h0llow said:
nvm.. i got it...can u tell me if i am right though?
TT THHT HHHH
HTT HTHT HTHHH
HHTT THHHT THHHH
HHHTT THTH HHTHH
THT HHTHT HHHTH
= 15 possibilities
chance player A (T) wins = 10/15 = 66.67%
chance player B (H) wins = 5/15 = 33.33%
am i right?
There are lots of other possibilities---infinitely many, in fact. So, we need a systematic way to approach the problem. Let us define two functions:
A(a,b) = probability that player A wins the game, given that A goes first and A must win a rounds before B wins b rounds,
and
B(a,b) = probability that A wins the game, given that B goes first and A must win a rounds before B wins b rounds.
Note that both refer to Mr. A winning; the different letters correspond to who starts.
If A starts and wins the round, the coin passes to B; at that point, B will be starting and A needs to win a-1 rounds before B wins b rounds. If A starts and loses the round, A passes the coin to B, at which point B starts and A must win a rounds before B wins b rounds. Thus
A(a,b) = (1/2)*B(a-1,b) + (1/2)*B(a,b).
Next, suppose B starts and A must win a rounds before B wins b rounds. If B wins the starting round, A starts next and must win a rounds before B wins b-1 rounds. If B loses, then A must win a before B wins b. Thus
B(a,b) = (1/2)*A(a,b-1) + (1/2)*A(a,b).
We can start with low values of a and b, and use these to get higher values (illustrated below). You want to compute A(6,4) and B(6,4), which are the probabilities that A will win, depending on who starts
What is A(1,1)? If A wins the first round he wins the game and B does not even get to play a round. This has prob = 1 /2. If A loses the first round, B starts next and must lose his first round, thus passing the coin back to A, and we are back to A(1,1) again. The probability of this is (1 /2)(1 /2) = 1/ 4. Therefore, A(1,1) = 1 /2 + (1/ 4)A(1,1), so A(1,1) = 2/3.
Similarly, B(1,1) = 1/3.
Where does the value A(1,1) = 2/3 come from? Well, when A starts and wins one round before B does, the possible outcomes are W (stop), LLW (stop) LLLLW (stop), LLLLLLW (stop) ,..., which has total probability
A(1,1) = \frac{1}{2} + \frac{1}{2}\left(\frac{1}{4}\right) + \frac{1}{2} \left(\frac{1}{4}\right)^2 + \cdots\\<br />
= \frac{1}{2} \left[ 1 + \left(\frac{1}{4}\right) +\left(\frac{1}{4}\right)^2 + \cdots \right] <br />
= \frac{1}{2} \frac{1}{1 – (1/4)} = \frac{2}{3}.
This accounts for infinitely many possible outcomes, but is still pretty simple.
Note that the above calculation is the same as what we would get if we used A(0,b) = B(0,b) = 1 for b > 0, and A(a,0) = B(a,0) = 0 for a > 0.
Anyway, you can now get A(2,1), B(2,1), A(1,2), B(1,2) and then A(2,2) and B(2,2). Then you can get even more values in a similar way. Just keep going until you arrive at A(6,4) and B(6,4). This will involve a lot of simple calculations; you could, for example, do it in a spreadsheet.
RGV
