Dice probabilities with re-rolls

I've looked at the Wizard of Odds dice page ( wizardofodds.com/ask-the-wizard/probability/dice/ ), and found the answers to a number of questions there, but not these particular questions.

The probability of rolling at least one 6 with four dice is 1-(5/6)^4.

1. What is the probability of rolling at least two 6s with four dice -- and is it the same as the probability of rolling at least one 5 AND at least one 6? And how do you build that equation? How would it change if you had more or fewer dice total; or were trying to get at least three 6s; or at least two 5s OR 6s?

2. What is the probability of rolling at least one 6 (or separately, at least two 6s) with four dice if you can sometimes re-roll? For example, what if you must keep one die each round, but can re-roll all of the rest (so the first round, you roll 4 dice; second round, 3 dice; third round, 2 dice; fourth round, 1 die)? How would that be different if you could keep re-rolling whether or not you achieved a 6 in some of the early rolls? Or what if you can re-roll all dice, up to 3 times? Again, how do you build that equation, and how would it change with more dice, or different re-roll options?

I'm asking in such detail because I would like to learn for myself how to create these equations and solve them, but I can't figure out the steps I need to take to get me there. Any good tutorials would be greatly appreciated as well.

Thank you in advance for your help!

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silly

Sally

Quote: kinead

1. What is the probability of rolling at least two 6s with four dice -- and is it the same as the probability of rolling at least one 5 AND at least one 6? And how do you build that equation? How would it change if you had more or fewer dice total; or were trying to get at least three 6s; or at least two 5s OR 6s?

There are two ways of calculating the probability of rolling at least two 6s with four dice:
(a) Calculate the probabilities of rolling exactly 2, exactly 3, and exactly 4 sixes, and add them up;
(b) Calculate the probabilities of rolling exactly 0 and exactly 1 six, add them up, and subtract the sum from 1.

One of the basic things you need to remember is:
The probability of something happening = 1 minus the probability of it not happening.

If you are trying to calculate getting at least 3 6s, it is easiest to calculate the probabilities of exactly 3 6s and exactly 4 6s, and add them.

Calculating the probability of getting at least one 5 and at least one 6 is a little harder. You could divide the "good" rolls into groups (one of each; one 5, two 6s; one 5, three 6s; two 5s, one 6; two of each; three 5s, one 6), count them, add them up, and divide by the total number of possible rolls (1296):
One of each is 4 (possible dice for the 6) x 3 (possible remaining dice for the 5) x 16 (number of ways the other two dice can be rolled so neither is a 5 or 6) = 192
One 5 and two 6s is 6 (possible pairs of dice for the 6s) x 2 (possible remaining dice for the 5) x 4 (number of ways the remaining die can be rolled so it is not 5 or 6) = 48
One 5 and three 6s = 4 (there are 4 possible dice for the 5, and for each one, only one group of three remaining for the 6s)
Two 5s and one 6 = 48
Two of each = 6 (there are 6 possible pairs of dice for the 6s, and for each one, only one pair remaining for the 5s)
Three 5s and one 6 = 4
The total = 192 + 48 + 4 + 48 + 6 + 4 = 302
The probability = 302 / 1296 = 151 / 648.