this is a random walk question( advanced math) can any genius help me?

Question

two teams are playing in a championship. they each have equal probability of winning the first game. for the remaining games, whoever won the last game has a psychological high, so they now have a 2/3 chance of winning the next game. the championship ends when one team has won W more games than the other. what's the expected number of games that will be played?

hint:Focus on the difference between the number of games won by team A and the number of games won by team B. Initially, this difference is zero, and with each game, it increases or decreases by one. The tournament ends when this difference reaches W or W

Answer 1

see my answer and explanation from the same question you posted (you double posted).

Number of games to play = 1 + 3(W-1)

Answer 2

A good problem and a good hint. I'll need to work on it a bit.

If the probability didn't change it would be fairly easy. That change in p(win) really makes a difference. I'll be back to see what people have done.

+ added
I don't think that intel-kni is right because the difference will increase by one with each game but the probability depends upon who won the last game:


Let l, w represent loss or win for one of the teams:
Example subsequences for one team:

ww , that second w has p=2/3 because it comes after a win

lw, that w occurs with p=1/2 because it comes after a loss

In both cases the number of wins increases by 1 but with different probability

++ added
a math guru: I took a look at that Wolfram stuff. Note that the probabilites didn't change

Maybe defining some itermediate RVs would help:
Let
A be number of wins of team A
B be number of wins of team B
A+B = N, number of games played (neglect ties)

Or, maybe these should be Bernoulli trials. Still the issue for me is how to handle the 'memory effect', that the probability of winning depends only on the result of the last game. In a way this might be able to be handled by 4 possibilities at each trial( after the first)

WW
WL
LW
LL

These patterns would be repeated at each game after the first.

Answer 3

Like modulo I need to think about it a bit but my first guess (really stretching that expected word to the limits of its definition) is 3W. The expected change in the current "balance" is a +1 with probability 2/3 and a -1 with probability 1/3. That is sounding like an expected step size of 1/3 which would mean 3W steps to expect to be at W.

<

Answer 4

The difference will increase by 1 with 2/3 chance, and decrease with 1/3. And you know it's 1 after 1st game.

the answer is a limit of the sum of an infinite series:

Sum for w=W,W+2, W+4, ...
of w * (2/3)^(W-1) * combinations(W-1 from w-1)

combinations(K from N) = K!(N-K!)/N!
N!=1*2*...*N

Answer 5

there's a typo. the last sentence should read "The tournament ends when this differences reaches W or -W". Good luck.

intel_knight, what is little w in your answer? thanks.