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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?