# The Two Envelope Game Show

This is an original riddle I made, heavily inspired by the Two Envelope problem. I have my answer to this problem, but (full disclosure) even I’m not 100% sure I’ve got it right. I post it half because it seems very interesting to me, and half because I want to see if others agree with my reasoning. Enjoy.

--

Jack is invited to a new game show hosted by his favorite game show host, Monty Hall. To start, Monty welcomes Jack and shows him the latest in revolutionary 1970’s technology, a slot machine! But not just any slot machine. When a “betting” envelope containing cash or a check is put in front of this device and the lever is pulled, it will analyze it and dispense a “reward” envelope. 50% of the time the reward envelope will contain double the value of whatever was in the betting envelope, and the other 50% of the time it will contain only half the value of whatever was in the betting envelope. Truly a marvel of the ages.

Monty hands Jack \$100 in \$10 dollar bills and asks if he wants to take the gamble and play the machine. Having studied some basic statistics in preparation of coming here, Jack is ready to win as much money as possible and runs through some math in his head. He realizes that if he bets $10, half the time he wins \$20, the other half \$5. This gives him an expected value for each game he plays of \$12.50.

(\$20 * 0.5) + (\$5 * 0.5) = \$12.50 This machine is practically printing money! Jack plays 10 times, winning 5 games and losing 5 games, giving him$125. Grateful that studying statistics has finally paid off, he’s about to leave when Monty announces that Jack has won the right to play the bonus game against Kirby, Jack’s hated rival from his high school days!

A cart containing 20 sealed envelopes is brought out. Monty explains each envelope contains a check with a value that is a multiple of \$2. The prizes are \$2, \$4, \$8, \$16, and so on doubling each time, up to the most valuable envelope containing \$1,048,576. Jack will chose one of these envelopes at random and let the slot machine analyze it.

The game is played as follows: Jack will receive whatever is in the original betting envelope, while Kirby will receive the reward envelope. Each of them will be allowed to open their own envelope and see what is inside, but not tell the other player. They then have the option to trade for the other person’s envelope. If both players agree to trade, the envelopes are swapped. If either of them does not want to trade, then they both just keep the envelope they started with.

Jack picks a random envelope from the 20, lets the machine analyze it, then opens it to find \$512. He remembers how he won money playing this machine earlier and eagerly asks to swap envelopes with Kirby, since he knows the reward envelope always has a better expected value. Meanwhile, Kirby opens the reward envelope and finds \$256. Kirby reasons that there are only two possibilities, either Jack has \$128 and the machine doubled it, or Jack has \$512 and the machine cut it in half. Under the rules of this game those two scenarios are equally likely to happen. In other words, it’s a 50/50 shot for which of those two scenarios has occurred. That means that his expected value for trading is \$320, which is more than the \$256 he has.

(\$512 * 0.5) + (\$128 * 0.5) = \$320 Kirby accepts the trade, and both men walk away feeling they had gotten the better of the other. Did either of them make a mistake? ## 5 Answers I think the answer is ... They are both wrong. because Normally Jack should offer to swap if he is sure the swap will be accepted either way. But this is not the case. Jack picks an envelope with an amount between $$2^1$$ and $$2^{20}$$. Kirks receives an envelope with between $$2^0$$ and $$2^{21}$$. If Kirby finds $$2^{20}$$ or $$2^{21}$$ he will know that he has the higher envelope and he will not propose a swap. If Jack sees $$2^{19}$$ or $$2^{20}$$ he can expect the other envelope to be more valuable on average and would propose a swap. But wait! If Kirk has indeed the higher envelope, $$2^{20}$$ or $$2^{21}$$, he will refuse to swap. The only way for the swap to be accepted is if Kirby has the lower envelope. Therefore Jack should not offer to swap a $$2^{19}$$ or $$2^{20}$$. Once Kirby figures that out he knows that if he gets a $$2^{18}$$ or $$2^{19}$$ Jack will not swap his $$2^{19}$$ or $$2^{20}$$. The only way for the swap to be accepted is if Jack has the lower envelope. And therefore Kirby should hold on his $$2^{18}$$ or $$2^{19}$$. And so on, all the way down. The result is that regardless of the amount (except maybe \$1) neither Jack nor Kirby should propose to swap.

It is not a paradox as in the unexpected hanging paradox because at the end, when one player sees \$1 or \$2, he knows that it is the smaller envelope, he would propose a swap but it would be refused. There is no paradox, we just have one player who knows he has the smaller envelope but can do nothing about it.

• I feel like you've fallen into the unexpected hanging paradox, and that it is only true for two of the values but none of the rest. en.wikipedia.org/wiki/Unexpected_hanging_paradox Nov 4, 2020 at 19:48
• And now I feel like I am wrong, so I'm going to +1 but leave my original comment. Nov 4, 2020 at 19:50
• I don't thing the unexpected hanging paradox works here because in the last step, when Kirby finds $1, he has every reason to swap but Jack refuses. And he is right. Nov 4, 2020 at 22:33 • +1 Good Answer, I agree that this is what would happen if both Jack and Kirby are Game Theory Rational agents. However once Kirby sees the$256 in his envelope and receives the offer to swap from Jack, he knows that Jack is not playing rationally. Therefore wouldn't he be rational to accept the offer along his original line of reasoning? Or maybe it would depend on an updated expectation of whats in Jacks envelope that would be impossible to calculate objectively, since Jack is not playing rationally. Hmmm... Nov 21, 2020 at 14:50
• The reasoning assumes both want to maximize their profit. If Jack acts irrationally then we are in the situation that you find in any game, you can depart from the equilibrum strategy in the hope of earning more, but at the risk of it backfiring if your opponent adjusts to it and gains an advantage. Nov 22, 2020 at 16:31

Both calculations are correct. The seeming paradox is not actually a paradox; it is merely a difference of estimates between two parties due to incompleteness of information.

Analysis:

Jack is given a uniformly random choice between 20 possible envelopes, and the slot machine can produce the double or half of the input with equal probability. In total, the entire situation will be one of 40 equally-probable outcomes: $$\begin{array}{c|cc}\text{Jack} & \text{Kirby(Halve)} & \text{Kirby(Double)} \\\hline 2 & 1 & 4 \\ 4 & 2 & 8 \\ 8 & 4 & 16 \\ 16 & 8 & 32 \\ 32 & 16 & 64 \\\hline 64 & 32 & 128 \\ 128 & 64 & \bbox[yellow]{256} \\ 256 & 128 & 512 \\ \bbox[yellow]{512} & \bbox[yellow]{256} & 1024 \\1024 & 512 & 2048 \\\hline 2048 & 1024 & 4096 \\ 4096 & 2048 & 8192 \\ 8192 & 4096 & 16384 \\16384 & 8192 & 32768 \\ 32768 & 16384 & 65536 \\\hline 65536 & 32768 & 131072 \\131072 & 65536 & 262144 \\ 262144 & 131072 & 524288 \\ 524288 & 262144 & 1048576 \\1048576 & 524288 & 2097152\end{array}$$

The only information available to Jack is that he's got \$512, so to Jack's view, the expected value of Kirby's envelope is indeed $$(256+1024)/2 = 640 > 512$$. The only information available to Kirby is that he's got \$256, so to Kirby's view, the expected value of Jack's envelope is indeed $$(128+512)/2 = 320 > 256$$. To be extra safe, we can evaluate conditional probabilities:

\begin{align}P(J=512 \mid K=256) &= \frac{P(J=512 \wedge K=256)}{P(K=256)} = \frac{1/40}{2/40} = \frac12 \\P(J=128 \mid K=256) &= \frac{P(J=128 \wedge K=256)}{P(K=256)} = \frac12\end{align}

Therefore, we conclude that both calculations were right; Jack was unlucky and Kirby was lucky this time. Note that there's nothing wrong with losing money in one trial when it is expected to earn money (the expected value is just an expected average over large number of trials). Also note that Kirby will refuse to swap if he's got \$1048576 or higher. • The fact that both player expect to earn money is also a seeming paradox, as this would suggest that "changing envelopes" would increase their total expectation. The explanation to this, is exactly the last sentence: Kirby will refuse to swap if he's got$1048576 or higher. In that case, changing envelopes will cause a huge decrease of Kirby's expectation. This decrease cancels exactly all the increases of other cases. Nov 4, 2020 at 20:52
• @WhatsUp (and Bubbler) Well put. The looming paradox is neatly resolved by the existence of a maximum possible amount.
– Bass
Nov 4, 2020 at 21:55

Revised to include a recursive path to a perspective of the result from a situation that did not occur.

Did either of them make a mistake?

That question is more interesting than the rest of this sentence but, simply put, . . . .

. . .no, neither player made a mistake, but only due to the luck of their envelopes having \$512 and \$256 along with the fact that . . . .

. . . both Jack (J) and Kirby (K) aren’t perfect logicians and both had suboptimal strategies.
As such, . . . .

. . . both were destined to offer a swap with those \$amounts and, further, both correctly calculated their probabilistically expected gains, . . . . . . which is clear in Bubbler’s explanation. The game would have a different outcome, however, if both players were perfect logicians and knew that about each other. Scenario: J and K are mutually-known perfect logicians In this scenario a swap would never occur as J would never offer one. A reduced example demonstrates how this would play out with 10 twinned envelopes that allow for clean calculations. Two betting envelopes each of \$2, \$4, \$8, \$16 and \$32 for Jack.

   Prob-     Jack      Kirby
ability     (J)       (K)

.1         $$2$$1     (only 1 way for K to have $1) .1 $$2$$4 .1 $$4$$2 (only 1 way for K to have$2)
.1         $$4$$8

.1         $$8$$4     (2 ways each for K to have $4,$8 or $16) .1 $$8$$16 .1 $$16$$8 .1 $$16$$32 (only 1 way for K to have$32)

.1        $$32$$16
.1        $$32$$64     (only 1 way for K to have $64) K will not offer a swap at $$32 or$$64 as such swaps would surely lose money.  Each player’s model of the other player includes uncertainty that recursively bubbles all the way through to the two possibilities that K has \$32 or \$64 and wouldn’t offer to swap. Had both players assumed that each other would see how that works, those two possibilities would have influenced the possibility that did occur. Suppose J has the minimal \$2 and considers two models of K, one model where K has \$1 and the other where K has \$4. J muses that the possible K who has \$4 would also consider two models of a possible J where J has either \$2 or \$8. Carrying this out produces a fractal-like recursive root system of models that overlaps itself to form an endless lattice whose right edge terminates at one of two models of a possible K who has either \$32 or \$64 and will not offer to swap.  J has$2
/\
/  \
/    \
/      \
$$2 J's model of K has$$1      $$2 J's model of K has$$4
\         /\
\       /  \
\     /    \
$$2 J's model of \ / \$$1/$$4 K's model of J has$$2      $$2 J's model of$$4 K's model of J has $8 /\ /\ / \ / \ / \ / \ / \ / \ ... model of K has $$1 ..K has$$4 $$2 J's model of$$4 K's model of \ /\ /\$8 J's model of
\       /  \        /  \              K has $16 \ / \ / \ \ / \ / \ ... of K's model of J has $$2 ..J has$$8 $$2 J's model of$$4 K's model of /\ /\ /\$8 J's model of
/  \          /  \         /  \         $16 K's model of / \ / \ / \ J has$32
/      \      /      \     /      \
... model of K has $$1 ..K has$$4    ..K has $$16$$2 J's model of $4 K's \ /\ /\ model of$8 J's
\          /  \          /  \              model of $16 K's \ / \ / \ model of$32 J's
\      /      \      /      \            model of K has $64 \ / \ / \ and won't swap \ / \ / \ ... of K's model of J has $$2 ..J has$$8 $$2 J's model of$$1/$4 K's model of
/\            /\            /\         $$2/$$8 J's model of
/  \          /  \          /  \       $$4/$$16 K's model of
/    \        /    \        /    \      $$8/$$32 J's model of
/      \      /      \      /      \        $16 K's model of / \ / \ / \ J has$32
/          \  /          \  /          \
... model of K has $$1 ..K has$$4    ..K has $$16$$2 J's model of $$1/$$4 K's
\         /\            /\               model of $$2/$$8 J's
\       /  \          /  \              model of $$4/$$16 K's
model of $$8/$$32 J's
\   /      \      /      \            model of $16 K's model of$32 J's
\  /          \          model of K has $64 and won't swap \  In the first rightmost branch J's model of \$4 K's model of \$8 J's model of \$16 K's model of \$32 J's model of K has \$64 and certainly won't swap. This lattice of a tree includes half of all possibilities for the \$amount that J might have while all terminal branches have K not swapping for the same reason. The other half of all possible J amounts are included in a similar lattice spreading from a supposed J having \$4, where all terminal branches consider a model of K who has \$32 and also wouldn’t swap. Thus J has no reason to offer a swap. The reverse reasoning from here was presented in this answer’s original posting and admittedly parallels Florian F’s previously-posted reasoning). Here are probabilistic gains and losses from each player’s perspective if the players swap at every chance other when K has \$32 or \$64.  Probabilistic expectation, Probabilistic expectation, offering to swap offering to swap only Prob J every time NET Prob K for lower amounts NET .2 $$2 (1+4)/2 =$$2.5 +$$.5 .1$$1 2 = $$2 +$$1 .2 $$4 (2+8)/2 =$$5 +$$1 .1$$2 4 = $$4 +$$2 .2 $$8 (4+16)/2 =$$10 +$$2 .2$$4 (2+8)/2 = $$5 +$$1 .2 $$16 (8+16)/2 =$$12 -$$4 .2$$8 (4+16)/2 = $$10 +$$2 .2 $$32 (16+32)/2 =$$24 -$$8 .2$$16 (8+32)/2 = $$20 +$$4 .1 $$32 32 =$$32 0 Weighted averages \_ _ _ _ _ _ _ _ .1 $$64 64 =$$64 0 $$12.4$$10.7 -$$1.7$$15.5 $$17.2 +$$1.7  The rows for J’s \$16 and \$32 and the rows for K’s \$32 and \$64 reflect K’s not offering a swap at \$32 or \$64. Prob J Expectation NET Prob K Expectation NET .2 $$16 (8+ 16 )/2 =$$12 -$$4 .1$$32 32 = $$32 0 .2$$32 (16+ 32 )/2 = $$24 -$$8 .1 $$64 64 =$$64 0 /\________/\ /\_____/\ No swap if K has $$32 or$$64 means that J’s $$16 and$$32 along with K’s $$32 and$$64 remain unchanged when J offers to swap but K does not.  Thus J can probabilistically expect to lose money if J recklessly offers a swap at \$16 or \$32 while K wisely doesn’t at \$32 or \$64. As each player is a well-known perfect logician J would not offer a swap at \$16 or \$32 and K would know that. This in turn produces negative expectations for K were K to offer a swap at \$8 or \$16. Prob J Expectation NET Prob K Expectation NET .2 $$16 16 =$$16 0 .2 $$8 (4+ 8 )/2 =$$6 -$$2 .2$$32 32 = $$32 0 .2$$16 (8+ 16 )/2 = $$12 -$$4 /\______/\ /\_______/\ No swap if J has $$16 or$$32 means that J’s $$16 and$$32 along with K’s $$8 and$$16 remain unchanged when K offers to swap but J does not.  Thus K also won’t offer a swap at \$8 or \$16. This seesaw of decreasing expectations cascades all the way down the \$ amounts to where K has nothing to lose by offering a swap only at \$1 or \$2, but why even bother as K knows that J won’t offer to swap at any \$amount. • Interesting perspective, and I believe this is the correct strategy to take, since the puzzle explicitly says that a trade happens only upon agreement of both players. Without this information, we wouldn't know if refusal of K at \$32+ will actually impact J's expectations at \$16+. Nov 4, 2020 at 23:24 I think there's an alternative, simpler solution here, based on the assumptions made by the existing answer. Note that the existing answer assumes perfect play and players can rationalize for the other (and each player assumes the other is doing perfect play). With these assumptions, we can simplify: If either person gets an offer, it is because the other has deduced that the swap will produce a higher expected value. Therefore they should never accept a swap! My answer is that they are both wrong My reasoning. There is a great deal of information provided and an interesting set up as in all magic tricks. But, to my mind the devil is in the details. Jack picks one envelope. Kirby gets EITHER half or double the contents of Jack's - at no point is a series of games played in the final round. There is a 50% chance that your opponent has a larger sum, and a 50% chance that they will swap hoping they have the smaller amount. Personally I see no advantage to a swap other than a coin toss on both sides - there is a 50% chance of losing half, 50% of doubling up. • Nobody ever says a "series of games" is played - I don't see how that's relevant exactly. Also, the question explains why there is an advantage, no? "I bet 10 dollars and can win either 20 or 5 if I swap, so I should swap because I make an average of 12.5". What's wrong with this logic? – Deusovi Nov 4, 2020 at 19:10 • Yes, the chance of getting doubling up and losing half are equal. But doubling up makes you more money than you'd lose by losing half, so it's still a net positive. A coin flip game where you earn \$10 on a win and lose \\$5 on a loss is certainly worth playing. Nov 4, 2020 at 21:23
• The set up is a series of games was my inference. I understand your point re double or half as opposed to double or lose - but this is a single instance - your maximum guaranteed return is to keep what you have. If you are happy to risk play. You can't achieve the maximum calculated value (which you can in a series) - it's double or half Nov 4, 2020 at 21:37