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# Does withdrawal without replacement change your expectation?

6 June 2011

Pretend you’re at a slot machine when a message pops onto the screen saying you and three other lucky patrons have been chosen to participate in a free drawing for cash prizes. You check with your host and learn that the game involves five envelopes, each containing one of \$5,000, \$2,500, \$1,000, \$500, and \$100. Each of the four players, in turn, selects an envelope and receives what’s inside. Envelopes aren’t replaced after being drawn; once an amount has been taken, it’s unavailable to whoever’s left. Your host says players will draw in the order they get to the lobby and sign-in; he urges you to rush down right away so you can get the first pick, to ensure the \$5,000 isn’t gone before your turn.

This seems to make sense. Or, does it? The contestant who draws first has a chance of one in five (20 percent) to win \$5,000. If this individual gets something else, the chance at \$5,000 is one in four (25 percent) for the second player. Likewise for the third, if neither of the first two gets the \$5,000, the chance improves to one in three (33-1/3 percent). And for the fourth, it’s one in two (50 percent). So, is it preferable to draw when the \$5,000 is sure to be there but the chance is 20 percent, or go later when the biggie might be gone but – if not – the probability you’ll get it will be greater?

A sophisticated solid citizen would do a calculation on the back of a convenient napkin and find the theoretical “expected win”or “expectation” for each of the players. The arithmetic involves multiplying each possible prize by the associated probability, and adding up all the products.
The first player has all five envelopes available. Take 20 percent of \$5,000 plus 20 percent of \$2,500 plus 20 percent of \$1,000 plus 20 percent of \$500 plus 20 percent of \$100. The sum is \$1,820.

For the second player, four scenarios are possible. If the \$5,000 is gone, the “conditional expectation” would be 25 percent of \$2,500 plus 25 percent of \$1,000 plus 25 percent of \$500 plus 25 percent of \$100. This equals \$1,025, not as good as going first. But, what if the first player took \$100, not \$5,000? The conditional expectation would be 25 percent of \$5,000 plus 25 percent of \$2,500 plus 25 percent of \$1,000 plus 25 percent of \$500. This is \$2,250, better than going first. Results for the first player winning \$2,500, \$1,000, and \$500 would yield conditional expectations for the second contestant of \$1650, \$2,025, and \$2,150, respectively. Analogously for being the third and fourth up to bat. Your expectation would be lower if the folks who go earlier take the big money, and is higher if their envelopes are at the low end.

The fly in the ointment of this logic is, when you’re assigned drawing order, you don’t know what those ahead of you will do. So conditional expectations don’t apply. The correct basis for your decision would be the “joint expectation” associated with your position in the line-up. That would involve the probabilities of various outcomes of earlier as well as of current rounds.

For the second spot, you’d have to multiply the probability that the \$5,000 is gone (20 percent) by the associated conditional expectation for the second player; you’d get 20 percent of \$1,025 or \$205. Likewise, the joint expectation for the second round with \$2,500 gone is 20 percent of \$1,650 or \$330. Do the same for the other first-round possibilities to get second round joint expectations of \$405, \$430, and \$450 for first-round draws of \$1,000, \$500, and \$1,000 respectively. Adding \$205, \$330, \$405, \$430, and \$450 gives \$1,850. The joint expectation for the second participant equals that for the person who goes first.

Joint expectations are also \$1,850 for the third and fourth players, although the arithmetic is messier. Figuring joint expectation for the second round, there were only five ways the first round could be resolved (\$5,000, \$2,500, \$1,000, \$500, \$100), each involving one probability. But by the third round, 10 combinations of envelopes could be out of action (\$5,000/\$2,500, \$5,000/\$1,000, \$5,000/\$500, \$5,000/\$100; \$2,500/\$1,000, \$2,500,/\$500, \$2,500/\$100; \$1,000/\$500, \$1,000/\$100; \$500/\$100) – involving the products of two probabilities each. After the third round, there are 10 combinations of envelopes that could be unavailable – involving the products of three probabilities each.

The bottom line is that joint expectation, which is what you’re facing before the game begins, is the same regardless of your place in the line. If you’re a blackjack buff, you might want to think about how this affects your action. How does your expectation differ according to whether or not you count cards, where in the dealing order you sit, and what the other players do. The answer should make you less intolerant of bezonians at third base who stand on 12 versus 2-up and feed the dealers winning nines. For, as the inimitable inkslinger, Sumner A Ingmark, intoned:
In those situations where randomness rules,
Apostles of order may well act like fools