A hat contains two blue marbles and three red marbles. If you draw a blue marble, you win $1. If you draw a red marble, you lose $1.
Question 1: Would you play this game if you get just one draw, i.e. is the game either fair or biased in your favor?
Question 2: Would you play this game if you may draw, without replacement, until you decide to quit or the hat runs out of marbles?
Question 3: What are the expected values in both cases?
Source: A problem that was sent by e-mail around our Math Department.
Hint: Gather a hat with two blue marbles (chips?) and three red marbles (chips?). Now, simulate the problem situation until you get a feel for a good way to approach a solution.
Solution Commentary: First, the game is not a good one if you only get one draw. Because of the higher probability of drawing a red marble, the expected value is (2/5)*($1) + (3/5)*(-$1) = -$0.20.
John W. of our department offered this solution for the second case: Make the first draw and then stop as soon as (number of blues drawn - number of reds drawn) is greater than or equal to 0. Your expected take in this case is $0.20. That is, stop when the sequence of draws shows B, RB, RRBB, RRBRB, or RRRBB.
Do you agree with his logic and claim?