Posted by: rjgage | October 10, 2008

## There’s no recruiting like e-recruiting!

Dear Math 152,

I thought that since this time of the year is interview season, I would post on a couple of interview-type probability questions that I have had asked to me in my e-recruiting adventures. Here goes:

Q1: You are given 50 black marbles and 50 white marbles and two empty jars. Divide all of the marbles between the jars in any way you like in order to maximize the probability of choosing a white marble when a marble is chosen randomly from either jar.

A1: Place one white marble in one jar, and all the rest in the other. That way you have a 50% chance of a white marble for certain, and a 49/99 chance for a white marble from the other. The probability of a white marble in this case is (1/2)*(1) + (1/2)*(49/99).

Q2: You play a version of Russian Roulette where two bullets are placed in consecutive chambers in a revolver. The revolver is then spun and the trigger is pulled. If at that point you are still alive, you can choose whether or nor to spin the chamber again before pulling the trigger the second time. Do you spin?

A2: No, you pull the trigger again without re-spinning. Because the bullets are placed next to each other in the chambers, of the four empty chambers three do not have an empty chamber after it, while one does. Therefore after we know that the current chamber is empty, we see that the probability of a bullet next is ¼ vs. 2/6=1/3 in the case of a re-spin.

Q3: You choose a number, any number from 1 to 6. Then 3 dice are rolled. If none show your number, you lose \$1. If one shows your number, you win \$1. If two show your number, you win \$3. If three show your number, you win \$5. Is this a fair game?

A:3 Yes. The expected value of playing the game is:

(\$5)*nCr(3,3)*(1/6)^3 + (\$3)*nCr(3,2)*(1/6^2)*(5/6) + (\$1)*nCr(3,1)*(5/6)^2*(1/6) +(-\$1)*nCr(3,3)*(5/6)^3 = \$0. It is a fair game.

Good luck on the e-recruiting!

Love,

Jack