# Probability of predicting, then throwing, a particular multiset for 5 dice.

My friend shared with me a story that after losing to his SO at Yahtzee, before they put the game away he just randomly predicted he would roll four 5's and a 1. He then got that roll and freaked out.

I wanted to calculate exactly how likely that was but my combinatorics knowledge isn't quite strong enough. My question is, for 5 dice, given that the order in which they're thrown doesn't matter, how many possible outcomes are there?

### My leads and Internet "research"

After a bit of searching it seems that what I'm looking for is combinations with repetition, which led me to this Wikipedia article. So it seems that

$$\binom {n+k-1} {n-1}$$

where $n=6$ and $k=5$ would give me the total number of possible rolls...

$$\binom {6+5-1} {5}=\binom{10}{5}=252$$

Assuming $252$ is the correct count for the number of possible rolls with 5 dice, this would lead to the probability

$${{1}\over{252}}^2 = {1\over63504}$$

or roughly 0.02% chance of predicting and then throwing a certain multiset of 5 dice.

If someone could verify this or point me in the right direction that would be appreciated. Thank you!

### Update after Hagen von Eitzen's answer

Hagen von Eitzen made it clear that I actually don't want combinations with repetition since not every roll is equally likely. For example, there is one way to roll $\left\{6, 6, 6, 6, 6\right\}$ but five ways to roll $\left\{6, 6, 6, 6, 5\right\}$. Because of this order cannot be discounted. There are a total of $6^5=7776$ ways to roll 5 dice, and the roll predicted by my friend can be thrown in 5 ways, giving it a probability of $5 \over 7776$ of occuring. So the probability of my friend randomly predicting this particular roll and then immediately rolling it is $${5 \over 7776}^2 \approx 0.000041\%$$

...Hopefully

Steven Stadnicki pointed out to me that the randomness of the guess doesn't play into the probability, leaving the conclusion that the odds of this happening were $${5 \over 7776} \approx 0.064\%$$

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The different outcomes do not have the same probability of $\frac1{252}$. For example "five sixes" has probability $\frac1{6^5}=\frac 1{7776}$. You can increase your chance of winning by predicting one of the more common outcomes. For example "1, 2, 3, 4, 5" has a probability of $\frac{24}{3125}$, about twice as much as $\frac1{252}$.
For a simple die, your calculation would produce $\frac 1 {6^2}$ (instead of $\frac1{252^2}$). That is the probability that I e.g. randomly pick "3" as prediction and then roll a "3". However, if you play the prediction game with one die, you will notice that the prediction is correct about $\frac16$ of the time, not $\frac1{36}$. –  Hagen von Eitzen Sep 28 '12 at 16:32
Actually, after thinking about it, it seems correct to square the probability. He chose his prediction seemingly at random and then rolled it immediately. Doing this with one die, where you make a new random prediction before every roll, you'd have a $1 \over 36$ probability of being successful on any given attempt, right? –  leo-the-manic Sep 28 '12 at 17:07