Number of total possible outcomes when calculating probablities in Yahtzee

So we have five six sided dice and we want to calculate the probabilities of getting results such as four of a kind...ie 55553.

So we find the total number of ways of getting four of a kind and divide it by the total number of possible outcomes from rolling 5 dice.

The number of total possible outcomes is $6^5$. ie..6 outcomes for each dice. But this is saying that a result of, say, 12345 is different from 54321. They are both getting counted as unique outcomes when we use $6^5$...yet they are the same result...So why do all the probability guides for Yahtzee say we use $6^5$ for the number of possible outcomes when that means we count many equivalent combinations as separate outcomes?

Imagine the dice as being tossed one at a time, or imagine them different-coloured. Then there are, as you point out, $6^5$ outcomes. They are all equally likely.

So for example the outcome $66666$ is just as likely as the outcome $12345$, which in turn is just as likely as $54321$.

But for the game, the order does not matter. Thus a reaasonable-sounding way to list the outcomes is just to list the results on the dice, say in non-decreasing order. Thus $12345$ is one outcome, and $22346$ is another, and $16666$ still another. But we do not list $66166$, since that's the "same" outcome as $16666$.

In principle, that's fine, we can define as we wish what we mean by outcome. The big problem is that if we do not consider order, then not all outcomes are equally likely. For example, five $1$'s is substantially less likely than getting a pattern of $5$ numbers that consists of the numbers $1$, $2$, $3$, $4$, and $5$. Indeed, the latter is $120$ times as likely as the former.

Thus if we use sequences as outcomes, or equivalently consider the dice to be of different colours, we get that all outcomes are equally likely, and we can obtain other probabilities by simply counting.

If we decide instead to not distinguish between the dice, then all outcomes are no longer equally likely, and computing probabilities gets more complicated. That's why Yahtzee books recommend using the $6^5$-element sample space. We can indeed work with a smaller sample space, but then analysis of probabilities becomes more difficult.

Using the $6^5$-element sample space, we can see fairly easily that four $6$'s and a $3$ can happen in $5$ of our $6^5$ equally likely outcomes, so has probability $\dfrac{5}{6^5}$. And we can see that getting the numbers $1$, $2$, $3$, $4$, and $5$ showing on our dice happens in $5!=120$ of our outcomes, so has probability $\dfrac{120}{6^5}$.

• I keep getting $6 \cdot \binom{5}{4}(\frac{1}{6})^4(1-\frac{1}{6})=\frac{5^2}{6^4}$. Whaere did I make a mistake? – Alex Oct 24 '12 at 10:16
• If you are computing the probability of $4$ of a kind, the expressions you gave in the comment above are perfectly correct. You used a procedure somewhat different from pure counting, namely adding binomial probabilities. But the two procedures are equivalent. – André Nicolas Oct 24 '12 at 15:17