# Expected Value on a roll of 3 die

You wager 1 dollar on the roll of three die, any doubles wins 2 dollars, any triples wins 3 dollars, and anything else loses. Find the expected value of this wager.

What I did was find the probability of each possibility. Probability of rolling a double with three die is $(1/6)*(1/6)*(5/6)=(5/216),$ prob of rolling a triple is $(1/6)*(1/6)*(1/6)=(1/216),$ and the prob of losing is $(6/6)*(5/6)*(4/6)=(120/216).$

Is this correct or did I do something wrong?

• Welcome! I took the liberty of formatting your question so that the problem you're working on is quoted. In order to get the best help, please edit the question to include the work you've done on the problem and where you are stuck. – Matthew Leingang Mar 15 '17 at 13:47

You are undercounting the winning cases severely.

Consider the triples. Of the $216$ possible sequences of three numbers that can come up, one of them is $1,1,1,$ which is a triple. But $2,2,2$ is also a triple. Already that's $2$ possible sequences out of $216,$ and there are more even than that, but you counted only one sequence to get a probability of $1/216.$

Essentially your "triples" are the chances of getting triple $1$s, not the chance of getting triples in general when any triple is good enough.

For the doubles, you have given the chance to roll $1,1,x$ where $x\neq 1.$ There are $5$ possible sequences like that, with a total probability (for just those five sequences) of $5/216,$ which you gave as the total probability for all doubles. So if we count the sequences $1,1,x$ first, then you didn't count $2,2,x,$ you didn't count $x,1,1$ or $1,x,1,$ and so forth.

Update: To find the correct probabilities, you have to take all possible favorable outcomes into account somehow.

For just three dice, "brute force" is possible to do by hand: write down all possible sequences, count the ones that have doubles, and count the ones that have triples. The disadvantage of this method is that there is a strong possibility you will overlook one or more of the doubles.

Another alternative is to work out a probability tree. Traveling down the branches of the tree, where each branch is labeled with the probability of going that way, you get products like the product $\left(\frac16\right)\left(\frac16\right)\left(\frac56\right)$ that you calculated earlier. To make things easier, however, rather than six branches for the roll of the first die, you can accept any roll as a single case with probability $1$; just remember when you are looking for a triple, you are trying to match whatever the first roll was, so for the second roll there is only one branch you can go down and it has probability $\frac16.$ The doubles will be the sum of the probabilities you find on multiple paths, however, because it is not possible to combine all possible doubles on one path through the tree: one way to get a double is if the second roll matches the first, but there are different ways to get a double if the second roll does not match the first.

A third alternative is to use combinatorics to count the outcomes and then divide by the total number of possible outcomes (which is $216$ in this question). This works nicely for problems like this since you have a well-organized set of possible outcomes that are all equally likely and cover the entire probability space. For triples, you need to count how many sequences of three numbers from $\{1,2,3,4,5,6\}$ (with repetition allowed) have all three numbers the same; for doubles, you need to count how many sequences have exactly two numbers the same. Think of things such as: how many different ways could I choose the number that gets doubled or tripled; how many different ways could I choose which places in the sequence have the double or triple; given a double or triple, how many ways can the remaining places be filled with numbers?

• So for triples it should be (6/216) and for doubles it should be (18/216), correct? – Jon Jones Mar 15 '17 at 16:07
• $6/216$ is good. $18/216$ is still too small--possibly you did not account for the $5$ different possible numbers for the "other" die (not in the matching pair of dice)? The three probabilities (triples, doubles, neither) should add up to $216/216,$ that's one way to check an answer. (You have the correct probability for the losing rolls already.) – David K Mar 15 '17 at 17:35