P(A|R_2) and P(B|R_2) in the white board do not mean the same thing for different players and in particular P(B|R_2) is different of yours. It was just bad notation... In the white board: P(A|R_2) ⇒ probability of A retaining his title of champ given given that a 2nd round was required P(B|R_2) ⇒ probability that B is the challenger of A given that a 2nd round was required However for you P(B|R_2) means: P(B|R_2) ⇒ probability of B becoming the champ given given that a 2nd round was required Your calculation is right and so is hers given the problem questions, it's just that her notation was a bit misfortuned.
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!
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?
No, it's not a mistake. Although the problem is indeed set up weirdly, Katie mentioned at 8:30 that P(B) is shorthand for P(Bo wins 1st round) [wins 2 games], instead of P(Bo wins eventually) [wins 4 games].
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
if B or C wins twice why is there a 2nd round required? Shouldn't it be the opposite?
so A has to win only once to retain his championship?
that's not fair
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
P(A|R_2) and P(B|R_2) in the white board do not mean the same thing for different players and in particular P(B|R_2) is different of yours. It was just bad notation...
In the white board:
P(A|R_2) ⇒ probability of A retaining his title of champ given given that a 2nd round was required
P(B|R_2) ⇒ probability that B is the challenger of A given that a 2nd round was required
However for you P(B|R_2) means:
P(B|R_2) ⇒ probability of B becoming the champ given given that a 2nd round was required
Your calculation is right and so is hers given the problem questions, it's just that her notation was a bit misfortuned.
would have been nice if you explained the rules of the game. How am i supposed to know what round 2 is...???
+Serozhah Milashuk ocw.mit.edu/courses/electrical-engineering-and-computer-science/6-041sc-probabilistic-systems-analysis-and-applied-probability-fall-2013/unit-i/lecture-3/MIT6_041SCF10_tut01.pdf
+Tatenda Chawanzwa thank you
@@tatendachawanzwa8439 yeah, really wish she would have clarified up front that each round has 2 games
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!
only leaf nodes represent outcomes of the experiment
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?
You are not considering that the chance of Ci even entering the second round is a lot lesser than Bo's.
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)
No, it's not a mistake.
Although the problem is indeed set up weirdly, Katie mentioned at 8:30 that P(B) is shorthand for P(Bo wins 1st round) [wins 2 games], instead of P(Bo wins eventually) [wins 4 games].
Cool explanation. By the way - you are adorable saying: "I'm just testing you guys" :)
What if the games end in a draw or stalemate?
they choose the wrong game to base a probability problem on ;)
the question said draws are not possible.
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
read the problem....one of the comments has a link to it....
Good to be reminded at the beginning of all of these videos that they are "fun".