Paradoxical Game Show Problem

Question

Here's a problem that has had me scratching my head for a long time:

Imagine you're in a game show, and are presented with 2 boxes. You are told that both boxes contain a sum of cash, but one of the boxes contains twice as much as the other. You do not know which box has the double prize. The game works in 2 phases:

1. Choose any of the boxes you want.
2. Look inside the box. At this point you can decide to keep the contents, or switch to the other box.

So imagine that you've chosen a box, and it contains \$100. From here, you can calculate the "expected value" of the other box to be $0.5 \times \$50 + 0.5 \times \$200 = \$125$ and therefore decide to switch.

But then it follows that you would have made the same decision for any value $x$ that you would have found in the first box! So then why not just pick the other box in the first place?

In other words, the strategy of "pick a box at random, and then switch, no matter what" is equivalent to "pick a candidate box at random, and then pick the other box, and keep it", which is also equivalent to "pick a box at random, and keep it". Which means that switching is the same as not switching.

But this seems like a paradox, because we just calculated that switching the box after your initial choice increases your expected winnings by a factor of 1.25!

Answer 1

Here is how you can view this in order to make it right for you:

• The total amount of money in both boxes is $3X$ dollars

• One box contains $X$ dollars and the other box contains $2X$ dollars

• Now you pick a box, and you're thinking "maybe I should pick the other box":

  • There is a $50$% chance that the other box contains $\frac{1}{3}$ of the amount
  • There is a $50$% chance that the other box contains $\frac{2}{3}$ of the amount

• So the expected amount of money that you'll get by picking the other box is $\frac{1}{2}\cdot\frac{1}{3}+\frac{1}{2}\cdot\frac{2}{3}=\frac{1}{2}$

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Note that $\frac{1}{2}$ of the amount of money is the expected (average) portion that you win.

In essence, you will win either $\frac{1}{3}$ of the amount of money or $\frac{2}{3}$ of the amount of money.

Answer 2

This simply because average in terms of probabilty doesn't work. Like if you use an averrage by getting the square root of the product of 200 and 50 you get 31.6... Using average in probabilty only works if you do it like 'barak manos' (right below this post).

Answer 3

If you give a bank $\$100$ the teller will flip a fair coin with sides labelled $\$50$ and $\$200$, and give you the result. Do you do this or keep your money?

This is how you're viewing the game show problem.

But the problem is more the case that: the teller has a coin with a side marked $\$100$ turned face up, and the bank guarantees that the other side is either labelled $\$50$ or $\$200$. If you give the bank $\$100$ the teller will turn that coin over and give you the result.

It's either one coin or the other, but how do you assess the probability of which?

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Just looking into the box and finding $\$100$ will only tell you that the game show divided up a total either $\$150$ or $\$300$ into the two boxes.   It won't tell you the probability of which.   You can't determine the expected return of switching boxes.

To find: $\operatorname{E}(Y\mid X=100)$

You need: $\operatorname{E}(Y\mid X=100) = 50 \Pr(Y=50\mid X=100) + 200 \Pr(Y=200 \mid X=100)$

You only know: $ \Pr(X=100, Y=50 \mid X+Y=150)=\tfrac 12, \\ \Pr(X=50, Y=100 \mid X+Y=150)=\tfrac 12, \\ \Pr(X=100, Y=200 \mid X+Y=300)=\tfrac 12, \\ \Pr(X=200, Y=100 \mid X+Y=300)=\tfrac 12$

You cannot get to there from that.

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You had an equal ($50\%$) chance of picking the box containing either the smaller or larger amount. You picked the box containing $\$100$. This does not necessarily mean that there is an equal chance that $\$100$ was the smaller or larger amount. That depends entirely on how the game show selects the amount to distribute.

$$\Pr(X=z \mid X+Y=3z) \not\equiv \Pr(X+Y=3z \mid X=z)$$

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@Bhoot No. You know, for instance: $\begin{align}\Pr(Y=50\mid X=100) & = \Pr(X+Y=150\mid X=100) \\ & = \tfrac{\Pr(X=100\mid X+Y=150) \Pr(X+Y=150)}{\Pr(X=100\mid X+Y=150) \Pr(X+Y=150) + \Pr(X=100\mid X+Y=300) \Pr(X+Y=300) } \\ & = \frac{\tfrac 12 \Pr(X+Y=150)}{\tfrac 12 \Pr(X+Y=150) +\tfrac 12 \Pr(X+Y=300)} \\ & = \frac{\Pr(X+Y=150)}{\Pr(X+Y=150) + \Pr(X+Y=300)} \end{align}$

However, you don't know what $\Pr(X+Y=150)$ or $\Pr(X+Y=300)$ are.

The game show could choose the total amount in any way (and they are not restricted to using those two amounts). All you know is how the amount is divided and one of the divisions.

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Basically, there are two independent random variables to this problem. $A$ the total amount, and $B$ the proportion of that amount in the chosen box.

We know that $B$ has a Bernoulii distribution. $(3B-1)\sim{\cal B}(1, \tfrac 12)$

Thus we can assess the expectation of the proportions of the amount in the boxes.

$$\Pr(B=b)=\begin{cases}\tfrac 12 & b\in\{\tfrac 13, \tfrac 23\}\\ 0 & \text{else} \end{cases} \\\begin{align}\operatorname{E}(B) & =\frac 13\frac 12+\frac 23\frac 12 \\ & = \tfrac 12\end{align} $$

However, we know nothing of the distribution of $A$. So the condition $AB=\$100$ cannot inform us of the state of either $A$ or $B$.

$$\begin{align} \Pr(B=b \mid AB=100) & = \frac{\Pr(B=b \cap AB=100)}{\Pr(AB=100)} \\ & = \frac{ \Pr(B=b)\Pr(A=\tfrac{100}{b}) }{ \Pr(B=\tfrac 13)\Pr(A=300)+\Pr(B=\tfrac 23)\Pr(A=150)} \\ & = \frac{\Pr(A=\tfrac{100}b)}{\Pr(A=300)+\Pr(A=150)} \\ \text{Likewise:} \\ \Pr(A=a \mid AB=100) = & \frac{\Pr(B=\tfrac{100}a)}{\Pr(A=300)+\Pr(A=150)} & a\in \{150, 300\} \end{align}$$