Expected payout after rolling a dice three times and getting the highest roll You can roll a dice three times. You will be given $X$ where $X$ is the highest roll you get. You can choose to stop rolling at any time (e.g., if you roll a $6$ on the first roll, you can stop). What is your expected payout? Any ideas about how to solve this?
(1/6)(1) ≈ $0.17$
(1/6)(2) ≈ $0.33$
(1/6)(3) = $0.50$
(1/6)(4) ≈ $0.67$
(1/6)(5) ≈ $0.83$
(1/6)(6) = $1.00$
≈ 3.50  this is what I get for the 1st roll but having trouble reasoning out how to calculate the conditional probability of 2nd and 3rd roll along with the 1st to come up with the overall expected payout of rolling the dice 3 times.
 A: You roll three dices. Every possible outcome is a triple $(A_1,A_2,A_3)$ where $A_i$ is the score of the $i$-th dice. Every one of these triples occurs with probability $P(A_1,A_2,A_3)=1/216$. The payoff $X$ is a function of the outcome: $X(A_1,A_2,A_3)=\max(A_1,A_2,A_3)$. Then the expected payoff is given by the formula
$$
E=\sum\limits_{(A_1,A_2,A_3)}X(A_1,A_2,A_3)P(A_1,A_2,A_3)=\frac{1}{216}\sum\limits_{(A_1,A_2,A_3)}X(A_1,A_2,A_3)
$$
The problem is now to answer the following question: how many triples give me a payoff of $X$? Say we want to count in how many cases we get 6 as a payoff: that happens whenever one of the three dices gives a 6, i.e. 36+(36-6)+(36-6-5)=91 (namely, there are 36 possible configurations in which the first dice is a 6. There are 36 in which the second is, but we don't want to count twice the configurations in which the first and second dice are both 6. Likewise, the third dice will be 6 in 36 configurations, 6+5 of which we already counted). Let's count how many times you get a 5: 25+(25-5)+(25-5-4)=61. To get a 4, we have 16+16-4+16-4-3=37. To get a 3 we have $9+9-3+9-3-2=19$. For a 2, we have $4+4-2+4-2-1=7$. The sum of these numbers is 215, the remaining possibility is to get $(1,1,1)$. Hence, the expected payoff is the result of 
$$
E=\frac{1}{216}(6*91+5*61+4*37+3*19+2*7+1*1)=\frac{1071}{216}=4.95.
$$
