Chance of rolling a seven is 6 out of 36, or .16667.

Chance of rolling any other number 30 out of 36, or .83333.

Say you roll any number other than 7, the chance of a second roll being anything other than a 7 is again .83333.

So is the chance of two rolls in a row being any number other than 7 .83333 to the second power?

And the chance of three rolls in a row no seven .83333 to the third power?

If my calculations are correct, the chance of 10 rolls without a seven is .83333 to the 10th power, or .16151.

Am I right?

Quote:HeadlockI'm hoping to get some math help from the experts.

Chance of rolling a seven is 6 out of 36, or .16667.

Chance of rolling any other number 30 out of 36, or .83333.

Say you roll any number other than 7, the chance of a second roll being anything other than a 7 is again .83333.

So is the chance of two rolls in a row being any number other than 7 .83333 to the second power?

And the chance of three rolls in a row no seven .83333 to the third power?

If my calculations are correct, the chance of 10 rolls without a seven is .83333 to the 10th power, or .16151.

Am I right?

You bet! What is the next question!?

So the chance of 10 rolls without a 7 is just slightly less than rolling a 7: .16151 compared to .16667.

If I'm playing crapless craps and get 10 rolls without a 7, what is the expected payout of those 10 rolls?

2 and 12 pay 11 to 2, 3 and 11 pay 11 to 4.

(Edited to change 2 and 12 pay 11 to 2, not 11 to 4)

. Not sure what you are asking? What were you betting on?Quote:HeadlockThanks Soopoo.

So the chance of 10 rolls without a 7 is just slightly less than rolling a 7: .16151 compared to .16667.

If I'm playing crapless craps and get 10 rolls without a 7, what is the expected payout of those 10 rolls?

2 and 12 pay 11 to 2, 3 and 11 pay 11 to 4.

(Edited to change 2 and 12 pay 11 to 2, not 11 to 4)

Quote:HeadlockSo is the chance of two rolls in a row being any number other than 7 .83333 to the second power?

That is right (5/6)^2 = 69.44%.

Quote:And the chance of three rolls in a row no seven .83333 to the third power?

Yes.

Quote:If my calculations are correct, the chance of 10 rolls without a seven is .83333 to the 10th power, or .16151.

Am I right?

Yes.

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.

So your question is, given the shooter threw 10 non-seven numbers, what’s the expected winnings of this ten throws?

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.

Quote:ThatDonGuyYes, 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.

Haven’t gotten to a computer yet but this doesn’t make sense to me. The least you could possibly win is 11.6667 if you rolled all 6s and 8s.

EV of one roll that didn’t 7:

2*(11/2*1/30 + 11/4*2/30 + 9/5*3/30 + 7/5*4/30 + 7/6*5/30) = 1.85556.

So 10 rolls EV is 18.5556.

Would love to understand why this is wrong.

Quote:unJonHere’s what I did after swapping 4 and 10 to place bets.

EV of one roll that didn’t 7:

2*(11/2*1/30 + 11/4*2/30 + 9/5*3/30 + 7/5*4/30 + 7/6*5/30) = 1.85556.

So 10 rolls EV is 18.5556.

Would love to understand why this is wrong.

I am assuming that once a bet is won, it is not bet again. The EV of the second roll < 1.85556 in this case as if you roll the same number as in the first roll, you don't win anything.

Quote:ThatDonGuyI am assuming that once a bet is won, it is not bet again. The EV of the second roll < 1.85556 in this case as if you roll the same number as in the first roll, you don't win anything.

Ah. Got it. Guess the OP just needs to clarify what he was asking.

But all craps players know that 7's come much more often than 10 roll hands.

What do I not understand here?

Quote:HeadlockWe all agree that the chance of rolling a 7 at any time is .166667, and the chance of rolling ten numbers without a 7 is .16151.

But all craps players know that 7's come much more often than 10 roll hands.

What do I not understand here?

The chance of rolling 7 on a particular roll is 1/6, but the chance of rolling at least one 7 in 10 consecutive rolls is 1 - 0.16151 = 0.83849. You are comparing one roll against ten.

Quote:HeadlockPlease explain further.

It's a little hard to put into words, but I'll try...

Here's one way to look at it:

Take any particular set of ten rolls. For each of the ten rolls, considered separately, the probability of rolling a 7 on that roll is 1/6. However, the probability of none of the rolls being 7 depends on all ten of the rolls taken together. If you roll any 7s, then that set of 10 rolls cannot have "zero 7s," but you also have the possibility that you will roll more than one 7 in the set.

Here's another:

Suppose you start rolling, and you roll 4 times without a 7 before rolling a 7.

You then roll 8 more times without a 7 before rolling a 7.

You then immediately roll another 7.

You then roll 3 more times without a 7 before rolling a 7.

You then manage 10 rolls in a row without a 7.

The probability of rolling 10 times in a row without a 7 is about 0.1615056 - but you have to remember that "10 in a row" starts over every time you roll a 7. Whenever you roll a 7, the probability of the next 10 not having any 7s is 0.1615056.