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Here's the question:

6 Dice are rolled. Which is more likely, that you get exactly one 6, or that you get 6 different numbers?

Here's what I've done:

The number of possible outcomes is $6^6 = 46656$.

The probability of rolling exactly one 6 = $\frac{1}{6}\times(\frac{5}{6})^5 = \frac{3125}{46656}$

The probability of getting six different numbers is:
$C(6,1)\times C(5,1)\times C(4,1)\times C(3,1)\times C(2,1)\times C(1,1) = \frac{720}{46656}$

Therefore if everything I've said above is true, then it is more likely that you will roll exactly one six. However I'm really not sure about the last part. Is this correct way to solve this type of problem and can the combinations part be simplified?

EDIT: Since I'm also looking for a better idea of how to solve this type of problem, rather than just this specific case, so could you please include how to solve this problem for rolling 5 dice, as well as/or instead of 6 dice in your answer, so that I can see the pattern of what is happening? Many thanks.

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    $\begingroup$ The probability of exactly one $6$ is $\binom{6}{1}$ times what you wrote down. $\endgroup$
    – André Nicolas
    Dec 7, 2014 at 22:11
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    $\begingroup$ The probability of exactly one six is six times as great as you have computed, because the six can appear in any of the six positions. $\endgroup$
    – Mark Bennet
    Dec 7, 2014 at 22:12
  • $\begingroup$ So if there were 5 dice, would it be $(\frac{5}{6})^4 \times \frac{1}{6}$ multiplied by C(6,1) or C(5,1)? $\endgroup$
    – JC2188
    Dec 7, 2014 at 22:32
  • $\begingroup$ After re-reading what @MarkBennet commented, it should be C(5,1), because there are 5 positions the 6 can be in? $\endgroup$
    – JC2188
    Dec 7, 2014 at 22:41
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    $\begingroup$ You can also use intuitive reasoning to find that exactly one six will be more frequent, simply because every case of each die being different also satisfies exactly one six, but not every case of exactly one six has every die being different. $\endgroup$
    – fluffy
    Dec 8, 2014 at 0:51

7 Answers 7

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The question is very easy to answer without computing probabilities. Every combination with six different numbers contains exactly one six. There are then additional combinations which contain exactly one six - e.g. $111116$. So the probability of exactly one six is greater.

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    $\begingroup$ Simple and obvious! +1. $\endgroup$
    – TonyK
    Dec 7, 2014 at 22:18
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    $\begingroup$ +1 Very elegant. I guess the OP edit was done after you posted your answer. $\endgroup$
    – user1853181
    Dec 8, 2014 at 3:04
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    $\begingroup$ Very nice, inspiring way of apporaching problems. $\endgroup$
    – jeremy radcliff
    Dec 8, 2014 at 5:04
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    $\begingroup$ real nice solution there! $\endgroup$
    – Math chiller
    Dec 8, 2014 at 6:15
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    $\begingroup$ I'm happy that OP accepted another answer. General methods solve lots of problems - and that's fine. $\endgroup$
    – Mark Bennet
    Dec 8, 2014 at 23:00
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The number of rolls with exactly one 6 is $$\binom{6}{1}5^5=18750$$ the number of rolls with all dice different is $$6!=720$$

For $5$ dice, the number of rolls with exactly one 6 is $$\binom{5}{1}5^4=3125$$ and the number of rolls with all dice different is $$6\cdot 5\cdot 4\cdot 3\cdot 2=720$$ so the number with exactly one 6 is still larger.

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  • $\begingroup$ So by $\binom{5}{1}5^4=3125$, what you are saying is, there are 5 possible places the 6 can be, therefore multiply all the places it could be, by the probability ($5^4$) that the other dices don't show a six? $\endgroup$
    – JC2188
    Dec 7, 2014 at 23:23
  • $\begingroup$ @JC2188 Here I am counting the number of possibities; to get the probability from that you divide by the total number of possibilities. Though if we replace the $5^4$ with $(\frac{5}{6})^4(\frac{1}{6})$ then we obtain the probability. In any case, there are 5 possible places the 6 can be, then the probability that the die at that place is a 6 and none of the others is a 6 is $(\frac{1}{6})(\frac{5}{6})^4$, which is essentially what you are saying. $\endgroup$
    – Matt Samuel
    Dec 7, 2014 at 23:26
  • $\begingroup$ Brilliant thanks, one last check; If I wanted to do this for 3 die, I would simply do $5^2 \times C(3,1)$, to count the number of possibilities? The total number of possibilities would be $6^3$ and the number of rolls with different numbers would be $\frac{6!}{3!} = 6\times5\times4 = 120$ $\endgroup$
    – JC2188
    Dec 7, 2014 at 23:34
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    $\begingroup$ @JC1288 Yes. Extra characters so I can send the message. $\endgroup$
    – Matt Samuel
    Dec 7, 2014 at 23:35
  • $\begingroup$ For another formula that's easy to compute, see Henry's comment on the question. $\endgroup$
    – Jon Coombs
    Dec 11, 2014 at 19:43
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then it is more likely that you will roll exactly one six.

Intuitively makes sense, because in each of the combinations where every dice is different, there is exactly one six. Therefore, there are more cases of one six than all different.

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  • $\begingroup$ You solution is the most clear & elegant :-) How many people tried to calculate the exact probabilities without any need. $\endgroup$
    – peterh
    Dec 10, 2014 at 10:24
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The trick is to compare the sets of possible outcomes for each condition. Whichever condition's result set has a higher cardinality -in other words, whichever set has more possible outcomes- is more likely to occur, and if the two sets are equal in size, then they are equally likely.

When the number of dice is the same as the number of sides per die -for example, 6d6- you can take a shortcut by showing that one condition is a proper subset of the other. This is how @MarkBennet's answer works. Because there are as many dice as sides, every outcome with all different numbers must show exactly one of each number. So as long as it's possible for your chosen number to come up at all (your example uses a 6 on a six-sided die), then every outcome with all different numbers must also have exactly one of the chose number. You can also show outcomes with exactly one six which don't have all different numbers, like (2, 2, 3, 4, 5, 6), but you can't show outcomes with all different numbers but not exactly one six. Therefore, the set of outcomes with exactly one six is larger, and therefore it's more likely to roll exactly one six.

When there are more dice than sides, you have a different shortcut: the pigeonhole principle. Every die must show some number, so since there are more dice than sides, there must always be at least one number that comes up twice. This means that the set of outcomes with all different numbers is zero, so as long as it's possible for your chosen number to come up at all, then the set of outcomes with exactly one of that number has a greater cardinality. This makes exactly one six more likely than all different numbers, because there are no possible outcomes with all different numbers.

But when there are more sides than dice, there are no shortcuts. You can still show outcomes which satisfy both conditions at the same time, like (2, 3, 4, 5, 6) on 5d6: all different numbers, and exactly one six. And you can still show outcomes with exactly one six but all different numbers: continuing our example, (3, 3, 4, 5, 6) on 5d6 is one of them. But now, you can also show outcomes containing all different numbers but not exactly one six: e.g. (1, 2, 3, 4, 5).

When there's no shortcut, you have to compare the cardinalities of both sets, which is what you did in the question (and @MattS did in his answer). Whichever set has a higher cardinality is more likely to occur.

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You can actually do this without any of the formulae that you've written. If you all six die are different, then there must be exactly one of each number; in particular, there is exactly one six. However, you can get one six and then not all of the others, eg $(6,1,1,1,1,1)$. Thus we see that the set of outcomes where you get exactly one six - call this result $A$ - is a strictly contains the set of outcomes where you get one of each - call this result $B$. Thus $$\Bbb P(A) > \Bbb P(B).$$ However, this doesn't generalise well: for example, change to five dice and you could have $(1,2,3,4,5)$ which is a member of $B$ but not $A$. As such, not very helpful for the additional bit that you put on, but I just thought I'd let you know about it anyway. :)

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  • $\begingroup$ Isn't it the other way round? Ie, the set of outcomes where you get one of each is a subset of the set of outcomes where you get only one 6. Therefore, only one 6 is more probable. $\endgroup$
    – Max Williams
    Dec 10, 2014 at 16:46
  • $\begingroup$ Haha, oops, yeah. Corrected now! Thanks. $\endgroup$
    – Sam OT
    Dec 10, 2014 at 20:18
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First, the obvious way:

Likelihood of exactly one 6: = 1/6 * (5/6)^5 = 5^5/6^6

Likelihood of exactly one of each number: = 6/6 * 5/6 * 4/6 * 3/6 * 2/6 * 1/6 = 6!/6^6

5^5 > 6!, so "exactly one 6" is the more probably result.

Now, the easy way to do it:

"exactly one of each number" is a subset (special case) of "exactly one six", therefore "exactly one six" happens more often, and is more likely.

Now, if you want to get n similar dice to each have a different value, they must have n sides, so, the pattern you're looking for is something like ...

(n-1)^(n-1)/(n^n) vs (n!)/(n^n)

Where n = 6:

1/6 * 5/6 * 5/6 * 5/6 ... vs 6/6 * 5/6 * 4/6 * 3/6 ...

Where n = 8:

1/8 * 7/8 * 6/8 * 5/8 ... vs 8/8 * 7/8 * 6/8 * 5/8 ...

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Since every "Diverse set" includes one "six", any set with one "6" that isn't a "Diverse set" increases odds of just one six, rather than getting a "Diverse set."

Your have better odds of getting one "6."

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