00:01
So for this problem, let's say that event a1 is the event that he is ahead on the first play.
00:12
So, and let's say that w is the event that he is ahead by $2, or in other words, he wins.
00:21
So we'd have that the probability of winning, by the law of total probability, the probability of winning would be equal to the probability of winning, given that he's ahead on play one, times the probability of being ahead on play one, plus the probability of winning, given that he loses play one, so a one complement, times the probability of a1 complement.
00:48
So, we know that then the probability of winning would be equal to one -third, the probability of winning, given that he's ahead on the first play, plus two -thirds, the probability of winning, given that he is not ahead on the first play.
01:13
Now, the way that we can think of this is say that a -i, a -sub -i, is the probability of winning with initial value i, or initial fortune i.
01:38
So, p of w, probability of winning overall, would be a -i, probability of winning, given that he's ahead, on the first play would be a .i.
01:53
Plus 1 over 3, then we'd have plus 2 over 3 times the probability of winning given that he is not ahead on the first play.
02:02
So that would be probability of winning or probability of a.
02:07
I minus 1.
02:07
The probability that he wins given that he starts at one less than he, or that he, one less than our i value is.
02:15
That is.
02:16
We know that a 0, the probability of winning, if he has zero dollars, is zero, and the probability of winning at i plus two, the probability of winning, given that we're ahead by two, well, we know that we've defined winning as being ahead by two.
02:33
So a .i.
02:33
Plus two is equal to one.
02:37
We now have a differential equation, or pardon me, a difference equation, though actually i'll do a little bit of rearranging here to put this into a more familiar form for difference equations.
02:49
If we multiply, through everything by three, then subtract a minus 1 from both sides.
02:55
We can write that a .i.
02:56
Plus 1 would be equal to 3ai minus 2, a .i minus 1.
03:04
So this is in the form.
03:07
We can try solving for this by making the onslaught solution that ai, or actually i'll put it into this form, let's say ai is equal to a to the power of i plus 1, which may seem a little bit confusing, but the reason why i write it that way is because now we can write this as a to the power of i plus 2 is equal to 3a to the power of i plus 1 minus 2a to the power of i to the power of i.
03:43
Or we can divide everything by a to the power of i which gives us a squared is equal to 3a minus 2 which now we can put into the typical form of a quadratic equation a squared minus 3a plus 2 is equal to 0 and we can solve this by factoring specifically we can factor this as a minus 1, a minus 2 equals 0.
04:15
And so we'd be able to find then that a must be equal to either 1 or 2...