# Newcomb’s Paradox And Probability

This article was suggested by reader JH who saw it discussed at Massimo Pigliucci’s site.

When he was at the Lawrence Livermore Laboratory, William Newcomb devised the following puzzle, which some say is a paradox.

You will play a game against an evil entity that can perfectly predict your future actions. Just for the sake of giving the Evil One a name, let’s call him “Bill”. Bill is evil because he likes to taunt you with the possibility of making money, only to jerk that chance away from you at the last moment.

Here’s what Bill does, as traditionally described. Later, we add clarifications and twists. Bill puts \$1,000 in a clear box, and then also presents you with an opaque box inside of which might contain \$1 million. You will be allowed to take both boxes or just the opaque one. To clarify: you may not take the \$1,000 without also taking the opaque box. Obviously, you may keep the contents of whatever box or boxes you take.

Now for the evil part. If Bill predicts—and remember, he’s never wrong—that you will take just the opaque box, he will leave it full of money. But if he predicts that you will greedily take both boxes, then he will put nada in the opaque box, which, after mandatory donations to our betters in Washington, nets you only about \$550.

Why Bill would ever want to play this strange game is never mentioned. Anyway, the easiest way to short his pockets by a cool million is to decide now and forevermore to take just the opaque box. Because he’s never wrong, Bill will figure you’d do just this, and thus you’d walk away to find new messages on your answering machine from your brother-in-law asking if he could drop by for a chat about this new fishing boat he’s had his eye on.

By the premises of the game, there is no way you can out-think old Bill and collect the million and the thousand simultaneously. You cannot, for example, declare to the world, “I’m only taking the opaque box!” and then, at the last moment, grab both, because old Bill will have anticipated this. Nor can you attempt a crash course in hypnosis to convince yourself that you’re only taking the opaque box, only to be awakened to discover yourself picking up both. Bill would have predicted this perfectly, too.

Not much of a paradox, really. (Though there’s always someone who thinks he’s discovered a way to cheat Bill). Using just brain power and no external devices, the best you can do is a million—which would be good enough for anybody but a politician. So, in honor of our money-thirsty betters, here’s what they can do when next confronted by Bill and his boxes.

When Bill lays out his choices, take out a coin and flip it. If it comes up Heads, take both boxes; tails, take just the opaque one. As long as you are not the sort of politician who would cheat, then your actions cannot be predicted ahead of time by Bill. Since there is only a 50% chance of you taking both boxes, this is what Bill must predict.

He must, of course, come to a decision and not continuously fly between the horns of the dilemma. How Bill does this, how he, that is, comes to a definite prediction is his business, but come to one he must if he is to play his weird game.

There are problems with this. Coin flips are only “random” because it is difficult to know what the initial conditions of the flip are. But if you did know them, then the coin flip is perfectly predictable. And since it’s you who will flip the coin, and old Bill has the lock on your mental actions, we could say that he knows how you will flip the coin. Which, of course, means he knows if you’ll take one box or two.

Regardless whether this is so—you can always have a pal flip the coin, removing Bill’s prognosticative abilities, because the rules do not say Bill can predict anybody’s actions, just yours—predicating your actions on a coin flip can actually be worse for you than just picking the opaque box, because you run the chance of ending up with only \$1,000. This happens when the coin lands Heads and this is what Bill predicted. The chance of this—assuming Bill follows the probabilities—is 0.5 x 0.5 = 0.25.

The chance Bill predicts Heads and you Tails is the same: meaning you will decide to take just the opaque box, but you’ll come away broken-hearted. The chance that Bill predicts Tails and you Heads is also the same, meaning there’s a 25% shot at making \$1,001,000. Finally, the chance that Bill and you both go Tails is the same, meaning a 25% chance of scoring a million.

Homework: use a quantum number-maker-upper (QNMU), a device whose outcome Bill cannot prefigure, and which spits out Heads with probability p. You can either just take the opaque box, or use the QNMU to decide what to do. What strategy is best and why? If you use the QNMU, what value of p are you picking, and why?

### 26 Comments

1. TomVonk

William

I didn’t know this one but there must be something that I missed or misunderstood .
I see neither paradox nor dilemma .
Where is the problem ?
As the game assumes that you are interested in money , you always take the closed box only . Seems trivial .
You get 1 M and walk happily away unless the suker wants to replay in which case you take again the closed box and have 2 M .
What am I missing ?
Is it that you are supposed to play the game without knowing the rules ?
Why would anybody even THINK of flipping a coin or its quantum mechanical equivalent when he can get 1 M with a probability of 1 ?

2. Briggs

Greed, Tom. Simple greed. Or maybe your mother needs an operation that costs \$1,001,000, and one million just won’t do.

On the other hand, I have phrased the problem dryly. Professional philosophers are better at making the simple complex, of turning what seemed reasonable into that which must be bizarre. Russell boasted of this, and even said that this the definition of philosophy.

3. mt

Always take just the opaque box for the 1M. If you use some device that Bill can’t predict to make calls, Bill can predict that you think you can do better than \$1000, and therefore would always leave the opaque box empty.

4. mt is close to it, but not quite. Yes, you should pick the opaque box unconditionally, but not because Bill might cheat. That’s not within the parameters of the game.
The reason you should pick the Opaque is that the expected value of using the QNMU ultimately boils down to 1000000 – 999000p. There’s no positive value of p which can give you an expected payoff better than 1 million, so just set p=0 (meaning you definitely pick the opaque box), and collect \$1 million, or save the trouble, pick the opaque box to begin with and don’t even touch the QNMU.

5. Doug M

Wrong, wrong, wrong.

You are supposed to take both boxes. Bill has placed the money in the boxes before you make your choice. There is nothing he can do once the puzzle has been presented to you, and nothing to be lost by taking both boxes.

That is the paradox.

6. SteveBrooklineMA

I don’t really get this either. Since Bill is “evil,” wouldn’t he want you to get as little money as possible? You are guaranteed to get at least \$1000 any way you play it. Bill can guarantee you get no more than \$1000 by leaving the opaque box empty. Isn’t that what an evil guy would do?

But the game says if he knows you will choose the opaque box only, he will put \$1,000,000 in there. Doesn’t seem evil to me at all. In fact, since I know he will do this, he’s given me a way to guarantee I get \$1,000,000. Thanks Bill!

7. mt

Doug: Bill knows what you are going choose before putting the money in the opaque box or not. So if your decision will be to take both, the opaque box will be empty when presented. If your decision will be to take only the opaque, it will have the 1M. Inserting randomness is supposed to defeat Bill’s predictive ability, and therefore give a chance at getting the 1M + 1K.

Tony: I don’t know about the full expected value calculation, but if Bill has a 50% chance of taking away the 1M no matter what you choose, the 1K isn’t going to make up for it.

8. Doug M

The more intesting discussion is not wheter to pick one box or two, but what is Bill? What is a highly accurate predictor? Is Bill ominicient? If Bill knows what you are going to do before you do, do you really have a choice in the matter? And if such a creature really exists, we are just rehashing Calvin’s arguements of Fate vs. Freewill.

9. SteveBrooklineMA

After reading the post at the blog Briggs links to, which uses a computer instead of Bill, I thought some more:

In the problem you describe, you assume a near perfect predictor (PP). This leads to a contradiction. Thus no such PP is possible. It’s not a paradox. It’s proof by contradiction, a standard math technique.

To see this non-existence of a PP more directly, suppose you choose by flipping a coin. It’s then clearly impossible for a computer program to nearly perfectly predict the outcome in advance. This is practically the definition of coin flip.

Note that the setup of the problem describes a computer that can nearly perfectly predict your choice. This is quite different from a computer
that correctly predicts you will choose based on a flip of a coin. The latter might be possible, but the former isn’t. If a person’s strategy is to choose based on the flip of a coin, then no computer program can predict the choice at better than 50%.

10. Jim Fedako

You cannot forget the premise that Bill is never wrong. Never wrong. Flipping a coin cannot violate that premise â€“ nothing can. So Bill MUST know the outcome of the coin flip (or any other random generator). No paradox allowed here. No probability either.

11. mt

Of course, the real solution is to put the clear box inside the opaque box and then take only the opaque box.

12. @Doug – The point isn’t what Bill has already done at the time of your decision. It is what he expects you to do that makes the difference. If I were Bill, and you were the player, I would know to withhold the money because I know that you would always take the other box. Likewise, if you were Bill and I were the player, you would leave the money because you know I would leave the other box.

13. Doug M

The point is exactly what Bill has done at the time of your decision. And how he think you think he thinks you will behave. And now that Bill has acted, what is ‘rational’ to maximize your payoff. Now, you have a problem that there are 2 boxes in front of you. One has \$1000, one may have \$1 million. You can take one box or you can take both boxes.

As the problem is traditionally stated Bill is a ‘highly accurate’ predictor. Which gets into the debate of what that means. If we assume that Bill makes the occaisional error, does that change the analysis?

14. Sander van der Wal

Who cares about the 1000 dollar. Put the million in the bank and collect the interest.

15. Of course if we assume Bill makes errors, it changes the calculations. But the expected value calculation is one step ahead of you: it accounts both for the probabilities and payoffs of you cheating Bill, Bill cheating you, both of you playing fair, and both of you cheating. That’s why it generates the strictly-dominant, pure strategy to take the \$1mil, because acting with any other intention can never exceed that expected value (despite the fact that it may, with some probability, decrease your maximum possible payoff).

You raise an interesting point, in that Bill has no options once you’re in the room, but you forget that he knows what you’re planning before you get there. He knows what you set p to. And by selecting to take both unconditionally, you set p to 1, thereby signaling to Bill that he MUST withhold the million.

Now, if you want to investigate exogenous factors (like Bill making a judgment error about you), you’re changing the parameters of the game. And while that may mean your solution is correct, it does not mean your solution generalizes to the as it was game presented to us. We cannot bend reality to our will, but we can make choices to optimize our expected outcomes.

16. TomVonk

William
On the other hand, I have phrased the problem dryly. Professional philosophers are better at making the simple complex, of turning what seemed reasonable into that which must be bizarre.
Yes that makes sense .
But in this particular case it fails for me . I still can’t see anything complex or bizarre .

Case 1 – “common sense” case
You take closed box , get 1 M , invest at 5% interest , get 50 k more .

Case 2 – “fuk Bill” case
You prevent him to know your choice by using an external process that we know as not being deterministically predictable . Any chaotic process will do (flipping a coin is an example) .
It is a demonstrated fact that chaotic processes are ABSOLUTELY unpredictable because the uncertainty on initial condition can’t be 0 (Heisenberg uncertainty relation) .
Even if Bill could perfectly predict my every single action , he couldn’t predict a chaotic system .
At least that’s how it works in our Universe .

What will Bill do ?
Well nothing because the rules of the game became undefined . Bill doesn’t know now if he should put 1 M in the closed box or not .
But the principle of the game is based on the fact that there is or is not money in the closed box .
So as Bill is caught in a logical impossibility to decide what he should do according to the rules of the game , the only logically consistent action left to him is to say “DIVIDE BY 0! ERROR IN 2545 . IMPOSSIBLE TO PLAY THE GAME . PLEASE PRESS RESET IF YOU WANT TO TRY AGAIN .” No complexity there .

P.S
Of course Bill could also in this case decide that as your decision became unpredictable , his decision will become unpredictable too . But that is a COMPLETELY different game . And a very trivial one at that . You flip a coin . He throws a die . Or not . Or he decides to never put money in the closed box untill he changes his mind . Etc . Bill’s divine predictive powers became irrelevant . No complexity here either .

17. Rich

Stripped of the paraphernalia, Bill offers you \$1,000,000 or \$1,000. You choose. Where’s the game or the paradox?

18. @Rich – Too oversimplified. The thing that makes the game interesting is that it is possible to get \$1,001,000, but dangerous to try.

19. @Tom – The game doesn’t become undefined just because we introduce randomness. The Universe doesn’t collapse in on us whenever we roll a die. Briggs has just adapted the pure form of the game to a game of vNM preferences. The idea of the stochastic choice (flip the coin, use the QNMU) simply denies Bill the comfort he had in predicting our behavior. Therefore, Bill adapts by following our example, as you mention in your postscript. I wouldn’t necessarily call the solution to that game trivial, though. You want to make this random chance coin flip? Fine, but now there’s 50-50 odds the mil will be there. And the point is that it is irrational to gamble for the \$1000. There’s no probability that will increase the expected payoff, so there’s nothing to gain from using the QNMU. I had to do two pages of math to prove it–not a difficult problem, certainly, but hardly a trivial one.

20. TomVonk

Tony

My point was just that it is ANOTHER game . The supposedly “perturbing” all knowing property of Bill becomes irrelevant . So we can as well forget it .
The problem is then trivial : A takes an action with probability Pa and B takes another action with probability Pb .
The logics are trivial , the maths to be done may be a bit lengthy but not difficult .
In any case there is nothing bizarre or complex .
Nothing to do with the initial game either – that’s what I meant with “undefined” .
The rule of the game as it is explicitely formulated here says : Bill puts or puts not money in the box function of his prediction of my action .
It is THIS rule that becomes undefined when Bill can’t make any prediction of my action .
So if we change THIS rule (in an infinity of possible manners) we obtain a completely different game .

21. Rich

Tony, there is no possibility of getting \$1,001,000. As formulated above, if you take both boxes you only get \$1,000. “If Bill predicts – and remember he is never wrong …”. If you take both boxes the opaque will be empty. Always.

Nor, in the game as described, is Bill’s prediction dependent on your own knowledge of your choice. He just knows. Always. So coin-tossing is irrelevant.

So you can take one box or both. In the first case you get \$1,000,000; in the second you get \$1,000. There is no choice that will give you \$1,001,000.

22. Rich – There is a possibility that you get \$1,001,000. If you commit to taking both, you’ll get \$1000. And if you commit to taking the opaque, you get \$1,000,000. But that’s why the randomness is introduced. Here’s the important distinction: Bill /can’t/ predict a random outcome the way he can predict your (sort of) deterministic behavior. So he has to guess, based on the probability you select.

The coin is the easiest probability distribution, so let’s use that as a basis for an example. You’ll get heads (take both) with a .5 probability. Remember, Bill’s optimal strategy is remove the million with the same probability that you take both, so he should flip a coin too. So there are four possible outcomes:

You heads, Bill heads – You get \$1000
You tails, Bill heads – You get nothing.
You tails, Bill tails – You get \$1,000,000
You heads, Bill tails – You get \$1,001,000

Each with probability .25. It’s totally possible.

There’s an important narrative to go along with this. Bill can’t remove the million once you’re in the room. He must make the decision before you get there, so you can’t see what follows. He knows what you will do, but if you commit to following the direction of a random process, knowing what you will do is not enough for him to definitively establish what his decision should be. He would need to know the outcome of a random coin flip (or some random process), and no matter how good he is at reading people, his odds of reading a coin toss are always going to be .5.

Also, Tom – Fair enough. My comments are directed specifically at the game in the question Briggs poses at the end of the post, not the pure form of the game.

Final Game theoretic observation – I think this resembles the Prisoner’s Dilemma (despite the information asymmetries) in that the Nash equilibrium is highly suboptimal. Doug is right in that a player has nothing to lose by taking both boxes once he is in the room. Therefore, I propose most players will take both boxes when actually put in this situation. (Nash equilibrium is (Remove, take Both)). However, this will lead players to the heartbreaking realization that for their greed, they have lost \$1,000,000. The optimal strategy is to leave the clear box, but clearly the math to back that notion up does not impress. I wonder what would?

23. If anyone is still reading this thread, I was curious about how a lot of people would respond, so I started a survey. It’s only 4 questions; Help me out with this little research project!

24. JJD

My personal choice would be to take the million bucks if I could, since the additional satisfaction I might get from an extra thousand or so would not be worth complicating the game. As Matt suggests though, maybe you have to pay tax on the million bucks or something, so you need more than the million and you must play the game. Or “kill Bill” (sorry) and rob him, if he is a wee defenseless but infinitely rich mortal creature.

If you decide to use the QNMU, Bill knows your chosen value of p, so if you can figure out Bill’s optimum strategy under the circumstances (building on Tony’s example with p=0.5) and thus his expected cost of playing, that should tell you your expected winnings as a function of p. Maybe then you can figure out what p maximizes that function. But I am going to take my million bucks and go to bed. Good luck with the solution. No pun intended.

25. JJD

Following up on my previous post, there is an information-theoretic way for Bill to calculate his best strategy and expected cost if he has perfect advance knowledge of the outcome of every press of the QNMU button and can stuff the boxes accordingly. The method is also applicable even if Bill has probability q > 0 of being wrong about the QNMU output in each case. You will still do one thing or another in each case based on the QNMU output, so the expected value of the game with omniscient Bill (or even Bill with partial inside knowledge) is a simple function of p (or p and q). It should be not too difficult to figure out whether there is a value of p that could yield an average of over a million dollars per play in the long run. I suspect that practically any nonzero probability of losing the million dollars would end up reducing your expected winnings to below a million. Apparently Tony has worked out the details, so he knows for sure.