- 09 Mar '06 22:05 / 1 editSuppose I play chess. I win my first game and lose my second. Thereafter, the probability that I win is equal to the proportion of previous games I have won. What is the probability that I win k of my first n games, for each possible value of k?

Now consider the same game, except that or my first a+b games, I win a and lose b. What is the probability of winning k of my first n games?

For an extra-credit rec, give a coherent mathematical account of how the title of this relates to the problem.

For an extra-credit PFCF rec, give a coherent account of any kind of how the title relates to RHP matters at large. - 09 Mar '06 22:27

The first statement is akin to saying if I flip a coin and I get ten*Originally posted by royalchicken***Suppose I play chess. I win my first game and lose my second. Thereafter, the probability that I win is equal to the proportion of previous games I have won. What is the probability that I win k of my first n games, for each possible value of k?**

Now consider the same game, except that or my first a+b games, I win a and lose b. What is the probabi ...[text shortened]... PFCF rec, give a coherent account of any kind of how the title relates to RHP matters at large.

heads in a row, the eleventh throw should be stastically tilted in the

direction of that throw also being a head. But in reality, the next throw

is still just 50-50. I think the same applies to your games. - 09 Mar '06 22:33 / 1 edit

No, by assumption, the probability of the next game being a win is dependent on how many I have won before,*Originally posted by sonhouse***The first statement is akin to saying if I flip a coin and I get ten**

heads in a row, the eleventh throw should be stastically tilted in the

direction of that throw also being a head. But in reality, the next throw

is still just 50-50. I think the same applies to your games.**regardless**of how it may actually be with real chess games. In fact, IIRC the rating system on RHP works in roughly the same way; a win adds to our estimate of a players likelihood of winning the next game. Thus, in the first part, the probability of winning my third game is 1/2, but the probability of winning my fourth game is... - 09 Mar '06 22:37

... 2/3 if you win the third game and 1/3 if you lose the third game.*Originally posted by royalchicken***No, by assumption, the probability of the next game being a win is dependent on how many I have won before, [b]regardless**of how it may actually be with real chess games. In fact, IIRC the rating system on RHP works in roughly the same way; a win adds to our estimate of a players likelihood of winning the next game. Thus, in the first part, the probability of winning my third game is 1/2, but the probability of winning my fourth game is...[/b] - 09 Mar '06 22:40

Now the second statement doesn't make sense. It seems to be the*Originally posted by royalchicken***No, by assumption, the probability of the next game being a win is dependent on how many I have won before, [b]regardless**of how it may actually be with real chess games. In fact, IIRC the rating system on RHP works in roughly the same way; a win adds to our estimate of a players likelihood of winning the next game. Thus, in the first part, the probability of winning my third game is 1/2, but the probability of winning my fourth game is...[/b]

same thing as your first, in both situations I think you are saying

win the first and lose the second in both. I read it several times and

the meaning still says both are the same.

Anyway, if if its not like flipping a coin, the third game based on

1 win and 1 loss should still be 50%, with two outcomes:

1 win, 2 losses, fourth game should be a loss, 1/3.

2 wins, 1 loss, fourth game, win, 1/3. So it looks like a series,

1/2, 1/3,1/4,1/5 etc. - 09 Mar '06 22:45 / 2 edits

You've got the correct answer, but for an essentially wrong reason which the second part of the problem should reveal, if I've understood your reasoning.*Originally posted by sonhouse***Now the second statement doesn't make sense. It seems to be the**

same thing as your first, in both situations I think you are saying

win the first and lose the second in both. I read it several times and

the meaning still says both are the same.

Anyway, if if its not like flipping a coin, the third game based on

1 win and 1 loss should still be 50%, w ...[text shortened]... 1/3.

2 wins, 1 loss, fourth game, win, 1/3. So it looks like a series,

1/2, 1/3,1/4,1/5 etc.

EDIT To clarify, after n games, the probability that I've won k of them is 1/(n-1), independent of k (for k between 1 and n-1 inclusive). The probability of winning a game on a given trial, however, is not independent of past trials in the way that coin tosses are.

In the second part, we don't care in what order the first a wins and b losses are accumulated, we just care that in the next trial, the probability of winning is a/(a+b) and all subsequent games follow the same rule as in the first part

Can you (or anyone else) prove what you've just said? - 09 Mar '06 23:03

Can you tell me the differance between the first statement and the*Originally posted by royalchicken***You've got the correct answer, but for an essentially wrong reason which the second part of the problem should reveal, if I've understood your reasoning.**

EDIT To clarify, after n games, the probability that I've won k of them is 1/(n-1), independent of k (for k between 1 and n-1 inclusive). The probability of winning a game on a given trial, however ...[text shortened]... the same rule as in the first part

Can you (or anyone else) prove what you've just said?

second? You say in your answer post you don't care the order of

wins or losses but that is not apparent in your second statement.

Can you clarify my muddled find? Also I got that result by intuition,

not solid math, which is how I usually do these things! I smelled a

pattern and went with it. - 09 Mar '06 23:34

The second question, if you like, can be rephrased:*Originally posted by sonhouse***Can you tell me the differance between the first statement and the**

second? You say in your answer post you don't care the order of

wins or losses but that is not apparent in your second statement.

Can you clarify my muddled find? Also I got that result by intuition,

not solid math, which is how I usually do these things! I smelled a

pattern and went with it.

In my first a+b games, I win a of them, and I win my next game with probability a/(a+b). Thereafter, the probability that I win a given game is equal to the proportion of games I have previously won. What is the probability that, in my first n games, I win precisely k of them? - 10 Mar '06 01:21 / 1 edit

For all k except 0 and n (which are both zero) the probability is 1/(n-1).*Originally posted by royalchicken***Suppose I play chess. I win my first game and lose my second. Thereafter, the probability that I win is equal to the proportion of previous games I have won. What is the probability that I win k of my first n games, for each possible value of k?**

Now consider the same game, except that or my first a+b games, I win a and lose b. What is the probabi ...[text shortened]... PFCF rec, give a coherent account of any kind of how the title relates to RHP matters at large.

This can be shown easily by a induction proof. For n=3 the probability of both WLW and WLL are 1/2 (obviously).

First it must be known that after n rounds the permutations of order for a k wins is (n-2)! / ((k-1)!*(n-k-1)!), this can be easily derived from the standard combination formula n!/(w!*l!) and is needed as the first two games do not contribute extra permutations of order.

Now assume that my answer holds true for the nth case (making the probability of k wins 1/n for all k). Therefore the probability of winning k games is 1/(n-1) [a], the chance of winning the next game k/n [b ] and the ways those k wins could be arranged (and hence the order at which k wins could be arrived at) (n-2)! / ((k-1)!*(n-k-1)!) [c]. It is then a simple algebra matter to show that a*b*c = 1/n for all values of positive integer values of n and 0 < k < n where k is an integer.

Therefore as I have shown that the first case is true and that if the nth case is true so is the n+1th it is proven. - 10 Mar '06 01:44 / 1 edit

Nicely done. It's more slick and first-principles-y than my solution, which used the total probability theorem to make a recurrence relation, although it's basically the same proof in disguise.*Originally posted by XanthosNZ***For all k except 0 and n (which are both zero) the probability is 1/(n-1).**

This can be shown easily by a induction proof. For n=3 the probability of both WLW and WLL are 1/2 (obviously).

First it must be known that after n rounds the permutations of order for a k wins is (n-2)! / ((k-1)!*(n-k-1)!), this can be easily derived from the standard combin ...[text shortened]... shown that the first case is true and that if the nth case is true so is the n+1th it is proven.

Maybe sonhouse can answer the second bit and you can prove it. What a team!