Free Two Envelopes Paradox Calculator

The two envelopes paradox is a classic puzzle about probability and expected value that continues to intrigue mathematicians and laypeople alike. This free online tool—the Two Envelopes Paradox Calculator—lets you experiment with the scenario, clarifying why swapping envelopes never improves your odds. Whether you are new to the paradox or want to solidify your understanding, the calculator’s simulation and calculation modes provide an intuitive, hands‑on experience.

The Setup

You face two sealed envelopes, each containing money. One envelope holds exactly twice the amount in the other. You pick one at random, say Envelope A. Before opening it, you are allowed to switch to Envelope B. Should you?

The naive reasoning runs as follows. Let the amount in your chosen envelope be $V$. Then the other envelope contains either $2V$ (with probability $1/2$) or $V/2$ (with probability $1/2$). The expected value of the other envelope is therefore:

$$E_{\text{other}} = \frac12 (2V) + \frac12 \left(\frac{V}{2}\right) = \frac54 V = 1.25\,V.$$

Since $1.25V > V$, it appears that swapping is always advantageous. But if you repeat the reasoning after switching, you would again find the other envelope’s expected value to be 25% higher, suggesting you should swap infinitely many times—a clear impossibility.

Where the Apparent Paradox Comes From

The flaw lies in using $V$ to denote two different amounts in the same equation. In the first scenario, $V$ represents the smaller amount; in the second, it represents the larger amount. Mixing them yields a meaningless result.

A correct analysis defines two fixed (but unknown) amounts: the smaller sum $s$ and the larger sum $2s$. The total money in play is $3s$. Because each envelope is equally likely to be the larger or the smaller, the expected value of either envelope is:

$$E = \frac12(s) + \frac12(2s) = \frac{3s}{2} = 1.5s.$$

Thus both envelopes have the identical expected value. There is no advantage to switching; your chance of picking the larger envelope was 50% and remains 50% regardless of swapping.

A Numerical Illustration

Suppose you peek into your envelope and find $20. Two possibilities exist:

1. Your envelope is the larger one. Then it holds $20, and the other envelope contains$10.
2. Your envelope is the smaller one. Then it holds $20, and the other envelope contains$40.

The expected value of the other envelope, given that you have seen $20, is the average of the two possible amounts:

$$E_{\text{other}} = \frac12 (10) + \frac12 (40) = 25.$$

This $25 is simply the midpoint between$10 and $40, not 25$20. The earlier mistake was to compute $1.25 \times 20 = 25$ by coincidence, but the reasoning behind that number was flawed.

Using the Two Envelopes Paradox Calculator

This online calculator offers two complementary modes.

Simulation Mode

Simulation mode recreates the paradoxical back‑and‑forth. You choose an initial envelope, and the calculator tells you the other envelope’s “expected value” is 25% higher. You may switch, only to be informed that the new other envelope again has a 25% higher expected value. After several swaps you return to your original choice, and the tool reveals the actual contents along with an explanation of the paradox. This hands‑on trial makes the fallacy tangible.

Calculation Mode

In calculation mode, you enter the amount you saw in your envelope (for example, $20). The calculator then displays both possible arrangements: one where your envelope is the larger, and one where it is the smaller. This gives you a clear picture of why the naive 1.25 factor does not represent a real gain.

Bottom Line

The two envelopes paradox highlights how misapplying expected value can lead to an apparent infinite switching incentive. The correct mathematics shows that the expected value of each envelope is the same—$\frac32$ of the smaller amount—so swapping is neutral. Use this free Two Envelopes Paradox Calculator to test the logic yourself and deepen your appreciation of probability.