# Probability and dice

Why, when we throw 2 dice, do we have 36 outcomes and not 21? This question may sound silly but the intuitive answer if you ask someone, would be that there are 21 possible outcomes if we throw both dice instantly.

• Why would someone think the answer was $21$? – lulu Oct 12 '16 at 23:58
• think about it,think for instance that you play backgammon. – Marios Gretsas Oct 12 '16 at 23:59
• we care about 21 outcomes – Marios Gretsas Oct 12 '16 at 23:59
• Not following. You can get six outcomes for the first die, and six for the second and $6\times 6=36$. I'd say that was the intuitive answer. i don't see $21$ however I look at it. – lulu Oct 13 '16 at 0:01
• For probability purposes, $36$ is easier to work with....that way each roll has equal probability. Doing it your way, $(1,2)$ has twice the probability of $(1,1)$. – lulu Oct 13 '16 at 0:06

Because the dice are independent. There are 36 outcomes of equal probability. You can sum the probabilities of the equivalent outcomes together to produce 21 outcomes of non-equal probability, but it matters how you decide they are equivalent. eg. the same pair of numbers shown by the dice, or the same total of numbers shown on both dice?

So there are 36 outcomes if the order of the dice matter (eg. if they were different colours).

Or 21 outcomes if you don't care about the order of the dice (they are the same color) - but these are not equally probable (matching numbers are half as likely as a pair of mismatched numbers).

There are 11 outcomes if you only care about the total of the dice.

Edit: Just for fun... if you're playing monopoly there are 15 outcomes - 6 cases of "doubles" (matching numbers on the dice, which in monopoly means you get another turn or possibly go to jail) and 9 different total numbers that can be rolled without rolling "doubles" (3 through 11). In this case (4,2), (2,4), (1,5) and (5,1) are equivalent but (3,3) has a different meaning. So it really does matter how you decide what "outcomes" are important to you when answering this question.

For two dice there are $11$ possible values for the sum: any value between $2$ and $12$. Most of those can occur in more than one way, so they are not equiprobable. To figure out the probability of each sum you need to think about all $6 \times 6 = 36$ ways the two dice can fall. Sometimes you care about something other than the sum - perhaps the probability of at least one even number. There too you have to look at all $36$ possibilities.

• 11, not 13 for the sum. – Luke Oct 13 '16 at 0:07
• @Luke Fixed it thanks. – Ethan Bolker Oct 13 '16 at 0:10