```Date: Jul 31, 2006 5:54 AM
Author: cuthbert1
Subject: Re: (difficult)Theoretical gambling puzzle

I can't do the math yet, but I'd like to know if I'm on the right track conceptually.I need to know the probability of drawing any sequence (length 1 - 52), such that it contains a given excess of red cards over black. For example, if I draw say 10 cards and have a profit of \$2, what is the probability that there is some sequence between 11 and 52 cards such that the excess is at least \$3? That is, for any given profit-level at any given sequence-length, is it rational to continue or to quit?My first problem is I can't calculate the denominator, that is, the total of all possible sequences. The total number of 52-card sequences is 52!/(26! x (52-26)!), i.e. 26 combined from 52. The total number of 1-sequences is 2, i.e. either a red card or a black card. I need the sum of 1-sequences, 2-sequences .... up to 52.Then I need to calculate the probability for each sequence of some excess of red cards over black cards occuring. For example, in all the 52-card sequences, there is an equal number of red and black cards. In the four 2-card sequences (BB, BR, RB, RR), there's a 0.25 chance of a 2-card excess, an even chance of no gain or loss, and a 0.25 chance of a \$2 loss. Given a sequence S of n cards with an excess e of red cards over black, what's the probability that S is a string of a longer sequence S' such that the excess  > e?Does this make sense?
```