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Suppose that there are three players A, B and C participating a round robin tournament with the following probabilities.

  • A wins against B = 0.7
  • A wins against C = 0.8
  • B wins against C = 0.6

Find the following:

  • A wins against B and C
  • A wins against B given B won against C
  • The probability that each person wins one match

Solution:

  • 0.7 * 0.8 = ~0.56
  • Not sure but ifeel it's 0.7 * 0.8 * 0.6 = ~0.34
  • 0.7 * 0.8 * 0.6 * (1 - 0.7) * (1 - 0.8) * (1 - 0.6) = 0.008064

Can someone check my solution?

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Guide:

We are not told if the rounds are independent of each other.

Suppose that they are independent and there is no draws.

Then the probability that $A$ wins against $B$ given $B$ won against $C$ is equal to the probability that $A$ wins against $B$, $0.7$.

For the third case, assuming it means a user win exactly $1$ match. Suppose the number of wins by them are represented as $(a,b,c)$ where $a,b,c \ge 0$ and $a+b+c=3$. The complement event is $(3,0,0), (0,3,0), (0,0,3)$.

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  • $\begingroup$ Thank you. I felt that there's something missing in the question. Thanks for clarifying this. $\endgroup$
    – Benj Cabalona Jr.
    Commented Jan 31, 2020 at 8:55
  • $\begingroup$ For the last case the complement is larger. You can have also (2,1,0) in various combinations. What we need to compute is the probability that a wins from b, b wins from c, c wins from a. Or the alternative b wins from a, c wins from b, a wins from c. And those probabilities are 0.7*0.6*0.2+0.3*0.4*0.8 = 0.18 $\endgroup$
    – Sextus Empiricus
    Commented Mar 24, 2023 at 12:19

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