# Simple Bayes rule question, how much to pay

I'm reviewing probability and this question is simple, not sure why I can't get the answer! How is this done?

There are 2 envelopes. One has 2 red beads, 2 black beads, and 1 dollar in it. The second has 1 red bead, 2 black beads, and 0 dollars in it. Suppose an envelope is selected at random and you are shown one bead, and it is black. How much should you pay for the envelope?

This seems like it could be solved with Bayes rule and expected values, but I'm trying to get Bayes rule stuck in my head, so I'm hoping for a solution with that :)

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I will assume that "we are shown one bead" means that we are shown one bead, chosen at random from the beads in the selected envelope.

Since intuition can sometimes lead us astray, we will use the formal conditional probability machinery. Let $W$ (for "win") be the event that the envelope contains $2$ red and $2$ black, and let $B$ be the event that the ball that was drawn is black. We want the conditional probability $P(W|B)$.

By the definition of conditional probability, we have $$P(W|B)=\frac {P(W\cap B)}{P(B)}.$$

We compute the two required probabilities on the right.

The probability of $W\cap B$ is $(1/2)(2/4)$. This is because we must pick the correct envelope (probability $1/2$) and draw one of the $2$ black beads from the $4$ beads in the envelope.

The probability of $B$ is $(1/2)(2/4)+(1/2)(2/3)$. This is because the event "we get a black bead" can happen in two disjoint ways: (i) We pick the envelope with $2$ black, $2$ red and draw a black bead or (ii) We pick the envelope with $2$ black, $1$ red and draw a black bead.

Divide and simplify. We get $3/7$.

The expectation of the amount of money in the envelope is therefore $(3/7)(1)+(4/7)(0)$, which is $3/7$.

As to how much one "should" pay for the envelope, the commonsense answer is as little as possible. But if we pay $3/7$ of a dollar for the envelope, then our expected total gain is $0$, making it a "fair" game. That is undoubtedly the expected answer.

A Somewhat Different Analysis: Let $a$ be the amount that we pay for the envelope, and let $X$ be our net gain.

If there is a dollar in the envelope, the net gain is $1-a$. If there is no money in the envelope, our net gain is $-a$. Thus $$E(X)=(1-a)(3/7)+(-a)(4/7).$$ Simplify. We get $$E(X)=\frac{3-7a}{7}.$$ This expectation is $0$ ("fair game") if $3-7a=0$, that is, if $a=3/7$.

Reality Check: The envelope with no money is (proportionally) richer in black than the envelope with money. So drawing a black bead means that there is a better than $1/2$ probability that we are dealing with the no money envelope. Thus the expected value of the money in the envelope is less than $50$ cents. And indeed $3/7$ is less than $1/2$. It is almost always a good idea to check whether a computed answer "makes sense."

Comment: If the bead we are shown is not chosen at random, the answer depends on the strategy that the opponent is using. For example, if the opponent always shows you a black bead, whatever envelope you choose, then the probability of winning a dollar is $1/2$, so a payment of $50$ cents makes the game fair.

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And it’s the expected answer that maximizes the expected value of one’s score. –  Brian M. Scott Aug 10 '11 at 1:36
@Brian M. Scott: When I wrote "expected answer," I wondered whether anyone would notice. –  André Nicolas Aug 10 '11 at 1:53
I suspected that it was deliberate. A nice touch to a nice answer. –  Brian M. Scott Aug 10 '11 at 1:56
+1. Great answer and complete with formal and intuition arguments. Like it very much! –  helios Aug 10 '11 at 7:16

When you say "at random", do you mean the two envelopes have equal probabilities of being chosen? If so then we can represent the prior distribution as $$(1/2, 1/2).$$ The likelihood function can be represented as $$(1/2,2/3)$$ (The probability of getting a black bead from the first envelope is $1/2$; from the second is $2/3$.)

Multiply the likelihood by the prior term-by-term: $$(1/2,1/2)\cdot(1/2,2/3) = (1/4,1/3),$$ then normalize to get a probability distribution, thus: $1/4 + 1/3 = 7/12$, multiply all entries by $12/7$: $$\frac{12}{7}\left(\frac14, \frac13\right) = \left(\frac37,\frac47\right).$$ That's the posterior probability distribution.

Bayes rule can be stated thus: Multiply the prior probbility distribution by the likelihood and then normalize, to get the posterior probability distribution.

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