[BruceZ]

[ QUOTE ]
Hey guys - got this question at work - would appreciate your help as my probability knowledge is a little rusty..

Basket with 5 balls - 3 black, 2 white... if you draw a black ball you have to pay 1 dollar, if you draw a white ball, you get a dollar... once a ball is drawn, it is not replaced, and you can stop drawing at ANY TIME.

What is the value to playing this game? My back of the envelope is coming to be about 17.5 cents - is that correct?

Thanks in advance and my apologies if this is not the right forum.

dibsy

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Answer in white: <font color="white">
20 cents. If you draw white on the first ball, stop. If you draw black on the first ball, keep drawing until you are even or until there are no more balls.

EV = 2/5*1 + 3/5 *[-1 + E(2,2)]

where 2/5 is the probability of drawing white first, at which point we win 1 dollar. Clearly we stop at that point since we would be a 3:1 underdog on the next draw, and if that is black, we would be a 2:1 underdog on the next draw, and then all we can do is break even on the last two. E(2,2) is the EV of 2 white + 2 black that remain after you draw black first with probability 3/5, at which point we gain -1 dollar plus E(2,2). Now we need to evaluate E(2,2).

Note that E(2,2) can be no worse than 0 since you can always draw all 4 balls. If you draw white first with probability 1/2, then stop since if you continue, the EV of the remaining balls is 0 anyway (if you continue there is a 1/3 chance of drawing white first = 1, a 1/3 chance of black,black,white = -1, and a 1/3 chance of black,white = 0). If you draw black first with probability 1/2, so there are 2 white and 1 black remaining, then you can gain a white by drawing 2 whites in a row with probability 2/3 * 1/2 = 1/3. So E(2,2) = 1/2*1 + (1/2 *1/3) = 2/3, and

EV = 2/5*1 + 3/5*(-1 + 2/3) = 1/5 or 20 cents.</font>