Probability of repeat numbers in craps

If I make 6 consecutive dont' pass/don't come bets, what is the probability that there will be a repeat number (4,5,6,8,9,10) without a seven-out? Two repeats? Show your work! And if you can calculate this, can you recommend a basic book on this type of probability problem? I will study your answer until I sort-of understand it. Thank you.

Quote: riverbed

If I make 6 consecutive dont' pass/don't come bets, what is the probability that there will be a repeat number (4,5,6,8,9,10) without a seven-out? Two repeats? Show your work! And if you can calculate this, can you recommend a basic book on this type of probability problem? I will study your answer until I sort-of understand it. Thank you.

Wow!
You are talking about a lot of states in a Markov Chain.
I am sure someone could do the math, but why?

I would sim this. Would be very easy to do.
Plus, we could get lots of info from it.
Maybe tomorrow if I'm slow at work.

Lots of Grad parties this week!

Thank you! I would sure appreciate it.

This sounds like something that needs to be solved by brute force.

For the "one-repeat" problem, determine each possible set of six point numbers with one repeat (e.g. {4,8,10,5,8,6}), then calculate each number's probability of both being a point number and not sevening out (for 4, it would be 1/12 (chance of rolling 4 as the point) x 1/3 (chance of rolling another 4 before a 7) = 1/36).
There are 6 possible numbers to be the duplicate; for each one, there are 5 numbers that would now not appear at all; for each of these 30 groups of numbers, there are 360 permutations, although each permutation within a group has the same probability. Figure them out, and then add them up.

This assumes that, by "a repeat number," no point appears more than twice. If a point can appear more than twice, you need to check the groups of 3-1-1-1, 4-1-1, 5-1, and 6 as well; it sounds like it's easier to check all 46,656 permutations of six point numbers (which would solve the "two repeat numbers" problem - and, for that matter, the "three repeat numbers" problem - at the same time.)

To help you out:
The probability of rolling, and then making, a point of 4 is 1/36, as shown earlier.
For a 10, it is also 1/36.
For a 5 (or a 9), it is 1/9 x 2/5 = 2/45.
For a 6 (or an 8), it is 5/36 x 5/11 = 25/396.

If, in your six rolls, you include the possibility of rolling a 2, 3, 7, 11, 12 as your first number, I get this:
1 duplicate point happens 1 time in 16.3
2 duplicate points happens 1 time in 27.55
3 duplicate points happens 1 time in 583.2

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silly
Sally

I think some terms need to be defined.

Do "repeat numbers" need to be consecutive? That is, if, for example, the six come-out rolls are 6, 4, 7, 6, 10, 11, are the two 6s considered "one repeat"?

What if the same number is rolled three times - is that considered one repeat or two?

My calculations were based on the answers being (a) no and (b) one repeat. For example, 6, 4, 7, 6, 10, 6 would be "one repeated number," as would 8, 8, 8, 8, 10, 4.

The OP is making, say $5, bets on the dont's for 6 rolls, so a repeat is a loss and per shooter before the 7out.
This can be solved exactly by a very large Markov Chain. Too much work for me.

By a simulation is faster and easier IMO.

For example, 6, 4, 7, 6, 10, 6.
$5 don't pass bet: roll 6 = point established
$5 don't come bet: roll 4 = bet moves to don't come 4
$5 don't come bet: roll 7 = 7out.
don't come bet loses
wins both don't pass and don't come 4.

The 7 is a 7out. So, no more rolls after the 7 for that shooter.
No repeating numbers to cause a dont pass or a dont come point number to lose
That is how I see the OP question

Sally

Yes, this is very helpful. I am a Mac user now, but I used to have WinCraps and I just loved it. I'm thinking about getting a used PC just for the purpose of using WinCraps. If I understand this, there is a 6l% chance of getting no repeat in six rolls, 28.6% of one, 9% of 2, and 1% of three. I once saw a guy come-out with 5 11s, so I guess he was more than 'one in a million'. Many thanks.

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silly
Sally

Mustang Sally,

I can't figure how to use the quote function on this forum, and it sure looks like a nifty function. So, your first table is for all shooters with hands of 6 or less, and 61% of them had no repeats of the 4,5,6,8,9, or 10. Irrespective of natural losers for the don'ts. That's just what I wanted to know. Thanks.