Intuition for sum of dice rolls Two dice. One 10-sided, one 6-sided. Guess a number between  2 and 16, if the sum of the dice shows that number, you get that number of dollars. What is the best number to guess to maximise winnings?
I was asked this question in an interview. My first thought was to add the two expectations with linearity $3.5 + 5.5 = 9$, but it turns out that if you actually enumerate the possibilities, $11$ gives the highest expectation.
I was wondering if someone can give me a good way to solve this problem without enumerating all possibilities. I wasn't allotted too much time to answer, so I think there must be some trick to it. I am still unable to find it even days after the interview. 
 A: There can't be more than $6$ possible outcomes for any given sum because one of the dice only has $6$ faces.  
Sums of $2$ through $6$ are too small to have six outcomes. Sums of $12$ through $16$ are too large.  Sums of $7$ through $11$ do give the six possibilities.
The way I first thought of it was to think of (but not draw) a $6$ by $10$ grid. Each upper-right to lower-left diagonal corresponds to a sum.  The longest diagonal in the grid is length $6$, and there are $5$ of these corresponding to the sums noted above ($7$ through $11$).
A: There are 60 possible outcomes.  Out of them, there are six possible outcomes where a 9 is rolled ($(1,8)$ through $(6,3)$ if you consider the six and ten sided dice as the first and second coordinates), and six possible outcomes where an 11 is rolled ($(1,10)$ through $(6,5)$).  Actually, any sum from 7 through 11 has an equal number of possible outcomes, but 11 is the best to get you the best amount of money.
Precisely, your expected winnings with 9 is $9\cdot\frac6{60}=\$0.90$ and the expected winnings with 11 is $11\cdot\frac6{60}=\$1.10$.  The expectation from 12, just to double-check, is $12\cdot\frac5{60}=$1.00$.
