# Thread: I have a real stumper here

1. ## 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!!!!!!!!!!!?????????????????//

2. ## Re: I have a real stumper here

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

3. ## Re: I have a real stumper here

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).

4. ## The Following User Says Thank You to Link For This Useful Post:

HoldenRg (03-30-2012)

5. ## Re: I have a real stumper here

btw.

I can dig it.........

6. ## The Following User Says Thank You to Link For This Useful Post:

HoldenRg (03-30-2012)

7. ## Re: I have a real stumper here

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...

8. ## Re: I have a real stumper here

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".

10. ## Re: I have a real stumper here

Originally Posted by Dason
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".
LOL. Thank you for clarifying Dason.

11. ## The Following User Says Thank You to Link For This Useful Post:

HoldenRg (03-30-2012)

12. ## Re: I have a real stumper here

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

13. ## Re: I have a real stumper here

Anyone? Anyone?

14. ## Re: I have a real stumper here

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.

HTH.

15. ## The Following User Says Thank You to Link For This Useful Post:

HoldenRg (03-30-2012)

16. ## Re: I have a real stumper here

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)).

17. ## The Following User Says Thank You to Link For This Useful Post:

HoldenRg (02-25-2012)

18. ## Re: I have a real stumper here

I am still a bit unsure but this is the best answer I have received yet. I appreciate it.

19. ## Re: I have a real stumper here

Thanks Link!!! So was my approach totally off or was I at least in the same area code of being right?

20. ## Re: I have a real stumper here

I hope I did not disrespect

21. ## Re: I have a real stumper here

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