Originally Posted by
Warwick44
John Galt: (<-- Who ARE you?) *hehe*
As to your question...you need to offer a little more information, unless I've misunderstood. Firstly, what is won with each roll? Is it some multiple of the number on the die? Or is the payoff the same so long as the number is 55 or less?
Clearly if it's the latter, then any roll you take is going to have 55% odds of landing on any of those numbers. And given the expected value of any roll of the dice is 50.5 (<-- remember there is no zero...so the expected value is NOT 50), you have a built in 4.5% advantage, and should simply bet the entire 10k on the first roll if losing it all is of no consideration.
HOWEVER, if the goal here is to calculate what value you would have to keep betting to give yourself even money odds at staying in the game and building a bankroll, then you would bet $4,784.70, which @ 1.045% payoff would be $5,000, which would keep you at a theoretical 50/50 odds of winning on each successive roll. (detailed as --> ((10,000 /2) / 1.045) = 4,784.70, which @ 4.5% advantage pays 5k)
So you just keep betting 4.5% less than 50/50 odds to keep yourself in the game. That of course, is subject to "bad runs", and will not necessarily keep you from losing it all after multiple rolls. I dont have Excel in front of me, but if I did I could work out a further optimization toggled out to whatever standard deviation one were comfortable with. Of course the more conservative you get in terms of sigma risk (say, beyond 1 sigma), the more you will have to roll many times.
I'm guessing your question is more advanced than this though, so please clarify. The answer gets more complex, and includes a game theory overlay to understand the optimal number of rolls, which would be partially determined by what each roll yielded -- IF the payoff is non-linear.
I can answer that with more info.