Ask the Slot Expert: Son of Monty Hall plays Survivor

Question: I just read your "survivor" article of today.

The odds don't work that way.

The odds ARE 50/50 once one of the losers is revealed. The odds change, just like in Blackjack once cards are revealed. Otherwise, counting cards would be meaningless, and we know that's not the case.

I thought of a couple of alternate scenarios, but I won't bore you with the machinations of having more than 3 choices, and the reality is that you had a 1 in 3 chance at the beginning, and that doesn't change, but there was also only a 1 in 3 chance that the particular remaining "non-choice" was the good one. Therefore, your chances are equal.

Answer: You're right that you had a 1 in 3 chance of selecting the good box. You're also right that that doesn't change, but then you argue that it changes to 1 in 2.

Well, actually you said that the "chances are equal" because each box had a 1 in 3 chance of being good in the beginning of the game. If the chosen box has a 1 in 3 chance of being good and the remaining unchosen box has a 1 in 3 chance, where did the other 1 in 3 chance go? Probabilities have to add to 1.

Remember how the game works.

• We are told that there is a prize in one of three boxes.
• We choose one of the boxes.
• The omniscient host reveals one of the boxes that we didn't choose. The box is bad.
• We are then asked if we want to stick with the box we chose or switch to the other box.

When we choose a box, we partition the boxes into two groups. Each box has a 1 in 3 chance of having the prize. The chosen group has a 1 in 3 chance of being good. The unchosen group has two boxes, so that group has a 2 in 3 chance of having the good box in it. These chances don't change.

Let's play Einstein and do some thought experiments. First, let's choose one box at random and see what's inside. Note the result. Close the box. Mix them up. Choose again. Repeat.

Won't we have chosen the good box one-third of the time? How does revealing what's inside one of the boxes we didn't choose affect that?

Let's play again, but remove the histrionics of opening one of the unchosen boxes. Choose a box and then open all of them at once.

Won't we have chosen the good box one-third of the time? And won't the unchosen group have the good box two-thirds of the time?

The following table shows the three possibilities. After the host reveals that one of the boxes in the unchosen group is bad, which row can we eliminate?

Chosen	Unchosen		Stick	Switch
G	B	B	Win	Lose
B	G	B	Lose	Win
B	B	G	Lose	Win

Einstein had to rely on thought experiments because he couldn't travel near the speed of light, but we don't have to rely on them. We can play the game ourselves.

Try it. I did.

I took three slips of paper. I wrote G on one of them and B on the other two. I "shuffled" the slips under a table so I couldn't see them and picked two of them at random. I then looked at what was on them. I hope you'll agree that not revealing what is on one of the slips has no effect on what is on the other slip.

Picking two slips is equivalent to switching to the unchosen group of two boxes in the game. I did this 100 times.

Here are my results: I won 65 times and lost 35.

The good box with the prize is in the unchosen group of two boxes two-thirds of time and the good box is in the chosen group of one box one-third of the time.

The game is confusing because we start with box probabilities, but the game is really about the probabilities between the two groups. There is a 1 in 3 chance the chosen box (group) has the prize and a 2 in 3 chance that the prize is in a box in the unchosen group. The order in which we open the boxes doesn't change either group's chances.

But consider this. Let's say we open the chosen box first and it is empty. Now each box in the unchosen group has a 50/50 chance of having the prize.

The difference in this situation is that we're comparing probabilities of boxes within a group, not between groups.

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