Sixteen players $S_1,S_2......S_{16}$ playing a tournament are divided into eight pairs at random. From each pair, a winner is decided on the basis of a game played between the two players of the pair. Assume all players are of equal strength.

Find the probability that exactly one of s1,s2  are present in the  winners. My attempt :

There are two cases possible 1) when S_1 and S_2 are not paired with each other In this case they can be paired in $14C1*13C1$ ways. now we want exactly one from S_1 and S_2 as winner.. This can be possible when S_1 wins and S_2 loses or S_1 loses and S_2 wins so the total number of ways comes out to be $14C1*13C1*2$!

2)when S1 and and S2 plays against each other..there are 2 cases possible and both are favourable..as we want exactly one from S1 and S2 as a winner

Sample space= 14C1*13C1*(both win,both loose,one wins one loose)*4+2 Probability =14C1*13C1*2+2/(14C1*13C1*4+2 ) I know the question has already answered before...but I don't what is wrong with this approach

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    $\begingroup$ Do you mean the probability that exactly one of $S1, S2$ is in the finals? Or that exactly one of them wins at least one game? Use the symmetry that says each player is equally likely have any given result. $\endgroup$ – Ross Millikan Feb 23 '18 at 19:53
  • $\begingroup$ @RossMillikan It seems like he wants the probability that exactly one wins in the first round, I think. $\endgroup$ – saulspatz Feb 23 '18 at 19:56
  • $\begingroup$ That's why I multiplied numerator by 2 and denominator by 4 as all contenders are of same strength..i have used the very basic principle of probability $\endgroup$ – Akul sharma Feb 23 '18 at 19:57
  • $\begingroup$ You should learn to format your questions in MathJax. This is very hard to read. Is your final answer equal to $1/2?$ $\endgroup$ – saulspatz Feb 23 '18 at 20:05
  • $\begingroup$ Sorry my mistake.. $\endgroup$ – Akul sharma Feb 23 '18 at 20:13

Consider $S_1$'s opponent. Either it is $S_2$ at $\frac{1}{15}$ chance or another player at $\frac{14}{15}$ chance.

In the first case the desired condition will be fulfilled (as you observed).

In the second case, there's a $\frac 12$ chance each that $S_1$ or $S_2$ will win and $\frac 12 $ chance each they will lose. So $\frac 14$ chance of each of ($S_1$ win, $S_2$ win), ($S_1$ loss, $S_2$ win), ($S_1$ win, $S_2$ loss), ($S_1$ loss, $S_2$ loss). This gives $\frac 14 + \frac 14 =\frac 12$ chance of ending up with exactly one of $S_1,S_2$ as winners.

The probability follows easily.

In your analysis you need to weight the two different cases appropriately, which is perhaps not so easy. But once $S_1$ is not playing $S_2$ it's certain that $S_2$ is not playing $S_1$, of course, so $14C1$ is enough to ensure they aren't playing each other. Then because there are 4 outcomes of interest when $S_1$ is not playing $S_2$, but only 2 when $S_1$ is playing $S_2$, we need to double the weight of the $S_1$ v $S_2$ option:

$(14\times 2+1\times 2\times 2)/(14\times 4 + 1\times 2\times 2) = 32 / 60 = 8/15$

Perhaps more generally it is better to weight the successful options down by the size of the option space for each case:

$(14\times 2 / 4 +1\times 2/ 2)/(14 + 1) = 8/15$

Which starts converging to the previous analysis.

  • $\begingroup$ Yes I know it should be like 14/15*(1/2)+1/15=8/15..but I am just confused what is wrong with my approach $\endgroup$ – Akul sharma Feb 23 '18 at 20:23
  • $\begingroup$ But tell me how you will ensure that when S1 is not playing with S2 then exactly one of S1 and S2 will win.. According to your answer.. $\endgroup$ – Akul sharma Feb 24 '18 at 4:00
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    $\begingroup$ There's a $\frac 12$ chance each that $S_1$ or $S_2$ will win & $\frac 12 $ chance each they will lose. So $\frac 14$ chance of each of ($S_1$ win, $S_2$ win), ($S_1$ loss, $S_2$ win), ($S_1$ win, $S_2$ loss), ($S_1$ loss, $S_2$ loss), giving the $\frac 14 + \frac 14$ probability of one win and one loss $\endgroup$ – Joffan Feb 24 '18 at 4:52

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