1
$\begingroup$

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.

$\endgroup$
2
  • 1
    $\begingroup$ In R: 1 - pbinom(2, 5, 1/6) equals 0.03549383. A quick simulation is in excellent agreement. $\endgroup$ Commented Feb 13 at 15:21
  • $\begingroup$ @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? $\endgroup$ Commented Feb 13 at 16:21

1 Answer 1

2
$\begingroup$

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.

$\endgroup$
7
  • $\begingroup$ 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? $\endgroup$ Commented Feb 13 at 16:19
  • 2
    $\begingroup$ @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. $\endgroup$
    – Stef
    Commented Feb 13 at 16:42
  • $\begingroup$ @Stef I didn't understand $\endgroup$ Commented Feb 13 at 17:17
  • 1
    $\begingroup$ ah sorry. Now I understand. So the outcomes with more than 3 ones are counted multiple times. $\endgroup$ Commented Feb 13 at 17:55
  • 1
    $\begingroup$ @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 $\binom 5 3 (1/6)^3$ $\endgroup$
    – Stef
    Commented Feb 13 at 18:02

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.