“tell me the chances of 6 on three dice (d6), 4 times in about 10 rolls”
EDIT: A non-ambiguous version of this question: what are the chances of getting a sum of exactly 6 on 3 six-sided dice exactly 4 times in 10 rolls?
And the answers fly! And they are all wrong.
So let’s look at an easy example. Given that you are rolling two d6, what is the probability that you will roll exactly one six?
- It’s the probability that you’ll roll a six plus the probability of not rolling a six? ( P(6) + P(~6) => 1/6 + 5/6 = 1 100% huh?
- It’s the probability that you’ll roll a six times the probability of not rolling a six? ( P(6) * P(~6) => 1/6 * 5/6 = 5/36 Nope, too low
- It’s one minus the probability of not rolling a six in the roll? ( 1 – P(~6) * P(~6) => 1 – 5/6 * 5/6 = 11/36 Nope, too high
To find this solution you have to remember the golden rule when dealing with dice probabilities, the probability of an event A happening is:
- P(A) = 1 – P(~A)
So lets apply that to the above problem. In what ways can we not roll exactly one six on 2d6? Well we could roll zero sixes or we could roll 2 sixes!
Thus our calculation becomes:
- 1 – P(~one six) = 1 – ( P(zero sixes) + P(two sixes) ) = 1 – P(zero sixes) – P(two sixes) = 1 – 5/6*5/6 – 1/6*1/6 = 1 – 25/36 – 1/36 = 10/36
Which is the correct answer!
Now obviously this is going to prove a little tedious when there are 10 or so dice/rolls since then you have to do 1 – P(ten sixes) – … – P(zero sixes) (excluding the one you are looking for)
So we have to come up with another way to do it.
Going back to our six example lets try another approach. We know that the probability is the # of successes out of the # of possible outcomes. So then let’s count up the # of successes and # of possible outcomes!
For successes we have (1,6); (2,6); … (5,6) and (6,1); … (6,5) which is 10 total successes.
How many possible outcomes are there? 6*6=36
Which gets us to our 10/36 answer.
There is one more way to do it which is what is the probability or rolling 6 , not six ( 1/6*5/36 = 5/36) and what is the probability of not six, six ( 5/6*1/6 = 5/36 ) now since those are the only two possibilities, add them together and get 10/36!
Let’s apply this to our big problem. How many ways are there on 3 dice to roll exactly a six? (There are 10) So the chances of rolling a six 10/216 and the probability of not rolling a six is 206/216. In ten rolls we’ll have 4 of the sixes and 6 of the not-sixes giving us P=( 10^4 * 206*6 ) / 216^10. Now we have to multiple that by the number of different ways we can arrange those 4 dice in the 10 trials (combination here 10 choose 4) = 210.
Finally 210* P = .07%
However I think the real number we’re looking for is what are the chances of rolling 6 or less, 4 or more times out of 10 times with three dice? The math on that is a bit more complicated, but boils down approximately 1%.
And that, dear readers, is how the dice rolls.
More on dice, statistics, and their place in WM/H later.