Chance of rolling at most one six with two dice in 20 rolls.

lo2

1. The problem statement, all variables and given/known data

Well as I said what is the chance of getting at most one six when rolling two dice twenty times?

2. Relevant equations

I know the probability of getting one six in one roll with two dice is:

11/36

And not getting one is:

25/36

3. The attempt at a solution

Then I figured out that the probability of not getting a 6 in [itex]n[/itex] rolls is equal to:

[itex]P = 1 - (\frac{25}{36})^n[/itex]

So is that so?

And how do you calculate the proability of getting at most one 6 in twenty rolls?

mfb

I can confirm your probability of getting no 6 in all rolls.

Your initial problem is unclear to me: What happens if you roll (6,6) one time and "no 6" the other 19 times? Your approach for a solution would include this in the probability, but the literal interpretation would not.

Anyway:
Can you calculate the probability to get exactly one 6 in 20 rolls?

If not, here is an easier subproblem: A 6 in the first roll, and no 6 in the other 19. How does that help to calculate the probability of exactly one 6 in 20 rolls?

Ray Vickson

Your probability P above makes no sense: it increases as n increases, with a limit of 1 as n → ∞, so you are claiming that for a very large number of rolls we are almost sure of not getting a six. That is, the more times we roll, the less chance we have of getting even one six. That is contrary to reality.

Anyway, what do you mean by 'getting a six'? With dice, the normal interpretation is that the total on the two dice is six.

You seem to be looking at the two dice separately and asking whether one of them shows a six. Since you are not adding the numbers, it would seem that 20 rolls of your two (separate) dies is the same as 40 rolls of a single die.