# Probability of an expected outcome

I'm in a class titled "Puzzle Based Learning" and we were given this problem:

There is a new game show and you are the participant. There are two doors, each has a suitcase with gold coins behind it. You know that these suitcases contain amounts from the set: 25, 50, 100, 200, 400, 800, 1,600, 3,200, and 6,400. You also know that one suitcase has twice as much coins as the other. You select the door and open the suitcase. You find 1,600 gold coins. Then the game show host offers you the opportunity to switch doors. Do you do it? Justify your answer.

In class, the professor tells us:

The two possibilities of the suitcase are 800 and 3200. The expected outcome of choosing one suitcase would be 800*.5 + 3200*.5 = 2000. Since 2000>1600, you should choose again because the expected outcome is greater than 1,600.

I was hoping someone could explain this a little more clearly. In my mind, it seems there is a 50/50 chance of getting 800 and 3200 and the "expected outcome" is meaningless because we shouldn't care about the payoff of choosing again. You have a 50% chance of losing.

Am I missing something? Is this a trick question? Is my professor pulling my leg?

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there is nothing wrong with any of that. whether you "care" about the payoff of choosing again is subjective. if you choose again you will either get 800 or 3200, and the expected value of that choice is 2000. it's still a coin flip. doubling up is just a bigger gain than the loss of losing half. – yoyo Mar 18 '11 at 18:42

There is little typo in your question. There should be $$800 \cdot \frac{1}{2} + 3200 \cdot \frac{1}{2} = 2000.$$ What your professor said is that the expected value of your winning is 2000 when you change the doors. In some way it means that when you play this game many times then your average winning is going to be closer and closer to 2000 (assuming that you always chose the door with 1600 first).

Formally, the Law of large numbers plays role here. If you take $X_i$ to be your winning in $i$'th game and $S_n$ be the sum of your winnings in $n$ games then $$\frac{S_n}{n} \to 2000.$$

Hence if you change the door then your average win will be greater then 1600. Here it is important that you either win 1600 more or 800 less then in your first choice with equal probability.

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So would it be a wise choice to choose again when you have 1600? How important is the 2000 figure if I'm only choosing once? – eternalmatt Mar 18 '11 at 22:09
In the mathematical theory, all that matters is that the expectation of switching is greater than not switching. In the real world, maybe not so. Suppose you had \$1M and were offered a 1/1oo shot at \$101M. Would you take it? Is \$101M really 101 times more valuable to you than \$1M? – Ross Millikan Mar 19 '11 at 14:52
Personally, I wouldn't take it. And I don't think anyone in their right mind would. But I really feel that psychology shouldn't play into this at all, as @Henry suggested. Perhaps I am just being stubborn? – eternalmatt Mar 20 '11 at 5:28

There is something missing, namely any statement that the 16 ways of putting the suitcases behind the doors are equally likely. If they are equally likely then you should always switch after opening a door unless you see 6400, for the reason given.

But suppose that you know each pair of suitcases $(n,2n)$ is three times as likely as the next pair up $(2n, 4n)$. In your particular example the choice would be between $1600$ and an expected $800\times \frac{3}{4} + 3200\times \frac{1}{4} = 1400$ so you would stay with what you see first; similarly for any amount you would keep what you see, unless you see 25 in which case you would switch doors.

If you simply don't know - it is a new game show - then you have to judge the psychology of the game designers. They may think more big winners will attract more viewers and so pay off in advertising; or they may just want to drive down prize costs. Your call.

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Okay, if you lose, you end up 800 down. If you win, you end up 1600 up. So you have a lot more to gain, than to lose. Of course you don't know which one is which. But you know each chance is equal, 50/50.

If the choice was between 800 and 2,400, then there'd be 800 to win or lose either way. So the choice wouldn't matter, it would be "evens". In that situation you may as well gamble or not.

If the choice was 800 and 2,401, then there's an extra dollar in winning over losing. So on average you're better off to gamble on switching doors.

Overall, the two mystery doors are worth 800 + 3200. That's 4,000. There's two of them, so the average is 2000. The average value of those cases is 2000. The value of the case you have is 1600. So would you swap 1600 for 2000? Yes, obviously!

Of course in a game show you only get one turn, so more important factors would be how much you need the money, etc. But mathematically we're working on the average best strategy. The one that, if everyone did it, would end up with more money overall. And that's take the chance.

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