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
Update after Steven Stadnicki's comments
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
$$
{5 \over 7776} \approx 0.064\%
$$
Best Answer
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 $\frac1{252}$.
And by the way, your calculation at best calculates the probability that someone first predicts one special outcome and then this outcome happens; what you want is that the prediction someone makes comes true.