In a previous thread, one poster suggested that after 10,000 or so rolls, your results would be "durn close" whether you take the odds bet or not. I disagree, and here is my attempt to explain why.
I'd like to modify the premise slightly from 10,000 rolls to 3,000 Pass Line bets resolved. This is a bit more than the number of expected Pass Line bets resolved over 10,000 rolls, and it is easier to solve.
I found in a Probability text book that for n independent trials with probability of success, p, and probability of failure, q, that the variance is n*p*q. Furthermore, I found that you can sum indpendent variances. The formula assumes a payout of 1 for success and 0 for failure. However, that is not the case in craps, so we need to define another variable "x" as the difference between the payout for winning and losing (e.g., if point is 8, a win is worth +7 and a loss -6, so x=13). When computing the variance we have to square x and multiply it to n*p*q.
For Pass Line bets only (no odds), n=3000, p~0.507071, q~0.492929, x=2 => Var=2,999.4 and standard deviation~54.8.
For the case of making Pass Line bets and taking 3x4x5x odds, the true calculation should be the sum for every possible outcome of the product of the square of the difference between that outcome from the mean and the probability of that outcome. It's hard to write in words and even harder to calculate. So I will instead approximate the variance by summing the variances of the possible outcomes of one trial and using their expected number of occurences in 3000 trials as the value for n.
For come out winners and losers, n=1000, p=8/12, q=4/12, x=2 => Var~888.89
For points 6 and 8, n~833.33, p=5/11, q=6/11, x=13 => Var~34,917.36
For points 5 and 9, n~666.67, p=4/10, q=6/10, x=12 => Var~23,040.00
For points 4 and 10, n=500, p=3/9, q=6/9, x=11 => Var~13,444.44
Summing the variances, we get Var~72,290.69, so the standard deviation for 3000 trials, taking odds ~268.9.
I would suggest that even after 3,000 trials 268.9 is significantly higher than 54.8, but I suppose that is a matter of interpretation.
If anyone knows a shortcut for calculating the true variance for the case of taking odds, or has any corrections to my math, please share.
The ratio of people to cake is too big.