Quote:HeadlockI realized last night when I was in bed reviewing the day that I did not give enough information. So lets say $260 across. $25 each except $30 on the six and eight.

Some questions:

1. Are they all place bets?

2. Are they turned off during subsequent comeout rolls?

3, Do you take all of them down after the tenth roll?

4. Are you sure you want to bet $25 rather than $24 on 2/3/11/12, since you're not going to get full odds on that extra dollar (in fact, you should get paid just as much on 3 or 11 for a $25 bet as a $24 if they don't have 25c chips)?

5. If a bet wins, do you repeat the bet, or just leave that number empty for the rest of the rolls?

I'm just wondering if I get ten rolls in crapless without a seven, what is the expected return if I have an equal amount on every number.

Quote:HeadlockI don't think your math is correct because the chance of hitting 2 or 12 is not the same as 6 or 8.

So you have $1 on all numbers with all place payouts but 4/10 paying 2/1 with a 5% upfront vig (so you actually have $1.05 on 4/10)?

With this those assumptions and assuming 10 rolls where the 7 doesn’t show you would win $18.76508 on average.

Quote:HeadlockI know. I was trying to keep it simple. Can we just assume $1 on each number, 2-12,? Can you calculate the estimated return for 10 rolls, no pressure, no take-downs?

I'm just wondering if I get ten rolls in crapless without a seven, what is the expected return if I have an equal amount on every number.

The exact number I get is 54,631,269,497,559,821 / 4,428,675,000,000,000, or about 12.3358.

Quote:ThatDonGuyThe exact number I get is 54,631,269,497,559,821 / 4,428,675,000,000,000, or about 12.3358.

That’s very different than my number. You sure you rolled with 30 permutations in denominator (given no 7s rolled)?

Quote:unJonThat’s very different than my number. You sure you rolled with 30 permutations in denominator (given no 7s rolled)?

Yes, and I also got this value in simulation.

Keep in mind that I placed 4 and 10 instead of buying them, although that shouldn't make the numbers that different.

I actually used a 10,241-step Markov chain (the first step is for the initial condition, then each combination of the number of rolls and which point numbers had already been rolled), although it turns out that I could have just done the following:

2 * ((1 - (29/30)^10) * 11/2 + (1 - (28/30)^10) * 11/4 + (1 - (27/30)^10) * 9/5 + (1 - (26/30)^10) * 7/5 + (1 - (25/30)^10) * 7/6)

For some reason, I didn't think that would work.