# Probability of 3 Increasing Numbers Dice Roll Problem [duplicate]

You roll three dice, one after another. What is the probability that you obtain three numbers in a strictly increasing order?

Intuitively, I want to say that this should be $$\frac{1}{6P3}$$, that is, 1 divided 6 Permutation 3. My reasoning is that there is only one specific ordering we desire: three strictly increasing numbers; and that there are 6P3 possible orderings. This would give a probability of $$\frac{1}{120}$$.

However, this is incorrect, as the actual solution is $$\frac{5}{54}$$. I've seen a couple different explanations, but none of them really make sense to me.

For example, the accepted answer in response to the same question posted here: Arranging items in strictly increasing order

In this solution, joriki writes that there are 6C3 = 20 strictly different ordered outcomes. This is hard for me to accept, since a combination is used to find the number of selections without regard to order. In any case, he also reasons that there are a total of $$6^3$$ total three number outcomes from having three dice rolled, which makes sense. Therefore, according to joriki, the probability must be $$20/216$$, or $$5/54$$.

While I can accept that this arrive at the right solution, I am having trouble understanding the reasoning of using a Combination as opposed to a Permutation in the numerator. Specifically, the concept of "ordered" vs "non-ordered" here, makes it hard to understand how it was appropriate.

• I don't follow your argument. Even if you were assured the three rolls were distinct (which you are not) there would be $3!=6$ ways to order the three rolls, only one of which would be strictly increasing, so the answer in that case would be $\frac 16$. Here, of course, you need to consider the possibility that some of the throws match.
– lulu
May 28 at 20:11
• Doesn't your solution ($1/120$) sound way too small to you? May 28 at 20:39

You need three different values. The number of ways this can be done is $$6 \cdot 5 \cdot 4 = 120$$.

The total number of rolls is $$6^3 = 216$$.

So the probability that you rolled three different numbers in three rolls is $$120/216 = 5/9$$.

But, the probability that the three numbers are in increasing order is $$1/3! = 1/6$$.

So, the answer is $$5/9 \cdot 1/6 = 5/54$$.

• I've seen this solution too. How do we justify multiplying these two probabilities together to arrive at the final solution? These aren't really independent events, are they? Because if they were independent events occurring together, that would make sense. May 28 at 20:13
• Maybe it's easier to think about the $120$ lists of distinct values, and then consider that only $1/6$ of them are in strictly increasing order. So, the list of increasing sequences is $120/6 = 20$, and then you can look at this over the entire sample space of $6^3$ to get $20/216 = 5/54$. May 28 at 20:16
• It hadn't occurred to me to think about it like that. Thanks. May 28 at 20:28
• @StanleyYu if you're satisfied with this solution, please accept it May 28 at 21:59