# Math Question on Probability...Help!!!!!

Neal Pan
Posers and Puzzles 29 Jan '08 03:47
1. Neal Pan
cya 2008 in Beijing
29 Jan '08 03:471 edit
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?]
2. AThousandYoung
All My Soldiers...
29 Jan '08 07:58
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?]
Is "the game last week" one of the 100?
3. Neal Pan
cya 2008 in Beijing
29 Jan '08 13:22
Originally posted by AThousandYoung
Is "the game last week" one of the 100?
nop
100 games from now on. doesn't include last week's game.
4. 29 Jan '08 14:48
OK. 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)
5. Neal Pan
cya 2008 in Beijing
01 Feb '08 05:10
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?
6. TheMaster37
Kupikupopo!
01 Feb '08 07:17
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?
Yes.

Sadly I have only 5 minutes, wich is not enough to type out that answer correctly.
7. 01 Feb '08 11:38
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?
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.
8. TheMaster37
Kupikupopo!
01 Feb '08 12:487 edits
Sorry 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.
9. uzless
The So Fist
01 Feb '08 18:01
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?]
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.
10. 01 Feb '08 18:38
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.
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...

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 ðŸ™‚)
11. 02 Feb '08 22:44
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.
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.
12. 02 Feb '08 22:48
OK, 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)
13. uzless
The So Fist
03 Feb '08 10:271 edit
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.
It only works when both probabilities are >50%...obviously.

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.
14. 03 Feb '08 17:48
Originally posted by uzless
try other probs....

70-60 = 45 to 55 (rounding)
80-65 = 42 to 58 (rounding
etc
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.
15. 03 Feb '08 19:45
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.
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 ðŸ™‚

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.