Two Envelopes Paradox Calculator - The classic envelope puzzle, resolved with a proper prior set by a known lower bound.
Use this two envelopes paradox calculator: enter the amount you see and the smallest amount the game can pay out, then compare the expected value of switching against keeping.
Two Envelopes Paradox Calculator
Results
What Is Two Envelopes Paradox Calculator?
The two envelopes paradox calculator begins with a familiar setup: you are handed one of two sealed envelopes, you open it and see an amount A, and you are offered the chance to swap for the other envelope. One envelope holds twice as much as the other. The puzzle is whether swapping can ever be worth more than keeping.
- • Students: Checking why the naive switching argument gives 1.25 times the held amount.
- • Curious readers: Wanting the resolved expected value of switching versus keeping.
- • Puzzle comparers: Weighing the two envelopes paradox against other conditional-probability puzzles.
- • Teachers: Demonstrating how a proper prior removes the incentive to switch.
The paradox is a classic in probability and decision theory because a simple expected-value argument appears to recommend switching forever, which is obviously absurd. The goal of this tool is to show where that argument breaks and what a correct expected-value calculation returns once you state the prior behind the amounts.
You only need two facts to run the numbers: the amount you see, and the smallest amount the game could ever have placed in an envelope. With those, the calculator returns the expected value of switching and the expected value of keeping side by side.
Like the two envelopes paradox, the Bertrand box paradox calculator shows how a naive counting argument hides a conditional probability that changes the answer.
How Two Envelopes Paradox Calculator Works
The calculator models the smaller amount as drawn uniformly from a known range, L to M. Once that prior is fixed, the two candidate contents of the other envelope, A/2 and 2A, no longer deserve equal weight, because seeing A tells you which one is more likely.
- A: The amount you observe after opening your envelope.
- L: The smallest amount the game can pay, which sets the lower end of the prior.
- M: The largest amount the game can pay, which sets the upper end of the prior.
- 2A and A/2: The two candidate amounts in the unopened envelope.
With a uniform prior over the smaller amount, the density of seeing A as the smaller envelope and as the larger envelope differ by a factor of two, because the larger value stretches the prior. Weighting 2A by two thirds and A/2 by one third gives the resolved expected value.
Compare that with the naive result of 1.25A. The naive version weights A/2 and 2A equally; the resolved version weights them by the actual chance each applies once a proper prior is set, which is what removes the self-contradiction.
Example: A equals 100 with L equals 1 and M equals 1000
The two candidate amounts in the other envelope are A/2 = 50 and 2A = 200. Because 100 sits inside the prior range, both are possible.
The naive average gives 0.5 times 50 plus 0.5 times 200, which is 125, or 1.25 times what you hold.
Under the uniform prior, the chance A is the smaller envelope is two thirds, so E[other] = (2/3) times 200 plus (1/3) times 50 = 150.
Switching is then worth about 50 more on average, because A is small enough that it is probably the smaller envelope.
According to Wikipedia.
According to Wikipedia.
The weighted average at the heart of the resolution is the same idea you apply in the expected value calculator when you weigh outcomes by their probabilities.
Key Concepts Explained
Four ideas carry the whole resolution in the two envelopes paradox calculator: the observed amount, the prior range that makes the problem well defined, the prior itself, and the way seeing A updates the probability that you hold the smaller envelope.
Observed amount A
The dollar value you see after opening your envelope. It is the number the whole argument hinges on, because the candidate contents of the other envelope are A/2 and 2A.
Prior range L to M
The smallest and largest amounts the game can put in an envelope. Fixing this range turns the puzzle into a normal expected-value problem with no ambiguity.
Proper prior
A probability distribution over the smaller amount that sums to one. Without a fixed range there is no proper prior, and the naive argument can be made to say anything.
Conditional update
After seeing A, the chance you hold the smaller envelope depends on where A falls in the prior range, not a flat one half; that update is what removes the self-contradiction.
Before assigning the conditional odds, you can check the post-update weights with the probability calculator to confirm the probability split.
How to Use This Calculator
Enter the three inputs and read the result rows. The calculator does the conditional weighting for you, so you only need the observed amount and the range of payouts the game allows.
- 1 Open your envelope and read A: Type the amount you see into the observed field. This is the value the candidate amounts A/2 and 2A are built from.
- 2 Enter the smallest possible payout L: Type the known lower bound the game can pay. It must be greater than zero and less than the largest possible amount.
- 3 Enter the largest possible amount M: Type the upper bound of the prior so that a very large A is treated as the larger envelope rather than the smaller one.
- 4 Read the expected value row: The tool reports the expected value of the other envelope after the conditional update, weighting 2A and A/2 by their actual chances.
- 5 Read the recommendation: When A sits inside the prior range the resolved expected value favors switching, because A is then probably the smaller envelope; far above the range it favors keeping.
Try A = 100 with L = 1 and M = 1000: the tool shows the other envelope is worth about 150 on average, while your own envelope is worth 100, so switching is the better choice here.
Both puzzles hinge on updating after you see information, which the Monty Hall problem calculator makes explicit by conditioning on the host's reveal.
Benefits of Using This Calculator
The two envelopes paradox calculator turns a confusing verbal puzzle into a concrete expected-value comparison you can change and inspect.
- • Shows the flaw directly: You can see the naive 1.25 multiplier next to the resolved result and watch why they differ.
- • Makes the prior explicit: By asking for L, the tool forces the usually-hidden assumption about how the amounts were chosen into the open.
- • Connects to other puzzles: The same conditional reasoning appears in the Monty Hall and Bertrand box paradoxes, so the lesson transfers.
- • Fast what-if exploration: Changing A or L immediately updates every row, which is easier than redoing the arithmetic by hand.
The two envelopes paradox is one of several surprises where intuition fails, much like the base rate trap shown by the false positive paradox calculator.
Factors That Affect Your Results
Two factors decide whether switching beats keeping: where A falls in the prior range, and whether the naive equal-weight assumption is used.
Prior range L to M
A fixed, finite range is what makes the resolution valid. Without it the conditional probabilities cannot be assigned, and the paradox stays open.
Observed amount A relative to the range
When A sits inside the range you are most likely holding the smaller envelope, so switching wins; when A is above the range you are certainly holding the larger one, so keeping wins.
Equal-weight assumption
Treating A/2 and 2A as equally likely is the source of the bogus 1.25 recommendation. Dropping it removes the paradox.
- • The standard version assumes one envelope holds exactly twice the other; a different ratio needs its own weighting.
- • The tool requires a known finite range; with no such bound the prior is improper and the answer is undefined.
- • It reports expected value only and says nothing about risk preference or how the amounts were actually generated.
If you enjoy counterintuitive probability, the birthday paradox calculator is another classic that rewards working the conditional math carefully.
Frequently Asked Questions
Q: What is the two envelopes paradox in simple terms?
A: You open one of two envelopes and see amount A. One envelope holds twice the other, so the unopened one is either A/2 or 2A. Averaging those with equal weight suggests 1.25A, which argues for switching forever. The paradox is that the same logic applies after you switch, so it cannot be right.
Q: Why does the naive switching argument give 1.25 times what I hold?
A: It averages the two possible other amounts, A/2 and 2A, with equal weights of one half each. That returns 0.5 times A/2 plus 0.5 times 2A, which equals 1.25A. The flaw is assuming those two states are equally likely after you have seen A.
Q: How does fixing the prior range fix the paradox?
A: Fixing a finite range from L to M gives the smaller amount a real probability distribution. Seeing A then tells you how likely you are holding the smaller envelope versus the larger one. Weighting 2A by that chance and A/2 by the other gives a definite expected value, so the answer stops being self-contradictory.
Q: Is switching envelopes ever the better choice?
A: With a uniform prior over the smaller amount, switching wins whenever A falls inside the prior range, because A is then probably the smaller envelope. If A is above the range it is certainly the larger envelope, so keeping wins. The naive argument that switching always wins comes from ignoring where A sits in the prior.
Q: Is the two envelopes paradox the same as the Monty Hall problem?
A: Both involve updating probabilities after seeing information, but Monty Hall has a host who deliberately reveals a losing door, which shifts probability onto the remaining door. The two-envelopes paradox has no such reveal; the update comes only from the value A you observe relative to the known lower bound L.
Q: Does the calculator assume the other envelope is always exactly twice mine?
A: Yes, it follows the classic version where one envelope holds exactly twice the other, so the two candidate amounts are A/2 and 2A. A different ratio would need its own weighting, and the calculator requires a known finite range because that range is what makes the posterior well defined.