David Blackwell, who died two weeks ago, was one of the first mainstream statisticians to “go Bayesian.” And for that and his unique skill in clearly explaining difficult ideas, we owe him plenty.
Blackwell handed in his slide rule and the grand age of 91. A good run!
He worked on cool problems. From his Times obituary, “His fascination with game theory, for example, prompted him to investigate the mathematics of bluffing and to develop a theory on the optimal moment for an advancing duelist to open fire.”
If that isn’t slick—and useful!—I don’t know what is. Of course it’s useful; because it doesn’t have to be two guys facing off with pistols, it can be two tank columns facing off with depleted uranium rounds.
One of the big reasons statisticians started the switch to Bayesian theory, or at least accorded it respect, is that it is aptly suited to decision theory, which Blackwell (with Girshick) explicated neatly in their to-be-read book Theory of Games and Statistical Decisions. I encourage you to buy this book: you can pick up a copy for as little as three bucks.
A classic decision analysis problem of the sort Blackwell examined is this.
St Petersburg Paradox
The estimable Daniel Bernoulli gave us this problem, one of the first creations of decision theory. You have to pay a certain amount of money to play the following game:
A pot starts out with one dollar. A coin is then tossed. If a head shows, then the amount in the pot is doubled. If a tail shows, the game is over and you win the pot, else the coin is re-flipped repeatedly until a tail appears. How much should you pay to play?
Suppose you pay ten bucks and the coin shows a tail the very first throw. You win the dollar in the pot, but it costs you a bundle. You won’t make any money unless a tail waits until at least the fifth throw.
The standard solution begins by introducing the idea of expected value. This is usually a misnomer, because the “expected” value is often one that you do not expect or is impossible. Its formal definition is this: the weighted sum of everything that can happen multiplied by the probability of everything that can happen.
For example, the expected value of a die roll is:
EV = 0.167*1 + 0.167*2 + 0.167*3 + 0.167*4 + 0.167*5 + 0.167*6 = 3.5,
where 0.167 is 1/6, the probability of seeing any result. This says we “expect” to see 3.5, which is impossible. The dodge we introduce is to turn the die roll into a game that can be played “indefinitely.” Suppose you win one dollar for every spot that shows. Then, for example, if a 5 shows you win $5.
If you were to play the die game “indefinitely” the average amount won per game would converge to 3.5, and seeing an average of 3.5 is certainly possible. For instance, you win $6 on the first roll and $1 on the second, for an average of $3.5 per roll. However, expected value is the average after you play a number of games that approaches infinity.
We can now apply expected value to the St Petersburg game:
EV = (1/2)*1 + (1/4)*2 + (1/8)*4 + … = infinity.
There’s a 1/2 chance of winning $1, a 1/2 * 1/2 = 1/4 chance of winning $2 (we see a Tail on the second throw), a 1/2 * 1/2 * 1/2 = 1/8 chance of winning $4 (we see a tail on the third throw), and so on. Those of you who have had a “pre-calculus” course will quickly see that this sum approaches infinity.
Yes, that’s right. The “expected” amount you win is infinite. Therefore, this being true, you should be willing to pay any finite sum to play! If you’re convinced, please email me your credit card number and we’ll have a go.
The classical solution to this “paradox” is to assume that your valuation of money is different than its face value. For example, if you already have a million, adding 10 bucks is trivial. But if you have nothing, then 10 bucks is the difference between eating and going hungry. Thus, the more you have, the less more is worth.
Through calculus, you can use a down-weighted money function to give less value to absurdly high possibilities in the St Petersburg game. So, instead of winning $2100 (a million billion billion) if a tail doesn’t show until the 100th toss, and which has the chance of 1/2100 = 8 x 10-31, you say that amount is worth only a vanishingly fractional amount.
Whatever down-weighting function is used (usually some form of log(money)), calculus can supply the result, which is that the expected value becomes finite. The results are usually in the single-dollars range; that is, the calculus typically shows the expected value to be anywhere from $2 to $10, which is the amount you should be willing to pay.
The real solution is to assume what is true: the amount of money is not infinite! Using only physically realizable finite banks, we know the pot can never exceed some fixed amount.
If that amount is, say, $1 billion, then the number of flips can never exceed 30. The expected value, ignoring down-weighting, of 30 flips is only $30 * (1/2) = $15. And we can, if we like, even include the down-weighting! (Even $1 trillion gives only a max 40 tosses with expected value $20!)
Thus, the St Petersburg “paradox”, like all paradoxes, never was. It was a figment of our creation, a puzzle concocted with premises we knew were false.
More on finitism, or mathematical constructivism here.