5 distinct dice results in 6 rolls

Question

Given we have rolls 6 iid dice, what are the odds of having 5 distinct rolls (where we have defined $N$ as the number of distinct outcomes)?

The work above is my professor's provided solution, but I don't think it's right? It's equivalent to $6!/6^6$, which would be equivalent to a situation where all the dice rolls are distinct? This can't just be a coincidence.

The way I pictured doing it was like so:

$$(6/6)(5/6)(4/6)(3/6)(2/6)(5/6)$$

where the first 5 terms are the 5 distinct dice rolls, and the last one is $5/6$ because it must be one of those distinct rolls. This gives me an answer of $0.077$. However, I don't think this is right because it doesn't account for the fact that the 2 non-unique outcomes can be placed in different positions among the 6 roll (6 choose 2 times actually).

Any help would be great, thanks!

Answer 1

You want the probability for selecting five distinct results when rolling six regular dice.

$\binom 65 5!/6^6$ counts the ways to select five from six results and then arrange them; against the ways to roll six six-sided die.   That's not right at all.   It is missing out on counting a sixth die, and its value and location.

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You have $\frac 6 6\frac 5 6\frac 4 6\frac 3 6\frac 2 6\frac 5 6$, which is the probability of rolling 5 distinct results on 5 dice then repeating one of those results with the sixth dice.   That's close, however, you don't require the repeated result to be in any particular place, and you want to select a place for it without over counting.

Well, $\frac 6 6\frac 5 6\frac 4 6\frac 3 6$ is the probability of rolling a sequence of four distinct results, and $\frac 2 6\frac 1 6$ is the probability of rolling a fifth distinct result, twice.   Finally $\binom 62$ counts the ways to merge these two sequences into one.

$$\frac{6!/2!}{6^4}\cdot\frac{2!}{6^2}\cdot\frac{6!}{4!\,2!}=\frac{25}{108}$$

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I'd count the ways to select dice to be a double and four singleton results, and ways to select the results for them from the six faces (1 for the double, 4 for the singles, and 1 not used); against ways to make a sequence of six choices with six independent options.

$$\dfrac{\binom{6}{2,1,1,1,1}\binom 6{1,4,1}}{6^6} = \dfrac{\frac{6!}{2!}\frac{6!}{4!}}{6^6} =\frac{25}{108}$$

Answer 2

Your professor's attempt of $\cfrac{\binom{6}{5}\cdot5!}{6^6}$ is almost correct, by a factor of 15: the answer is $$\cfrac{\binom{6}{5}\cdot5\cdot\frac{6!}{2}}{6^6}.$$

Here's how: choose five distinct dice rolls from six, and designate one of the just chosen five to appear twice. Arrange the six numbers, with one repeated, in $\frac{6!}{2!}$ ways.