# The Two Envelope Paradox

Created | Updated Mar 13, 2007

The Two Envelope Paradox is a puzzle in which an argument based on probability seems to conflict with common sense. Unlike some other such puzzles, this time common sense wins! Many discussions of the paradox can be found on the Web, but they tend to quickly shift into mathematical equations. This Entry discusses the paradox in plain English as far as possible.

The Set-up

In the Two Envelope Paradox, a stranger offers you a chance to play a game and win some money. He offers you a choice of two envelopes containing sums of money, and says one contains twice as much as the other. After you pick one envelope, you may open it.

'Now', he says, 'you can keep that money, or swap it for the other envelope'.

What should you do?

The Common Sense Answer

The common sense answer is that since you have two identical envelopes, there is no way of knowing which contains more money, and one guess is as good as another. Keep the one you've got or swap, it's a toss-up.

An Argument Based on Probabilities

Call the envelope you picked envelope A and the other one B. Suppose A contains £20. If you stick with A, you keep that £20. The rules tell us that B must contain £10 or £40. If A was the high envelope and you swap , you end up with £10, or £10 less than if you stick. If A was the low envelope and you swap, you end up with £40, £20 more than if you stick.

The amount you gain by swapping when A is low (£20) is double the difference between sticking and swapping when A is high (£10), and since there's nothing else to decide between the envelopes, swapping is a good bet.

The stranger hasn't offered to play more than once yet, but if he were prepared to play all day, this argument suggests that by always swapping you would gain and lose equally often, but gain twice as much as you lost on average. This would mean swapping will win 1.25 times as much as sticking, on average.

So even given just one chance at the game, the odds say you win more on average by swapping, so you should swap, right?

Not So Fast!

This argument has some strange implications.

At first glance it suggests that given a random choice between two identical envelopes, you can beat the odds and do better than a random choice. Someone picking randomly should win the average of all the amounts offered, but according to this argument, switching lets you win more money than you lose. How could that be?

A second oddity is that if you add a second player who always chooses the other envelope, he can apply exactly the same argument to the amount in his envelope, and he swaps when you do. According to the argument he should also gain, but he opens the envelopes you reject: if one of you does better, surely the other must do worse than the average?

Worst of all, notice that it doesn't matter how much is in envelope A, the argument implies that you will always gain by switching. Knowing this, you should switch envelopes before naming your choice. Now you choose A secretly, and switch to B. Say it contains £40, and the stranger now asks if you want to switch.

Oh-oh, the argument says A now contains £20 or £80, and you gain by switching back. Next time, you switch twice in your head before picking A, yet the argument still tells you to switch to B. No matter which envelope you are about to pick, the argument says you will gain by picking the other! How can you stop swapping and choose one?

There is no correct choice. It's a paradox!

An Alternative Answer Using Simple Logic Without Probability

Call the envelopes A and B, containing £x and £2x. If you pick A and stick, you win £x. If you pick B and stick you win £2x. If you pick A and swap you win £2x. If you pick B and swap you win only £x.

This table shows the possible outcomes:

Pick | Stick | Swap |
---|---|---|

A | £x | £2x |

B | £2x | £x |

If you always stick, you win £x or £2x. If you always swap you win £x or £2x. The situation is symmetrical, and there is no benefit to switching compared to sticking.

For any particular pair of envelopes, you do not gain more by switching when you have chosen the low envelope than the difference between sticking and swapping when you choose the high envelope, the differences when swapping are equal, and equal to the difference between the two possible outcomes when you always stick.

If the x's are confusing you, try it with any real amounts. For example, if A contains £10 and B contains £20, picking A and sticking wins £10, picking A and swapping wins £20. Picking B and sticking wins £20, picking B and swapping wins £10. The amount you gain when swapping and gaining (£10) is the same as the the difference between sticking and swapping when you choose the high envelope (also £10) so you cannot win more by swapping and gaining than you forego when you swap to the smaller envelope.

Since this is true with any particular pair of envelopes, it is clear that repeating the game with any other particular pair of envelopes will give the same result. Swapping every time gives the same result as sticking every time in the long run, the results are perfectly symmetrical.

This argument shows that the original common sense answer was correct: there is no way to choose between the envelopes other than random chance, sticking or swapping give equal chances of winning or losing the same amounts of money.

What's Wrong With the Probability Argument?

That probability-based argument still looks good, doesn't it? Yet it disagrees with both common sense and the simpler logical answer and it implies weird results, finally contradicting itself and falling into a paradoxical infinite loop. So what's wrong with it?

The answer is that, as stated above, it contains a faulty hidden assumption. This part:

'The amount you gain by swapping when A is low (£20) is double the difference between sticking and swapping when A is high (£10), and since there's nothing else to decide between the envelopes, swapping is a good bet..'

contains the hidden assumption that no matter what amount of money is in the envelope you choose, the other envelope is equally likely to be half or double that amount. This may not look wrong even when stated explicitly, but it is.

In the general set-up outlined at the start of the entry, the stranger doesn't say that the game is limited, but even if he is extremely rich, he cannot give away an infinite amount of money. Players are not told what that limit is, but is still affects the game. In fact it affects the odds in such just such a way that it makes this assumption wrong and balances the game, removing the advantage of swapping every time.

Suppose the stranger limits the maximum amount in any envelope to £100. Now if you pick an envelope and it has more than £50 in it, it cannot be the small envelope, as the large one would have to be more than £100.

The actual probability of gaining by switching in this case is zero, but the faulty argument says it's 0.5 (or 50%, or even odds, or a toss-up, in other words). The argument tells you to always switch, but when your first pick is more than half the limit, you always lose. Losses like this balance out gains made by switching when your first pick

happens to be less than half the limit.

A Demonstration

One clear way to look at the probabilities is to write out a table of possible outcomes in a particular game. This will show how the faulty assumption affects the odds. To keep the table small, let's suppose that the stranger limits the game to paper money amounts up to £40. For each pair of envelopes in the table, the first of the pair is the amount in the envelope the player picks, and the gain the player makes by swapping is shown.

(5,10) | +5 |

(10,5) | -5 |

(10,20) | +10 |

(15,30) | +15 |

(20,10) | -10 |

(20,40) | +20 |

(30,15) | -15 |

(40,20) | -20 |

Adding all the gains, the result is zero. It balances exactly. There is no benefit to swapping at all, on average.

The error in the paradoxical argument is to say that when the player chooses 30 or 40, they can still win by switching, but they can't. For 30 and 40, the probability that the other envelope contains 60 or 80 is precisely zero. The player always loses, and loses a lot. The faulty argument assumes that the pairs:

(30,60) | +30 |

(40,80) | +40 |

can be considered, but since those amounts are higher than the limit, those pairs aren't in the game.

No matter how the limit is, it affects the game in the same way as in this demonstration. Whenever a player picks an envelope containing an amount which is more than half the limit, they cannot win by switching, and their losses when switching balance gains they may make when they pick an envelope with less than half the limit.

Conclusion

Common sense does not always handle questions of probability well (for an example see the Entry on The Monty Hall Problem). The Two Envelope Paradox demonstrates that a plausible-sounding argument based on probabilities can be faulty in ways which are not obvious, and that common sense can sometimes win against bad odds.