the forth player, but not than the fifth player. So, permute these five numbers as follows: give the highest to the last person, the second highest to the first, and permute remaining numbers on the remaining people. Hence  

$P(X=3)=\frac{1}{20}$

Event X = 4 means basically the first player has won all the battles i.e, he has got the greatest number. Hence

$P(X=4)=\frac{1}{5}$