I have two quick questions about probabilities and dice.

  • What is the probability that that you can roll AT LEAST two fives in four rolls.
  • What is the probability that you roll no more than one five in four rolls.

I'm pretty sure I have been over thinking this, but I just want to check to make sure.

For the first problem the sample set should be ${6^4}$. Then the probability of rolling one 5 should be $\frac{1}{6}$. There are 6 ways to roll two fives. Should the answer to this problem then be $\frac{6}{6^4}$?

The second problem... so the probability of rolling a five is $\frac{1}{6}$ then the probability of not rolling a five would be $\frac{5}{6}$. Therefore, the answer should be $\frac{5^4+4}{6^4}$?

Any help would be much appreciated. Thanks!

  • $\begingroup$ For problem 1 you have two consider two things: a) they ask for at least 2 5s; b) you have also to include the fact that if you have two 5s out of 4 rolls, you also have 2 non-5s, each of them having a $\frac{5}{6}$ probability. $\endgroup$ – nicola Aug 31 '18 at 7:47
  • $\begingroup$ For your second question, it might be interesting how this complements the probability of your first question. Can you see the relation? $\endgroup$ – Jan Aug 31 '18 at 9:13

Let's model the first problem in the following way; let's check if at each roll 5 has come out or not. The probability of 5 is $1/6$.

Now the problem is asking you at least 2 times in four launches. Because the probability sums to $1$, we can say the solution would be 1 - "the probability of rolling the dice four times and not having any 5" (that would be $(5/6)^4)$ - "the probability of rolling the dice four times and just having one five" $((5/6)^3*(1/6)*4)$

The result is $19/144$ = $13.19$%.

For the second problem you have already the answer, because again we know that probability have to sum to 1. Specifically we have already computed the probability of at least two 5. So the probability of one 5 or less is $1-19/144$ = $86.8$%


In the first problem consider that you could roll 2, 3, or 4 fives. How many combinations are there for 3 or 4?

For the second problem you're right in considering the probabilities of 5s and not 5s. But I'm not sure why you have $5^4 + 4$ as your numerator.

The probability that the first dice is a 5 and the others aren't is $\frac{1}{6} \frac{5}{6}^3 = \frac{125}{1296} $

This is the same as the probability that the second only is 5, and third, and fourth. But also remember that "No more than 1 five" also includes no fives at all, so this will need to be considered.


The probability is the number of correct events, divided by the number of possible events. The number of possible events is $6^4 = 1296$. The number of correct events can be assessed in the following way:

There is exactly $1$ possibility to have four $5$'s.

For three $5$'s, there are 4 ways to distribute them among the four rolls. The remaining dice has $5$ different options, so there are $4\cdot 5 = 20$ possibilities to achieve this.

For two $5$'s, there are $\binom{4}{2} = 6$ ways to distribute them among the four rolls. The remaing dices have 5 different options each, so there are $6\cdot 5\cdot 5 = 150$ possibilities to achieve this.

That makes the total number of correct events $1+20+150 = 171$ and the probability thus is:

$$\frac{171}{1296} \approx 13.2\% $$

Generally, if you ask for the probability $p$ that in $n$ independent rolls an event with probability $q$ happens at least $k$ times, you have the formula:

$$p = \sum_{i=k}^n \binom{n}{i}q^i(1-q)^{n-i}$$

If you apply $n=4$, $q=\frac{1}{6}$, and $k=2$, you get the result above.

If you ask for an event happening no more than $k$ times, you can calculate the probability that the event is happening at least $k+1$ times, and subtract this from $100\%$. So the probability that you have no more than one $5$ is $100\%$ minus the probability that you have at least two $5$'s. As we have seen, the latter is $13.2\%$, so the former must be:

$$100\% - 13.2\% = 86.8\%$$


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