In hiring process of Data Scientists positions, it is quite common to include some interview questions about probabilities. This post is about a probability quiz. Let’s consider the following question:

**Question**

” *You have the option to throw a die up to three times. You will earn the face value of the die. You have the option to stop after each throw and walk away with the money earned. The earnings are not additive. What is the expected payoff of this game? *“

**Answer**

This is a tricky questions, since the expected payoff depends on player’s strategy. We will provide a solution assuming that we are dealing with a “rational” player who tries to maximize their expected payoff.

**One Roll**

Let’s assume that the player has only one roll. Then, the expected return would be:

\(Roll_1= (1+2+3+4+5+6)/6=3.5\)

Hence, the expected payoff of one roll is **3.5**

**Two Roll**

Now, let’s assume that the player has the option of up to 2 rolls. The best strategy for them would be to roll again if their result is less than the expected value of the “One Roll” which is 3.5. In other words, if they got **4, 5, or 6** should keep it and if they got **1, 2 or 3** should try again. Thus, the expected payoff in this case is:

\(Roll_2= ( Roll_1 + Roll_1 + Roll_1 +4+5+6)/6= ( 3.5+ 3.5 + 3.5 +4+5+6)/6 = 4.25 \)

Hence, the expected payoff of two roll is **4.25**

**Three Roll**

Now, let’s assume that the player has the option of up to 3 rolls. The best strategy for them would be to roll again if their result is less than the expected value of the “Two Roll” which is 4.25. In other words, if they got **5 or 6** should keep it and if they got **1, 2, 3** **or 4** should try again. Thus, the expected payoff in this case is:

\(Roll_3= ( Roll_2 + Roll_2 + Roll_2 +Roll_2+5+6)/6= ( 4.25+ 4.25 + 4.25 +4.25+5+6)/6 \cong 4.67 \)

Hence, the expected payoff of three roll is **4.67**, which is the answer to our problem! Recursively, we can answer this question for n>3. Clearly, as **n** is getting larger, the expected return will converge to the maximum value which is **6**.