- 29 Jan '08 03:47 / 1 editBobby and Crystal meet every Thursday for a game of chess. They find that after winning a game, Bobby has a 65% probability of winning the next game. Similarly, Crystal has a 60% probability of winning after she has won a game. Crystal won the game last week.

Q: If Crystal and Bobby play 100 games, how many games is each player likely to win???

[answer is 53 and 47, but how?] - 29 Jan '08 07:58

Is "the game last week" one of the 100?*Originally posted by Neal Pan***Bobby and Crystal meet every Thursday for a game of chess. They find that after winning a game, Bobby has a 65% probability of winning the next game. Similarly, Crystal has a 60% probability of winning after she has won a game. Crystal won the game last week.**

Q: If Crystal and Bobby play 100 games, how many games is each player likely to win???

[answer is 53 and 47, but how?] - 29 Jan '08 14:48OK. For a start, it seems to be assuming there are no draws (not enough information given otherwise). That's unlikely, but let's ignore that for now. I've a suspicion there will be more than one approach that works - here's mine.

Let P(n) be the probability that Bobby wins match n.

Then P(n + 1) = 0.65P(b) + 0.4(1 - P(n))

=> P(n + 1) - 0.25P(n) = 0.4

This is a difference equation - it has a general solution P(n) = A + Bx^n

Plug it in to the equation, and we get:

x = 0.25

A = 8/15

And since P(0) = 0 (because we know Bobby lost last week), B = -8/15

So P(n) = 8/15[1 - 0.25^n]

Bobby's expected number of wins is SUM[1, 100] P(n)

You can get the SUM[1, 100] 0.25^n term from the formula for a geometric series, if you want it. But if we're only interested in the nearest integer this term is small enough to ignore (it's less than 1/3, which is what you get if you sum it to infinity).

So Bobby is expected to win 8/15 * 100 = 53 (to the nearest integer) - 01 Feb '08 11:38

There probably is. It's a long time since I studied Markov chains (which I think this is an example of) though, and I can't remember how to do it that way.*Originally posted by Neal Pan***I think that's the best answer...so far i have ever seen**

so, you used the geometric series method, right?

what about Matrix then? is there a way to solve in matrix? - 01 Feb '08 12:48 / 7 editsSorry for the dots, but it's the only way i could align my matrices nicely :p

Put the probabilities in a matrix (B for Bobby, C for Crystal):

......................will win

....................B.......C

.............B / 0,65 0,35 \

last win.....|.................|

.............C \ 0,40 0,60 /

100 is large enough to find the answer in the equilibrium state;

( b c )/ 0,65 0,35 \ = ( b c )

........|.................|

.........\ 0,40 0,60 /

This gives a set of equations;

b = 0,65 * b + 0,4 * c

c = 0,35 * b + 0,6 * c

Solving gives one condition: 8c = 7b

Thus B wins 8/15 of the games and C wins 7/15 of the games;

B wins 53,33... games and C wins 46,66... games.

Rounding gives that B wins 53 games, and C wins 47 games. - 01 Feb '08 18:01

Simpler way:*Originally posted by Neal Pan***Bobby and Crystal meet every Thursday for a game of chess. They find that after winning a game, Bobby has a 65% probability of winning the next game. Similarly, Crystal has a 60% probability of winning after she has won a game. Crystal won the game last week.**

Q: If Crystal and Bobby play 100 games, how many games is each player likely to win???

[answer is 53 and 47, but how?]

The difference between the chance of winning is 5%. That means bobby will win 5 more games if they play 100 games. Odds are he'll win the first game so give him an extra win.

That means the total difference after 100 games will be 6 wins.

For a 6 game difference out of 100 games it must therefore be 47-53. - 01 Feb '08 18:38

the statement "odds are he'll win the first game" is false. the problem states that crystal has a 60% chance of winning if she won the prior week, AND crystal won the prior week. So odds are only 40% that he'll win the "first" game in the problem...*Originally posted by uzless***Simpler way:**

The difference between the chance of winning is 5%. That means bobby will win 5 more games if they play 100 games. Odds are he'll win the first game so give him an extra win.

That means the total difference after 100 games will be 6 wins.

For a 6 game difference out of 100 games it must therefore be 47-53.

i haven't read the answers closely but do they apply Bayes' theorem for conditional probability? or would there be an elegant method using Bayes' theorem? I imagine it would turn into something like the geometric series answer (or maybe something EXACTLY like that - like i said i didn't read very closely ) - 02 Feb '08 22:44

That's just coindidental that it works (even apart from the problem that Aetherael pointed out).*Originally posted by uzless***Simpler way:**

The difference between the chance of winning is 5%. That means bobby will win 5 more games if they play 100 games. Odds are he'll win the first game so give him an extra win.

That means the total difference after 100 games will be 6 wins.

For a 6 game difference out of 100 games it must therefore be 47-53.

By that argument, if the probabilities were equal, then they'd be expected to win 50 games each.

Now consider an extreme case where both probabilities equal 1. In that case, Crystal will win all 100 games. - 02 Feb '08 22:48OK, I
*think*this is the general solution. If the probabilities are p and q respectively (for the 0.65 and 0.6 in the original problem), then the expected number of wins for Bobby is:

[(1 - q) / (2 - p - q)]{100 - (p + q -1)[(1 - (p + q - 1)^100)/(2 - p - q)]}

(as long as p and q aren't both 1) - 03 Feb '08 10:27 / 1 edit

It only works when both probabilities are >50%...obviously.*Originally posted by mtthw***That's just coindidental that it works (even apart from the problem that Aetherael pointed out).**

By that argument, if the probabilities were equal, then they'd be expected to win 50 games each.

Now consider an extreme case where both probabilities equal 1. In that case, Crystal will win all 100 games.

As for the first game, it's statistically irrelevant when the total number of games is greater than say 50 since any individual game becomes a rounding error for when you state the answer in whole numbers.

try other probs....

70-60 = 45 to 55 (rounding)

80-65 = 42 to 58 (rounding

etc

By giving you the percentages of the next win, the "work" has already been done for you. The trick is to recognize that the percentage given is just another way of stating total wins since they play 100 games (since percentages are based in 100 units)

It's just a ratio expressed differently. - 03 Feb '08 19:45

thank you for doing the work - i was too lazy to work it out, but wanted to say something... i didn't think a "you're wrong" would be productive without some sort of actual information to back me up*Originally posted by mtthw***Those examples prove my point. The answer to the second one***isn't*58 to 42. It's 64 to 36. Work it out the long way.

and as earlier stated, if you look at the extremal case where crystal's probability to win if she won last week is 100%, then you could make bobby's % to win if he won last week any number you want. he's still going to lose every game if she won last week.