Bait and Switch

A new game show with doors has been invented and you are invited. In this case, there are two doors. Each has a pile of gold Krugerrands behind it, but one has twice as many as the other. You point to a door. The game show host opens it for you, then offers to allow you to switch your choice.

Picture it. You are on stage, the TV lights glaring. The host has just opened the door. Before you lie the coins, glistening. The audience lets out a moan of admiration and envy. For a moment you feel faint. Then the game show host, Monty Hall style, interrupts your reverie and offers you the opportunity to switch doors. (He does this to every contestant every time.) Do you do it?

Before you decide, you have to know the possibilities. Half the time, there are 100 coins behind one door and 200 coins behind the other. One quarter of the time there are 200 behind one door and 400 behind the other. One quarter of the time there are 400 coins behind one door and 800 behind the other.

You want to maximize your expected winnings.

Warm-Up

Suppose you see 100 coins behind your initially chosen door. Would you switch then?

Solution to Warm-Up

Duh. Clearly yes, because you cannot lose. If there are 800 coins, just as clearly you don’t change.

1. Would you switch if you see 200 coins behind the door?

2. Would you switch if you see 400 coins behind the door?

3. Suppose that the probabilities were 1/3, 1/3, 1/3 instead of 1/2, 1/4, 1/4. Then would you switch if there were 400 coins?

It’s a new season and the game show producers are afraid of a drop of popularity. For that reason the producers set the range of coins to be much larger: 25, 50, 100, 200, 400, 800, 1600, 3200, and 6400. All the door possibilities - 25-50, 50-100, and so on up to 3200-6400 - are equally likely.

4. Assuming your first selected door opens to any number of coins other than 25 or 6,400, should you switch?

Another season passes and the TV ratings are flagging just a little. The producer decides to extend the range from 1 coin to 1,048,576 coins. All x-and-2x possibilities are equally likely, he promises. But now that the extremes are so unlikely, the host won’t bother opening the door that the contestant first guesses. When the contestant makes his final choice, the curtain opens to a large bucket of gold, the sight of which causes predictable shrieks among the audience.

5. Should the contestant switch even when the host doesn’t open the contestant’s initial door?

Solution to Bait and Switch

1. Would you switch if you see 200 coins behind the door?

Let’s find the equi-probable events. Because the 100-200 situation is twice as likely as any other, we’ll write that one twice. The following are the equi-probable door pair situations:

100	200
100	200
200	400
400	800

Therefore, if you see 200 coins, you are twice as likely to be in the 100-200 situation as in any other. Switching to the other door, therefore, is twice as likely to cause you to lose 100 coins as to win 200. So, your expected winning by switching is zero. It’s your call.

2. Would you switch if you see 400 coins behind the door?

In this case, you are as likely to be in the 200-400 as in the 400-800 situation. In the first case, switching costs you 200 coins. In the second, switching enables you to win 400. So you expect to benefit by switching.

3. Suppose that the probabilities were 1/3, 1/3, 1/3 instead of 1/2, 1/4, 1/4. Then would you switch if there were 400 coins?

In this case, the equi-probable situations are:

100	200
200	400
400	800

Nothing has changed for the 400 coin case - you should switch. By the way, you should switch in the 200 coin case as well. Half the time, you’ll win an extra 200 coins.

4. Assuming your first selected door opens to any number of coins other than 25 or 6,400, should you switch?

Switch every time. It’s the same reasoning as in question 3.

5. Should the contestant switch even when the host doesn’t open the door to which the contestant first points?

Switching doesn’t help on the average. By symmetry it can’t. If it did, then the contestant could have simply chosen the other door to begin with. As the game show host opened no door, the contestant has gained no information. But how do we show this analytically? The equi-probable situations are:

1	2
2	4
4	8
…
1,048,576	524,288

In each case, if the contestant switches in one direction in the x-and-2x situation he will win x; if he switches in the other, he will lose x. So switching as a strategy yields no net gain (or loss). Why is this protocol different from the one where the host opens the door that the contestant first points to? In that protocol, the only time the contestant doesn’t switch is when he sees the bucket with 1,048,576 coins. That happens only one time in 40, but in that case, the contestant saves 524,288 coins by not switching. Therein lies the contestant’s edge.