Craps Formula

What is the formula for calculating the probility that the next toss will be a specific box number? How about the formula for numbers repeating?
Thanks for your help
K8

Real basic first question, you sure need to grasp this quickly.

2 dice have 36 possible outcomes. There is only one way to roll a 2, so the chances are 1 in 36. There are 6 ways to roll a 7, chances are 6 in 36 [or 1/6]. Etc.

https://wizardofodds.com/gambling/dice/

As for a number to repeat, you need to rephrase your question more specifically.

Quote: K8

What is the formula for calculating the probability that the next toss will be a specific box number?

Say the #6
From the Wizards page lined to above, we see there are 5 ways a #6 can roll from the 36 total

Probability that the very next roll will be a #6 = 5/36 or 13.89% if you need to see it that way
Same for the #8 = 5/36
The other numbers you can do and check if you are right
The formula being n/36
n= the total ways a specific number can roll

Quote: K8

How about the formula for numbers repeating?

Scenario #1) A #6 just rolled. You want to know the chance of a #6 rolling AGAIN, on the very next roll. (repeating 6s)
That should be 5/36
(n/36)

Scenario #2) You hold 2 dice in your hand.
What is the chance that on the very next 2 rolls, you will roll 2 6s?
5/36 * 5/36 = 1.93%
(roll#1 times roll #2)
n/36 * n/36
or (n/36)^2

Scenario #3) What is the chance that ANY number will roll twice in the next 2 rolls? 11.27%
= (1/36)^2 + (2/36)^2 + (3/36)^2 + (4/36)^2 + (5/36)^2 + (6/36)^2 + (5/36)^2 + (4/36)^2 + (3/36)^2 + (2/36)^2 + (1/36)^2
https://wizardofodds.com/ask-the-wizard/probability/dice/
^2 means to the power of 2
or multiply the number 2 times (1/36)*(1/36)

Hope it all helps

thank you, Very Much!

Well, almost OK. Next question. To get the probability of a a multiple bet such as inside place bets, do you add the number of combinations then divide by 36?

Quote: K8

Well, almost OK. Next question. To get the probability of a a multiple bet such as inside place bets, do you add the number of combinations then divide by 36?

This is often handled purely as an odds question:

Ignore outcomes that don't win or lose the bet. On the 6 or 8, there are 5 ways to win [the very number] and 6 ways to lose [a 7]. Thus the fair odds are 6 to 5. This method is very familiar to old timers going 'way back'. Note that it is odds, not probability.

Quote: K8

To get the probability of a multiple bet such as inside place bets, do you add the number of combinations then divide by 36?

Another nice question.
exactly.
You got it.

as long as they are mutually exclusive or disjoint you add the probabilities for the different bets.
(meaning they can not happen at the same time. A Place5 and a Place8 can not show on the same roll)

Place 5 and P9 = 4/36 each
P6 & P8 = 5/36 each
4+4+5+5 = 18/36 = 50% (surprised??)

Advanced Question
The probability that the next roll is either a #6 or any of the Hardways {22, 33, 44, 55}?
We know 5/36 is for the #6 and
1/36 is for each hardway.
So 9/36 would be the wrong answer because the 5/36 for the #6 already includes the Hardway 6{33}
(they are not mutually exclusive, meaning they could happen at the same time)

So we must subtract the probability that is common to both sets that being 1/36 for the 33
This was a simple example.
Some examples can get very confusing, but just keep the basic rule in mind.

No, not really suprised. Just confirmed what i firgured manually.
Thank You very Much
k8