1.1 Some Puzzles and Problems

The second-ace puzzle A deck has four cards: the ace and deuce of hearts, and the ace and deuce of spades. After a fair shuffle of the deck, two cards are dealt to Alice. It is easy to see that, at this point, there is a probability of 1/6 that Alice has both aces, a probability of 5/6 that Alice has at least one ace, a probability of 1/2 that Alice has the ace of spades, and a probability of 1/2 that Alice has the ace of hearts: of the six possible deals of two cards out of four, Alice has both aces in one of them, at least one ace in five of them, the ace of hearts in three of them, and the ace of spades in three of them. (For readers unfamiliar with probability, there is an introduction in Chapter 2.)

Alice then says, "I have an ace." Conditioning on this information (by discarding the possibility that Alice was dealt no aces), Bob computes the probability that Alice holds both aces to be 1/5. This seems reasonable. The probability, according to Bob, of Alice having two aces goes up if he learns that she has an ace. Next, Alice says, "I have the ace of spades." Conditioning on this new information, Bob now computes the probability that Alice holds both aces to be 1/3. Of the three deals in which Alice holds the ace of spades, she holds both aces in one of them. As a result of learning not only that Alice holds at least one ace, but that the ace is actually the ace of spades, the conditional probability that Alice holds both aces goes up from 1/5 to 1/3. But suppose that Alice had instead said, "I have the ace of hearts." It seems that a similar argument again shows that the conditional probability that Alice holds both aces is 1/3.

Is this reasonable? When Bob learns that Alice has an ace, he knows that she must have either the ace of hearts or the ace of spades. Why should finding out which particular ace it is raise the conditional probability of Alice having two aces? Put another way, if this probability goes up from 1/5 to 1/3 whichever ace Alice says she has, and Bob knows that she has an ace, then why isn't it 1/3 all along?

The Monty Hall puzzle The Monty Hall puzzle is very similar to the second-ace puzzle. Suppose that you're on a game show and given a choice of three doors. Behind one is a car; behind the others are goats. You pick door 1. Before opening door 1, host Monty Hall (who knows what is behind each door) opens door 3, which has a goat. He then asks you if you still want to take what's behind door 1, or to take instead what's behind door 2. Should you switch? Assuming that, initially, the car was equally likely to be behind each of the doors, naive conditioning suggests that, given that it is not behind door 3, it is equally likely to be behind door 1 and door 2, so there is no reason to switch. On the other hand, the car is equally likely to be behind each of the doors. If it is behind door 1, then you clearly should not switch; but if it is not behind door 1, then it must be behind door 2 (since it is obviously not behind door 3), and you should switch to door 2. Since the probability that it is behind door 1 is 1/3, it seems that, with probability 2/3, you should switch. But if this reasoning is correct, then why exactly is the original argument incorrect?

The second-ace puzzle and the Monty Hall puzzle are the stuff of puzzle books. Nevertheless, understanding exactly why naive conditioning does not give reasonable answers in these cases turns out to have deep implications, not just for puzzles, but for important statistical problems.

The two-coin problem Suppose that Alice has two coins. One of them is fair, and so has equal likelihood of landing heads and tails. The other is biased, and is twice as likely to land heads as to land tails. Alice chooses one of her coins (assume she can tell them apart by their weight and feel) and is about to toss it. Bob knows that one coin is fair and the other is twice as likely to land heads as tails. He does not know which coin Alice has chosen, nor is he given a probability that the fair coin is chosen. What is the probability, according to Bob, that the outcome of the coin toss will be heads? What is the probability according to Alice? (Both of these probabilities are for the situation before the coin is tossed.)

A coin with unknown bias Again, suppose that Alice has two coins, one fair and the other with a probability 2/3 of landing heads. Suppose also that Bob can choose which coin Alice will flip, and he knows which coin is fair and which is biased toward heads. He gets $1 if the coin lands heads and loses $1 if the coin lands tails. Clearly, in that case, Bob should choose the coin with a probability 2/3 of landing heads.

But now consider a variant of the story. Instead of knowing that the first coin is fair, Bob has no idea of its bias. What coin should he choose then? If Bob represents his uncertainty about the first coin using a single probability measure, then perhaps the best thing to do is to say that heads and tails are equally likely, and so each gets probability 1/2. But is this in fact the best thing to do? Is using a single probability measure to represent Bob's uncertainty even appropriate here?

Although this example may seem fanciful, it is an abstraction of what are called exploitation vs. exploration decisions, which arise frequently in practice, especially if it is possible to play the game repeatedly. Should Bob choose the first coin and try to find out something about its bias (this is exploration), or should Bob exploit the second coin, with known bias?

The one-coin problem Suppose instead that both Bob and Alice know that Alice is using the fair coin. Alice tosses the coin and looks at the outcome. What is the probability of heads (after the coin toss) according to Bob? One argument would say that the probability is still 1/2. After all, Bob hasn't learned anything about the outcome of the coin toss, so why should he change his valuation of the probability? On the other hand, runs the counterargument, once the coin has been tossed, does it really make sense to talk about the probability of heads? The coin has either landed heads or tails, so at best, Bob can say that the probability is either 0 or 1, but he doesn't know which.

A medical decision problem On a more serious note, consider a doctor who is examining a patient Eric. The doctor can see that Eric has jaundice, no temperature, and red hair. According to his medical textbook, 90 percent of people with jaundice have hepatitis and 80 percent of people with hepatitis have a temperature. This is all the information he has that is relevant to the problem. Should he proceed under the assumption that Eric has hepatitis?

There is clearly some ambiguity in the presentation of this problem (far more than, say, in the presentation of the second-ace puzzle). For example, there is no indication of what other options the doctor has. Even if this ambiguity is ignored, this problem raises a number of issues. An obvious one is how the doctor's statistical information should affect his beliefs regarding what to do. There are many others though. For example, what does it mean that the doctor has "no other relevant information"? Typically the doctor has a great deal of information, and part of the problem lies in deciding what is and is not relevant. Another issue is perhaps more pragmatic. How should the doctor's information be represented? If the doctor feels that the fact that Eric has red hair is irrelevant to the question of whether he has hepatitis, how should that be represented?

In many cases, there is no quantitative information, only qualitative information. For example, rather than knowing that 90 percent of people with jaundice have hepatitis and 80 percent of people with hepatitis have a temperature, the doctor may know only that people with jaundice typically have hepatitis, and people with hepatitis typically have a temperature. How does this affect things?