Question

You have 100 dollars, and there is a dollar bill behind each door. You roll a 100 sided die 100 times, and you take the dollar behind the door on the die roll if the bill has not been taken already (e.g. you roll 16, then you take the dollar behind door 16 if you haven’t already taken it). What is your expected payoff?

Answer 1

In the scenario where you roll a 100-sided die 100 times to collect a dollar behind each numbered door, the expected payoff is $100, since each number has an equal likelihood of being rolled once.

Step-by-step explanation:

The question you're asking involves probability and expected value, a concept from mathematics specifically relevant in understanding outcomes in scenarios involving randomness and repetition, like the one described with 100-sided dice and dollars behind doors. When rolling a 100-sided die 100 times for dollars behind 100 doors, your expected payoff is relatively straightforward to calculate. Since each roll has an equal chance to land on any number between 1 and 100 and no number will be chosen more than once, the expected outcome is that you will roll each number once.

Thus, on average, you are expected to open each door exactly once, so the expected payoff would be neatly the sum of all the dollars behind every door. As there are 100 doors, each with a dollar behind it, your expected payoff is simply $100.

Answer 2

63.21.

Step-by-step explanation:

You have 100 dollars, and there is a dollar bill behind each door. You roll a 100 sided die 100 times, and you take the dollar behind the door on the die roll if the bill has not been taken already (e.g. you roll 16, then you take the dollar behind door 16 if you haven’t already taken it). What is your expected payoff?

X=Σ100i=1 1Ai

where Ai is the event that door i is opened at least once, and 1Ai is the indicator function for event Ai.

Thus the expected payoff is:

E[X]=Σ100 as i=1 Pr[Ai].

to calculate Pr[Ai].

Ai∁ is the probability of the event that after 100 rolls, door i is not chosen, which is:

Pr[Ai∁]=(99/100)^100

Thus:

Pr[Ai]=1−Pr[Ai∁]=1−(99/100)^100.

E[X]=Σ100i=1Pr[Ai]=100×(1−(99/100)^100).

Also based on the following approximation for large n's:

(1−1n)n≈1e

we have:

(99/100)^100=(1−1/100)^100≈1/e.

The expected pay off is

E[X]=100×(1−(99/100)^100)≈100×(1−1/e)≈63.21.