In a 4 round chess tournament with 16 players (where the loser of each two player match is eliminated and the winner moves on to the next round), the pairings for all matches are decided randomly. The competing players all have different rankings and in any given match the higher ranked player always wins. What is the probability that the player with the second highest ranking is the losing finalist?
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1 Answer
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The answer is:
8/15
The reason is:
The only way for the rank 2 player to get to the finals is if the rank 1 player is in a different half of the tournament. There are 16*15 ways of arranging these two players, and 16*8 of those put the two in different halves. So the final answer is (16*8)/(16*15) = 8/15.
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$\begingroup$ wow dude, hold on, we need to find probability that second ranked player will be in the final, 1)probability that he gonna pass 1st round is 14/15(because there is only 15 players with whom he can play) 2) probability that he gonna pass 2nd round is 6/7 3) probability that he gonna pass 3rd round or be a finalist is 2/3. So I want to know what is the total should I add this 3 probability , or myltiply by each other? $\endgroup$– RanonCommented Feb 22, 2015 at 20:11
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$\begingroup$ If you want to do it that way, you'd multiply the probabilities at each step, giving you (14 * 6 * 2) / (15 * 7 *3). Simplifying that gives, again, 8/15. These are just different ways of asking the same question. You'll get the same answer however you do it. $\endgroup$ Commented Feb 23, 2015 at 0:07
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$\begingroup$ now I see, thank you Joshua I appreciate yout help $\endgroup$– RanonCommented Feb 23, 2015 at 1:11