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On May 15 2009 06:28 Muirhead wrote:
qrs I don't understand what's wrong (except in case k=0 which doesn't make any sense anyways)
No, that's all I meant: k=0. It makes sense, just it's a special case. It wasn't a serious error in your answer, of course, just a technicality.
I assumed that our answers worked out to the same thing otherwise. Let me recheck that.
Edit: Yeah, I assume it's the same: I wrote
2*[(2n - k - 1)! * n! * n! ]/ [2n! * (n - k)! * (n-1)!]
We can simplify that a little bit, at least:
2*[(2n - k - 1)! * n! * (n-1)]/ [2n! * (n - k)!]
obviously not the most readable notation, and you probably looked at the bottom version and assumed that the last bit of the numerator was (n-1)!, like yours, so the denominator made no sense. But it was just (n-1) and it was coincidental. My method was a bit different from yours: I let the man keep selecting until both boxes were finished and then counted how many of those satisfied the problem's requirements.
Your way is a little nicer, probably: it seems clumsy to be counting picks that he didn't make. Still, it should work out to the same.
Edit2: Anyway, both of us were wrong, I think. Pliers and aznmathfreak were right.
We didn't pay attention to the line "(after absentmindedly placing the box back in his pocket after using the last match)".
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+ Show Spoiler +
(2C1)(nCn)(nCn-k)/(2nC2n-k)
=
2(nCn-k)/(2nC2n-k)
Explanation: There are 2 boxes, choose one to be empty at the end. In that box, there are n matches, choose n of them. In the other box, there are n matches, choose n-k of them. Finally, in total, there were 2n matches, and you chose 2n-k of them. This problem is very similar to a deck of cards, if you think of splitting the deck into two piles beforehand (ex, red and black).
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i wonder if kdog was capable of solving this
but sometimes human mind evolves at a rapid rate
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This is Banach match problem isnt it? Way to make your homework kiddo!
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This is more a matter of being careful than being difficult.
I'm wondering if there's a simplified answer to a more difficult version:
what's the probability of winning n games before the opponent wins k games.
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+ Show Spoiler +i have not read other answers because i wanted to try it myself. so if this has been said already, i apologize i would try this a binomial probability problem, where pulling from the empty box is a failure, and pulling from the non empty box is a success, so you find probability of n-k sucesses in 2n-k trials, where the probability of success and failure is .5 (2n-k)!/((n-k)!(n!)) * (.5)^(n-k) * (.5)^(n) that looks really ugly in text, but its just this with n= 2n-k and k = n-k
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On May 15 2009 09:12 igotmyown wrote:
This is more a matter of being careful than being difficult.
I'm wondering if there's a simplified answer to a more difficult version:
what's the probability of winning a best of 2n-1 series (whoever wins n games first) if you're down k games.
It depends how many games are left, of course.
So you're asking, in essence, if player X needs x games to win and player Y needs y games to win (where y = x - k), what is the probability of Y winning.
OK, changed everything.
We need to know how many arrangements there are of y wins by Y and up to x -1 wins by x. The up-to bit is annoying: I'd rather not use a sigma. Fortunately, it does not actually make a difference. So let's count the arrangements of y wins by y and x -1 wins by X. There are (y + x -1)! / [y! (x-1)!] of those. (If you want to know what the match result looks like, just ignore the final x's in each case.) By symmetry, of course, there are (y + x -1)! / [x! (y-1)!] cases where x wins. To make the answer look simpler, let's just call the first number A and the second number B.
Probability of Y winning the match is
A / (A + B)
Let's see if this simplifies:
+ Show Spoiler +
(y + x -1)! / [y! (x-1)!] / (y + x -1)! / [y! (x-1)!] + (y + x -1)! / [x! (y-1)!]
(y + x -1)! / [y! (x-1)!] / (y + x -1)! * (1/ [y! (x-1)!] + 1/ [x! (y-1)!] )
1/ [y! (x-1)!] / (1/ [y! (x-1)!] + 1/ [x! (y-1)!] )
1/ [y! (x-1)!] / ([x! (y-1)!] + [y! (x-1)!] / [y! (x-1)!] * [x! (y-1)!] )
[y! (x-1)!] * [x! (y-1)!] / [y! (x-1)! * ([x! (y-1)!] + [y! (x-1)!)]
[x! (y-1)!] / ([x! (y-1)!] + [y! (x-1)!)]
[x! (y-1)!] / (x! y!/y + y! x!/x)
[x! (y-1)!] / (x! y! (1/y +1/x))
(x! y!/y) / (x! y! (1/y +1/x)
1/y / (1/y + 1/x)
1/y / [(x + y)/yx]
yx / y(x + y)
x / (x + y)
and I am back to the same answer. Unbelievable.
well, someone will correct me if I have erred, I am sure. Right now, it's looking to me like arrangements of wins are actually irrelevant, although I am not sure why.
+ Show Spoiler [original method] +We need to know how many arrangements there are of y wins by Y and up to x wins by x. The up-to bit is annoying: I'd rather not use a sigma. So let's count the arrangements of y wins by y AND x wins by X. There are (y + x )! / [y! (x)!] of those. In how many of those does y win the match? Well, in all cases where the last win (the guaranteed irrelevant one) is by x. So then, assuming that each player has an equal probability of winning, your answer is:
(y + x- 1)! / y! (x-1)! / (y + x )! / [y! (x)!] We can do some canceling:
x / (y + x)
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On May 15 2009 09:49 qrs wrote:Show nested quote +On May 15 2009 09:12 igotmyown wrote:
This is more a matter of being careful than being difficult.
I'm wondering if there's a simplified answer to a more difficult version:
what's the probability of winning a best of 2n-1 series (whoever wins n games first) if you're down k games. It depends how many games are left, of course. So you're asking, in essence, if player X needs x games to win and player Y needs y games to win (where y = x - k), what is the probability of Y winning. We need to know how many arrangements there are of y wins by Y and up to x wins by x. The up-to bit is annoying: I'd rather not use a sigma. So let's count the arrangements of y wins by y AND x wins by X. There are (y + x )! / [y! (x)!] of those. In how many of those does y win the match? Well, in all cases where the last win (the guaranteed irrelevant one) is by x. So then, assuming that each player has an equal probability of winning, your answer is: (y + x- 1)! / y! (x-1)! / (y + x )! / [y! (x)!] We can do some canceling: x / (y + x)edit: and the simplicity of the reduced form suggests a better way of doing this: Ignore the arrangement of wins--that's a red herring: Just think about the possible ways that the match can end. For example, if you need 3 wins and I need 5, it could finish with: 3:0 5:0 3:1 5:1 3:2 5:2 3:3 3:4 (y + x) possibilities (one of us gets all his wins: the other gets from 0 to all but one of his wins). x of them are good for me (the ones where you don't get all your wins). That's it.
Different arrangements do not have equal probabilities of occurring.
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I know, that doesn't make sense to me. So maybe I did make a mistake in the first calculation...
edit: redid everything a different way, still got same answer
Edited my answer above now.
_______
OK, maybe this is a better way to think of it: Suppose we have two equally skilled players, Bob and Mike, just to give them names, competing against each other to win 3 games. It stands to reason that since both are equally skilled, and both are trying to do exactly the same thing, their tasks are equally hard. Therefore we split the chances between the two of them and say that each has a 50% chance of winning.
Now suppose that Bob has to win 3 games while Mike has to win 5. This time, Mike's task is 5/3 as hard as Bob. Bob is the 5:3 favorite, making the odds of his winning 5/8 and the odd's of Mike's winning 3/8.
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You're omitting the 1/2^(n+r) values. Take the 2004 Boston Red Sox being down 3-0 to the yankees. Boston does not have a 1 1+4) chance of winning the series.
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On May 15 2009 17:02 igotmyown wrote:You're omitting the 1/2^(n+r) values. Take the 2004 Boston Red Sox being down 3-0 to the yankees. Boston does not have a 1 1+4) chance of winning the series.
You're right. I think my mistake was in counting only the outcomes that actually happen (with meaningless games not played) and assuming they all had the same probability, whereas really they don't, because "3-0 and wins" is shorthand for all the possible (unplayed) outcomes that begin with 3-0. If I'm going to take the count-the-outcomes approach, I had better count them all. So let me redo this as if it were a showmatch, where all games are played:
P= 2^(y + x - 1) outcomes possible.
F= (y + x - 1 CHOOSE y) + (y + x - 2 CHOOSE y + 1)...(y + 0 CHOOSE y) favorable outcomes. (There's probably a slicker way to count them but I can't think of one at the moment.)
Total = F/P.
So the Red Sox would have a 1/16 chance (but that was your point).
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EPT
