Medal to correct answer!!
A game of “Doubles-Doubles” is played with two dice. If a player rolls doubles, the player earns 3 points and gets another roll. If the player rolls doubles again, the player earns 9 more points. How many points should the player lose for not rolling doubles in order to make this a fair game?

3/5
27/35
9/10
1

Question

Medal to correct answer!!
A game of “Doubles-Doubles” is played with two dice. If a player rolls doubles, the player earns 3 points and gets another roll. If the player rolls doubles again, the player earns 9 more points. How many points should the player lose for not rolling doubles in order to make this a fair game?

3/5
27/35
9/10
1

anonymous · Answer

Probability of two doubles : 1/36; payoff : 12.
Probability of just one double : 5/36; payoff : 3 - x.
Probability of no doubles : 5/6; payoff : - x. 

Add the products of the probabilities and payoffs, assign to 0, and solve:
12/36 + 5*(3-x)/36 - 5*x/6 = 0

x = - 27/35

Which is to say, the player should lose 27/35 points for not rolling a double.

anonymous · Answer

May I ask how you got your fractions: 1/36, 5/36, and 5/6?

anonymous · Answer

My pleasure!

When I tackle questions like this one I make a tree graph. See the attached diagram. The first branch represents whether the first roll results in a double or not. If the first branch results in a double then there is a second branch and it too represents whether the roll results in a double or not, this time for the second roll.

When you roll a pair of dice you can enumerate all of the possible combinations: (1,1), (1,2), (1,3), ..., (6,6). If you do this you will find that one out of six of them are doubles. This explains the labelling on the first branch. (The probabilities must add to one; if one branch has a probability of 1/6 then the other must be 1-1/6.) The probabilities for the second branch are worked out in the same way. 

Now, to get the overall probability of reaching the branch representing two doubles in a row it's necessary to notice that this involves two independent events, the first roll and the second. But that's no problem; you know their probabilities, they're both 1/6 and therefore the probability of the compound event is 1/36. (This is the great thing about using the tree graph.) In the same way, the probability of getting one double then one non-double is 1/6 x 5/6; just multiply the probabilities that you find following one branch of the tree through to the end. Finally one branch remains which has the probability 5/6.