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.