MIT 6.041SC Probabilistic Systems Analysis and Applied Probability, Fall 2013
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Instructor: Katie Szeto
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A Chess Tournament Problem
MIT 6.041SC Probabilistic Systems Analysis and Applied Probability, Fall 2013
View the complete course:
Instructor: Katie Szeto
License: Creative Commons BY-NC-SA
More information at
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if B or C wins twice why is there a 2nd round required? Shouldn't it be the opposite?
What if the games end in a draw or stalemate?
so A has to win only once to retain his championship?
that's not fair
Cool explanation. By the way – you are adorable saying: "I'm just testing you guys" 🙂
Part C is unintuitive to me. If we know that Al beat someone in the first game he played, then intuitively it seems more likely that it would be Ci as opposed to Bo, but the result from C claims precisely the opposite. Can someone explain where my reasoning has broken down?
would have been nice if you explained the rules of the game. How am i supposed to know what round 2 is…???
If you check at 12:29 P(A|R_2) + P(B|R_2) > 1
I think correct answer is
P(B|R_2) = 0.1731; P(C|R_2) = 0.0277; P(A|R_2) = 0.7992 ; P(A|R_2)+P(B|R_2)+P(C|R_2) =1
There is a subtle mistake, each node of the tree does not correspond to an outcome, we reserve the word outcome for the overall outcome at the end of the overall experiment. As said By Professor John Tsitsiklis in Lecture 1. So each node is a result not an outcome please Clarify!
what i don't get :
1st point : if B beats C once then C beats B, then why does A retain his champion title ? Shouldn't B and C just match one last time before getting to the next round ? Computed probabilities would be different for sure
2nd point : If A wins only one game, does the assignment say he keeps his title ? for B or C needs to beat him twice doesn't seem to be very fair in terms of rules
Is it a mistake in solution for Part b (i) task P(B | R2)?
P(B intersection R2) should be 0.09 (i.e. in numerator)
Good to be reminded at the beginning of all of these videos that they are "fun".