Originally posted by eldragonflyI didn't make any assumptions, I wrote the problem. 😉
No the ambiguity was on my end, it was properly worded. Statistics has its own language, but you did make too many assumptions. It was not in error for me to limit the sample space the the two card experiment, but a little more thought would have yielded the same result.
A new problem! Another simple application of Bayes theorem, although I hope this one won't generate as much discussion:
The NY Mets and the LA Dodgers have just played their final game of the season, which was a make-up game due to a scheduling error. The venue was decided on a coin flip (50/50 chance of the game being held in NY or LA), but you missed the sports report so you don't know where it was held. During the season, the following stats were compiled:
1. When playing in NY, the Mets won 7/9 times and the Dodgers won 2/9 times.
2. When playing in LA, the Mets won 6/10 times and the Dodgers won 4/10 times.
(The statisticians who complied these stats assure you that the stats are perfectly predictive of the result of this make-up game.)
You overhear the result of the game on the radio on your way home from work, and the Mets won 11-4. A friend of yours, a compulsive gambler who doesn't follow baseball and doesn't know where the game was held either, says to you "I'll bet you even money that the game was in NY (i.e. I, your friend, will pay you even money if the game was held in LA, and you will pay me even money if the game was held in NY)."
(a) Is this a fair bet?
(b) If not, what odds should your friend give you to make it fair?
Originally posted by PBE6this is one of your careless assumptions, i find it a bit tedious of you to keep making these derogatory statements.
No, the dealer doesn't know what's on the other side of the card anymore than the contestant does. I should have made that more clear in the question.
Originally posted by eldragonflyThat's not an assumption, it's a premise of the problem. But I do take exception to your accusation that I made derogatory statements.
this is one of your careless assumptions, i find it a bit tedious of you to keep making these derogatory statements.
Try the new problem!
Originally posted by eldragonflyNo, I simply had to interpret the question the way that the author had intended it to be interpreted.
Oh but you are. You simply had to reword the problem or admit that is wasn't quite as random as you pretended. 🙁
If somebody brings up the Monty Hall problem at this point, I think I'll probably scream :-)
Originally posted by PBE67 to 3 that they played in NY
A new problem! Another simple application of Bayes theorem, although I hope this one won't generate as much discussion:
The NY Mets and the LA Dodgers have just played their final game of the season, which was a make-up game due to a scheduling error. The venue was decided on a coin flip (50/50 chance of the game being held in NY or LA), but you missed the ...[text shortened]... Is this a fair bet?
(b) If not, what odds should your friend give you to make it fair?
Originally posted by PBE6i don't understand the correlation between playoff game scores and final game location, i don't understand how that system works. Ultimately the location of the final playoff game was decided randomly by the toss of the coin. So i am assuming i am missing something here, the only other choice being that the randomly decided location and the final score are independent events.
Well then just answer me this, do you think Bayes theorem applies?
Hmm. Strange we're so close, yet different. I must be missing something subtle (or you are!). Maybe you can spot the flaw in this reasoning:
(M = Mets win)
We want to know P(NY | M) = P(NY & M)/P(M)
But also:
P(NY & M) = P(M | NY)*P(NY)
and
P(M) = P(M | NY)*P(NY) + P(M | LA)*P(LA)
We know:
P(M | NY) = 7/9
P(M | LA) = 6/10
P(NY) = P(LA) = 1/2
=> P(NY & M) = 7/9*1/2 and
P(M) = 7/9*1/2 + 6/10*1/2
So P(NY | M) = (7/9)/(7/9 + 6/10) = 70/124.
Originally posted by eldragonflyIt was decided on the toss of a coin initially. But we've been given some additional information (that the Mets won), and that allows us to make a further inference, which is where conditional probability comes in.
i don't understand the correlation between playoff game scores and final game location, i don't understand how that system works. Ultimately the location of the final playoff game was decided randomly by the toss of the coin. So i am assuming i am missing something here, the only other choice being that the randomly decided location and the final score are independent events.
It might be more obvious if you consider an extreme case. Let's say that instead of the win ratios given, the Mets always win in NY, and never win in LA. In that case when we hear that they win, we know the match must have taken place in NY.
[The formula I gave still gives the correct answer in this case - the answer resolves to 1 in the case where P(M|NY) = 1 and P(M|LA) = 0].
The case given isn't that extreme, but as long as the win percentages are different in the two cities we can refine the initial probability of 0.5.