Solving the two envelopes problem

Sup­pose you are pre­sented with a game. You are given a red and a blue en­velope with some money in each. You are al­lowed to ask an in­de­pen­dent party to open both en­velopes, and tell you the ra­tio of blue:red amounts (but not the ac­tual amounts). If you do, the game mas­ter re­places the en­velopes, and the amounts in­side are cho­sen by him us­ing the same al­gorithm as be­fore.

You ask the in­de­pen­dent ob­server to check the amounts a mil­lion times, and find that half the time the ra­tio is 2 (blue has twice as much as red), and half the time it’s 0.5 (red has twice as much as blue). At this point, the game mas­ter dis­closes that in fact, the way he chooses the amounts math­e­mat­i­cally guaran­tees that these prob­a­bil­ities hold.

Which en­velope should you pick to max­i­mize your ex­pected wealth?

It may seem sur­pris­ing, but with this set-up, the game mas­ter can choose to make ei­ther red or blue have a higher ex­pected amount of money in it, or make the two the same. Ask­ing the in­de­pen­dent party as de­scribed above will not help you es­tab­lish which is which. This is the sur­pris­ing part and is, in my opinion, the crux of the two en­velopes prob­lem.

This is not quite how the two en­velopes prob­lem is usu­ally pre­sented, but this is the pre­sen­ta­tion I ar­rived at af­ter con­tem­plat­ing the origi­nal puz­zle. The origi­nal puz­zle pre­scribes a spe­cific strat­egy that the game mas­ter fol­lows, makes the en­velopes in­dis­t­in­guish­able, and pro­vides a para­dox­i­cal ar­gu­ment which is ob­vi­ously false, but it’s not so ob­vi­ous where it goes wrong.

Note that for sim­plic­ity, let’s as­sume that money is a real quan­tity and can be sub­di­vided in­definitely. This avoids the prob­lem of odd amounts like $1.03 not be­ing ex­actly di­visi­ble by two.

The flawed argument

The flawed ar­gu­ment goes as fol­lows. Let’s call the amount in the blue en­velope B, and in red R. You have con­firmed that half the time, B is equal to 2R, and half the time it’s R/​2. This is a fact. Surely then the ex­pected value of B is (2R * 50% + R/​2 * 50%), which sim­plifies to 1.25R. In other words, the blue en­velope has a higher ex­pected amount of money given the ev­i­dence we have.

But no­tice that the situ­a­tion is com­pletely sym­met­ric. From the in­for­ma­tion you have, it’s also ob­vi­ous that half the time, R is 2B, and half the time it’s B/​2. So by the same ar­gu­ment the ex­pected value of R is 1.25B. Uh-oh. The ex­pected value of both en­velopes is higher than the other?...

Game mas­ter strategies

Let’s muddy up the wa­ter a lit­tle by con­sid­er­ing the strate­gies the game mas­ter can use to pick the amounts for each en­velope.

Strat­egy 1

Pick an amount X be­tween $1 and $1000 ran­domly. Throw a fair die. If you get an odd num­ber, put X into the red en­velope and 2X into blue. Other­wise put X into blue and 2X into red.

Strat­egy 2

Pick an amount X be­tween $1 and $1000 ran­domly. Put this into the red en­velope. Throw a fair die. If you get an odd num­ber, put 2X into blue, and if it’s even, put X/​2 into blue.

The differ­ence be­tween these strate­gies is fairly sub­tle. I hope it’s suffi­ciently ob­vi­ous that the “ra­tio con­di­tion” (B = 2R half the time and R = 2B the other half) is true for both strate­gies. How­ever, sup­pose we have two peo­ple take part in this game, one always pick­ing the red en­velope and the other always pick­ing the blue en­velope. After a mil­lion rep­e­ti­tions of this game, with the first strat­egy, the two guys will have won al­most ex­actly the same amounts in to­tal. After a mil­lion rep­e­ti­tions with the sec­ond strat­egy, the to­tal amount won by the blue guy will be 25% higher than the to­tal amount won by the red guy!

Now ob­serve that strat­egy 2 can be triv­ially in­verted to favour the red en­velope in­stead of the blue one. The player can ask an in­de­pen­dent ob­server for ra­tios (as de­scribed in the in­tro­duc­tion) all he wants, but this in­for­ma­tion will not al­low him to dis­t­in­guish be­tween these three sce­nar­ios (strat­egy 1, strat­egy 2 and strat­egy 2 in­verted). It’s ob­vi­ously im­pos­si­ble to figure out which en­velope has a higher ex­pected win­nings from this in­for­ma­tion!

What’s go­ing on here?

I hope I’ve con­vinced you by now that the in­for­ma­tion about the like­li­hood of the ra­tios does not tell you which en­velope is bet­ter. But what ex­actly is the flaw in the origi­nal ar­gu­ment?

Let’s for­mal­ize the puz­zle a bit. We have two ran­dom vari­ables, R and B. We are per­mit­ted to ask some­one to sam­ple each one and com­pute the ra­tio of the sam­ples, r/​b, and dis­close it to us. Let’s define a ran­dom vari­able called RB whose sam­ples are pro­duced by sam­pling R and B and com­put­ing their ra­tio. We know that RB can take two val­ues, 2 and 0.5, with equal prob­a­bil­ity. Let’s also define BR, which is the op­po­site ra­tio: that of a sam­ple of B to a sam­ple of R. BR can also take two val­ues, 2 and 0.5, with equal prob­a­bil­ity.

The flawed ar­gu­ment is sim­ply that the ex­pected value of RB, E(RB), is 1.25, which is greater than 1, and there­fore E(R) > E(B). The flawed ar­gu­ment con­tinues that E(BR) is 1.25 too, there­fore E(B) > E(R), lead­ing to a con­tra­dic­tion. What’s the flaw?


The ex­pected value of RB, E(RB), re­ally is 1.25. The puz­zle gets that one right. E(BR) is also 1.25. The flaw in the ar­gu­ment is sim­ply that it as­sumes E(X/​Y) > 1 im­plies that E(X) > E(Y). This im­pli­ca­tion seems to hold in­tu­itively, but hu­man in­tu­ition is no­to­ri­ously bad at prob­a­bil­ities. It is easy to prove that this im­pli­ca­tion is false, by con­sid­er­ing a sim­ple counter-ex­am­ple cour­tesy of Vin­cen­tYu.

Con­sider two in­de­pen­dent ran­dom vari­ables, X and Y. X can take val­ues 20 and 60, while Y can take val­ues 2 and 100, both with equal prob­a­bil­ity. To calcu­late the ex­pected value of X/​Y, one can enu­mer­ate all pos­si­ble com­bi­na­tions, mul­ti­ply­ing each by its prob­a­bil­ity. The four pos­si­ble com­bi­na­tions of X and Y are 202, 20100, 602 and 60100. Each com­bi­na­tion is 25% likely. Hence E(X/​Y) is 10.2. This is greater than 1, so the if the im­pli­ca­tion were to hold, E(X) should be greater than E(Y). But E(X) is (20+60)/​2 = 40, while E(Y) is (2+100)/​2 = 51. Hence, the im­pli­ca­tion E(X/​Y) > 1 ⇒ E(X) > E(Y) does not hold in gen­eral.

So there you have it. The pro­posed ar­gu­ment re­lies on an im­pli­ca­tion which seems true in­tu­itively, but turns out to be false un­der scrutiny. Mys­tery solved?… Al­most.

Im­pre­cise lan­guage’s con­tri­bu­tion to the puzzle

The ar­gu­ment con­cern­ing the origi­nal, in­dis­t­in­guish­able en­velopes, is phrased like this: “(1) I de­note by A the amount in my se­lected en­velope. (2) The other en­velope may con­tain ei­ther 2A or A/​2, with a 50% prob­a­bil­ity each. (3) So the ex­pected value of the money in the other en­velope is 1.25A. (4) Hence, the other en­velope is ex­pected to have more dol­lars.”

Depend­ing on how pedan­tic you are, you might say that the state­ment made in the third sen­tence is strictly false, or that it is too am­bigu­ous to be strictly false, or that at least one in­ter­pre­ta­tion is true. The ex­pected value 1.25A is “of the amount of money con­tained in the other en­velope ex­pressed in terms of the amount of money in this en­velope”. It is not “of the amount of money in the other en­velope ex­pressed in dol­lars”. Hence the last sen­tence does not fol­low, and if the state­ments were made in full and with com­plete ac­cu­racy, the fact that it does not fol­low is a lit­tle bit more ob­vi­ous.

In clos­ing, I would say this puz­zle is hard be­cause “in terms of this en­velope” and “in terms of dol­lars” are typ­i­cally equiv­a­lent enough in ev­ery­day life, but when it comes to ex­pected val­ues, this equiv­alence breaks down rather counter-in­tu­itively.