# Confusion about probability question

Some friends are sitting together playing a game that involves rolling dice. On one turn, a player rolls six 6-sided dice and gets one of each number showing. Another player sees this and asks "what are the chances of that?" If the probability of rolling six different numbers on six $6$-sided dice can be expressed as $\frac{a}{b}$ where $a$ and $b$ are coprime positive integers, what is the value of $(a + b)$?

My solution: There's something I really can't drill into my head for the memo given. But this is how I went:

If the first die shows a number, then the probability of that number being from $1$-$6$ is 1 and the probability of the second die having a different number is $\frac{5}{6}$ and so on . I.e. $$\frac{1}{1} . \frac{5}{6}. \frac{4}{6} .\frac{3}{6} .\frac{2}{6} .\frac{1}{6}$$

But there are 6 different dice, so I multiplied that product by $6$ to get $\frac{5}{54}$ and thus $a+b = 59$. When I checked the memo, the answer was obtained without multiplying by $6$ again, i.e $\frac{1}{1} . \frac{5}{6}. \frac{4}{6} .\frac{3}{6} .\frac{2}{6} .\frac{1}{6}$ Why is this? I don't get the logic behind this. Can someone please explain to me why my answer is incorrect?

• There is no reason to multiply the answer by $6$. Your reasoning before you multiplied by the extra factor of six was correct and complete. – paw88789 Apr 27 '15 at 10:47
• Why though? There are $6$ different dices. – Aspiring Mathlete Apr 27 '15 at 14:08
• What makes you think you should multiply by $6$? – paw88789 Apr 27 '15 at 15:02
• Because there are $6$ different dices. This means it can happen on 6 different dices. – Aspiring Mathlete Apr 27 '15 at 16:14
• But you already took into account that the dice can come up 1, 2, 3, 4, 5, 6 in any order. – paw88789 Apr 27 '15 at 16:32

## 1 Answer

The probability of rolling any of six numbers is: $\tfrac 6 6$. (That is: $\tfrac 1 1$.)

The probability of rolling any different number than that one is: $\tfrac 5 6$.

The probability of rolling any different number than those two is: $\tfrac 4 6$.

And so forth, as you correctly reasoned to this point.

Thus the probability of rolling six different number on six dice is: $\colour{blue}{\frac{6}{6}}\cdot\frac{5}{6}\cdot\frac{4}{6}\cdot\frac{3}{6}\cdot\frac{2}{6}\cdot\frac{1}{6}\cdot= \frac {6!}{6^6} = \tfrac {5!}{6^5}$

That is all that you need. You have worked out the product of probabilities that each subsequent die will have a different result from the ones you had previously examined. There is no reason that having six different dice would require multiplying this by six.

A different approach to the problem is looking at the probability of obtaining results 1,2,3,4,5,6 in any order.

The probability of one die obtaining 1 is: $\tfrac 1 6$

The probability of another die obtaining 2 is: $\tfrac 1 6$.

And so forth for each number.

So the probability that each of six die obtain the different numbers, in consecutive order, is: $\tfrac 1 {6^6}$

As you know, the set $\{1,2,3,4,5,6\}$ has $6!$ permutations (that is the number of distinct rearrangements).

Thus the probability of obtaining different numbers in no set order, is: $\tfrac {6!}{6^6} = \tfrac {5!}{6^5}$

In the first approach we do not care about the order of the results because we are calculating probabilities based on arbitrary results for the favoured cases. The first die is required to be any number, the second to be any other number, and such.

In the second approach we care about order because we first calculate the probabilities of particular results - first 1 followed by 2 and so on -, and then count the possible rearrangements of that which still give the favoured case.