# Probability to get a 1 in 5 rolls 3 or more times

If I roll a dice 5 times, what is the probability to get 3 or more ones?

That is, to roll a one three or more times?

Because I think it's near 0.05:

$$\binom{5}{3} \left(\frac{1}{6}\right)^3 = 0.046$$

But this guy sustains it is near 0.03.

• In R: 1 - pbinom(2, 5, 1/6) equals 0.03549383. A quick simulation is in excellent agreement. Commented Feb 13 at 15:21
• @COOLSerdash The correct formula is 1 - pbinom(2, 5, 1/6) - pbinom(1, 5, 1/6) - pbinom(0, 5, 1/6), right? We don't want 1 one and we don't want 0 ones either, right? Commented Feb 13 at 16:21

Your math is not right. You are missing part of the binomial probability formula. The correct calculation would be $$P(X = 3) = \binom{5}{3}\left(\frac{1}{6}\right)^3\left(\frac{5}{6}\right)^2 \approx 0.03215.$$

Additionally, this is the probability of exactly three ones. If you want three or more, you also need to add the probability of getting 4 ones and 5 ones:

$$P(X \geq 3) = \sum_{x = 3}^5 \left[\binom{5}{x}\left(\frac{1}{6}\right)^{x}\left(\frac{5}{6}\right)^{5-x} \right] \approx 0.03549.$$

You can use a calculator like this one to check those numbers.

• My reasoning was: I get a one 3 times in a row. The probability is (1/6)^3. Then I can get a one in whatever roll. There are $\binom{5}{3}$ to get 3 ones out of 5 rolls. Since the events are mutually exclusive, (if I get 1,1,1 I can't get 1,1,2,1) I add them. Then, since I want 3 or more ones, I don't care what I obtain in the other rolls. Why you add the probability of getting 4 or 5 ones, instead of ignoring what you obtain in the other rolls? Commented Feb 13 at 16:19
• @robertspierre To avoid double-counting. Consider the roll 1,1,1,1,5. With your method, you're counting that roll 4 times because there are 4 ways to form a triplet of ones using the 4 ones in this roll.
– Stef
Commented Feb 13 at 16:42
• @Stef I didn't understand Commented Feb 13 at 17:17
• ah sorry. Now I understand. So the outcomes with more than 3 ones are counted multiple times. Commented Feb 13 at 17:55
• @robertspierre Yes. The 25 combinations with exactly 4 ones will be counted 4 times each and the combination with 5 ones will be counted 10 times. You can check that in wzbillings's formula, if you add a factor 4 on the x=4 term, and a factor 10 on the x=5 term, you'll find the same wrong value 0.046 that you found with the formula $\binom 5 3 (1/6)^3$
– Stef
Commented Feb 13 at 18:02