The question is: \"Fivedistinct numbers are randomly distributed to players numb

Question

The question is: "Fivedistinct numbers are randomly distributed to players numbered 1through 5. Whenever two players compare their numbers, theone with the higher one is declared the winner. Initially,players 1 and 2 compare their numbers; the winner then compares hernumber with that of player 3, and so on. Let X denote the number oftimes player one is the winner. Find P(X=i),i=1,2,3,4."

I can solve the problemjust be working out all of the outcomes, but is there a better wayto solve (i.e. some conditional or binomial way to answer thequestion). The answer up on cramster provided littleexplanation.

Please help.

Explanation / Answer

Question: Five distinct numbers are randomly distributed to players numbered1 through 5. Whenever two players compare their numbers, the one with the highest one isdeclared the winner. Initially, players 1 and 2 compare their numbers; the winner thencompares with player 3, and so on. Let X denote the number of times player 1 is a winner. FindP (X = i), i = 0, 1, 2, 3, 5. Answer: Let’s start with the case that player 1 never wins. That is,player 1 loses to player 2. If we construct the sample space, we would see that P (X = 0) = 1 becauseexactly one-half of the 2 5! permutations have the ?rst number (player 1) greater than thesecond number (player 2). 1 2 5! 1 P (X = 0) = = 2 5! Player 1 wins exactly 1 game if player 3 has a larger number thanplayer 1, but player 1 has a larger number than player 2. The number of ways this can happen isthe same as the number of ways that player 2 loses to player 1 and player 3(hypothetically), etc. Per the solution I used, P (X = 1) = P (Y2 Y4 > Y5 . SPlayer 1 wins 4 games. = {(Y1 , Y2 , Y3 , Y4 , Y5 )} = {(5, , , , )} where denotes a wildcard meaning these players have someother number besides 5 here. For brevity, I do not list all of the arrangements, since we haveenough information to solve the problem once we restrict Y1 = 5. Thus, 4! 1 P (X = 4) = = 5! 5 Finally, 1 1 17 P (X = even) = P (X = 2) + P (X = 4) = + = 12 5 60

Related Questions

Navigate

• Browse (All)
• Browse T
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.