Probability game show

Question

Consider a quiz game show where you are given two questions to answer. Q1 will be answered correctly with probability 0.8 and Q2 with probability 0.5. Answering Q1 correctly carries a reward of 1000 while answering Q2 correctly carries a reward of $ 2000. You get to choose the order in which you answer the questions.The catch is that if you answer your first question incorrectly then the quiz terminates; you do not get a chance to attempt the second question. For a given order of answering the questions, let X be the random variable that is the number of correctly answered questions and let g(x) be the total reward received.

(a) Suppose you answer Q1 first and then Q2. Find the expected reward, that is, E(g(X)).

(b) Suppose you answer Q2 first and then Q1. Find E(g(X)).

(c) In what order should you answer the questions to maximize your prize money?

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For parts a and b, I understand the logic behind what you are likely to receive for each option, but can't come up with the exact reward sum one would be expected to walk out with..

For a, probability wise you are likely to get at least 1000$ and then a half-half toss between getting either an additional 2000 or 0.

For b, right off the bat in the very first question, you are equally likely to get either 0 or 2000$. If you get 2000, then you are likely to get an additional 1000 during the second question.

For c, I think the more profitable option would be a since you are very likely to walk out with at least 1k whereas in the other you have a 50% shot of leaving with 0.

Answer 1

Let $C_1$ be the event that you get question $1$ correct and $F_1$ the event that you get question $1$ incorrect. Similarly for question $2$.

Strategy (a): The possible outcomes are $F_1,~~C_1F_2, ~~C_1C_2$

The probability of arriving at the outcome $F_1$ is $0.2$ (as it is $1-0.8=1-Pr(C_1)$)

The probability of arriving at the outcome $C_1F_2$ is $Pr(C_1\cap F_2)=Pr(C_1)\cdot Pr(F_2)=0.8\cdot (1-0.5) = 0.4$

Similarly, the probability of arriving at the outcome $C_1C_2$ is $Pr(C_1)\cdot Pr(C_2)=0.8\cdot 0.5=0.4$. As a sanity check, by adding these together it should add up to one, and indeed $0.2+0.4+0.4=1$ as expected.

Each of these outcomes carries with it a value in terms of money earned. $F_1$ earns the player zero dollars. $C_1F_2$ nets the player only 1000 dollars, and $C_1C_2$ nets the player the full 3000 dollars (the 1000 for the first question plus the 2000 for the second)

To calculate the expected value, for each possible amount of money earned we take the money earned and multiply by the probability of getting that amount and add all of these results together.

So, we have $E[X]=0\cdot 0.2 + 1000\cdot 0.4 + 2000\cdot 0.4 = 0+400+800=1200$ when following the first strategy.

I leave it to you to calculate the expected value and corresponding probabilities for attempting to answer the second question first followed by the first question and comparing the two results to see which is the smarter strategy.

Answer 2

Your expected return on just Q1 is $0.2(0) + 0.8(1000)$
Your expected return on just Q2 is $0.5(0) + 0.5(2000)$

Your expected return taking Q1 first under the given rules is:

$$E(g_1(X)) = 0.2(0) + 0.8(1000+0.5(0)+0.5(2000))$$

And so on...