I have a real stumper here

I have a question that has stumped everyone I have asked. I
wonder if anyone can help. Here is the scenario.

Every time that team A plays on Tuesday they win 8/10 or 80% of
the time. Every time that team B plays on Tuesday they lose 6.6/10 or 66% of the time. This Tuesday they play each other, does the
probability of team A winning increase and the probablity of
team B losing increase?

Many tried to say that you multiply, but in this case the
situations are not independent, they are playing each other.

Can you dig it!!!!!!!!!!!?????????????????//
FYI this is not my homework. I am trying to find someone that can answer this question. I have asked several people in the so called math department in my district and everyone says let me get back to you on that. An answer would be greatly appreciated


Ninja say what!?!
You're right in stating that the events are dependent on each other. If A wins, then B has to lose (and vice versa). However, you haven't given us enough information.

You've said P(A=win | Tuesday) = .80 & P(B=win | Tuesday) = .34. To make this easier to explain, lets take the day out. Then P(A=win) = .80 and P(B=win) = .34.

Now, what you're asking is for P(A|B). That is, what is the probability of team A winning given that they're playing team B? This can be found by:

P(A|B) = P(A and B) / P(B)

Likewise, P(B|A) = P(A and B) / P(A).

You've given us P(A) and P(B). However, we don't know what P(A and B) is. Therefore, we can't find P(A|B).

Mean Joe

TS Contributor
Hey Link, if that is indeed your real name, I was thinking that P(A and B) most likely = 0. Since they play on Tuesday, it cannot be American football. And most other American sports (baseball, basketball, American Idol) cannot end in a tie/two winners. Unless we're talking aboot hockey...


Ambassador to the humans
Yeah I think there was some confusion when Link used B to refer both to "team B wins" and "the team that is being played against is team B".
So yeah I am not sure why I would take out Tuesday. There can't be a tie. So what I have come up with is combining both percentages and dividing them by 200. Gives the total percentage of one outcome. That is .80 + .66 = 1.4. Then 1.4 divided by 2= .73. Does this make sense to anyone or better yet can someone verify if this is right or wrong and why?

Still trying to get this one


Ninja say what!?!
Alright...I felt bad since no one was responding and will set up something formal and quick for you.

Let's assume that: A|Tuesday ~ Ber(p=.80) and B|Tuesday ~ Ber(p=.66). Let Team A "win" if A=1 and B=0, Team B "win" if A=0 and B=1, and a rematch occur otherwise (on an identical Tuesday).

Then P(Team A = "win") = P(A=1|Tuesday)*P(B=0|Tuesday) = (.80)*(.34) = .272
P(Team B = "win") = P(A=0|Tuesday)*P(B=1|Tuesday) = (.20)*(.66) = .132
P(Rematch) = (.80)*(.66) + (.20)*(.34) = .596

Now, since rematches don't count towards the overall probability of teams A or B winning, we can drop it and just calculate based off the prior two probabilities. This results in an "overall" P(Team A wins) = .272/(.272 + .132) = .67 and P(Team B wins) = .132/(.272 + .132) = .33.



Ninja say what!?!
please note that this approaches the problem with an independence and stochastic assumption (i.e. that it doesn't matter who team A plays...it could be team C, D, E, etc. and that there is randomness (no skill involved)).
I think I was a bit close all of my results seem to be about ten percent lower that than those of this formula. Interesting. Any more input is greatly appreciated
Well the way that I was coming up with the probability of each team winning. It is an earlier post . Also can someone explain why I am adding the probability that Team A wins plus the probability that Team B wins and then dividing. It seems to me that those two are different "units" shouldn't I be adding the probability that Team A wins plus the probability that Team B loses and then dividing? I am talking about the last part of the formula, the very end. Not questioning the answer just want to understand it better