Two envelopes paradox

Two envelopes: one contains x dollars, the other contains 2x. You pick one at random and find y dollars inside. Should you switch to the other envelope?

The argument for switching: The other envelope contains either y/2 (if you have the larger) or 2y (if you have the smaller), each with probability 1/2. So:

E [other] = \frac{1}{2} \cdot \frac{y}{2} + \frac{1}{2} \cdot 2 y = \frac{5 y}{4} > y

Since 5y/4 > y, you should always switch. But then the same argument applies to the other envelope, and you should switch back — forever. Clearly something is wrong.

Always switch

48.6%

Always stay

51.4%

The flawed argument:

You see value y in your envelope. The other holds y/2 or 2y with equal probability, so E[other] = (y/2 + 2y)/2 = 5y/4 > y — always switch!

The error:

Once you see y, it's no longer equally likely that the other is y/2 or 2y — that depends on the prior distribution of x. The "equal probability" assumption bakes in a contradiction. Both strategies win exactly 50% of the time.

games 500

Blue: always-switch win rate. Orange: always-stay win rate. Both converge to 50% regardless of the envelope amounts.

What the simulation shows

Both strategies win exactly 50% of the time. “Always switch” is not better than “always stay.” The paradox is entirely in the argument, not in the math.

Where the argument breaks

The flaw is subtle: the “1/2 probability” assumption is wrong once you’ve seen y.

If y is odd (or in general, if y could not be the smaller amount given the prior on x), then with probability 1 you hold the larger envelope.
The correct conditional probabilities P(you hold larger | you see y) and P(you hold smaller | you see y) depend on the prior distribution of x.

For most reasonable priors on x, these probabilities are not 50/50 at the value y you observe. The argument incorrectly treats y as a fixed unknown while simultaneously assuming symmetric uncertainty — a contradiction.

The formal issue

Let f(x) be the prior density on x. Then:

P(you hold smaller | you see y) = f(y) / [f(y) + f(y/2)/2], not 1/2 in general.

Only if f(x) ∝ 1/x (an improper prior) does this work out to 1/2 — but that prior cannot be normalized and doesn’t represent a valid probability distribution. The paradox exploits an implicitly assumed improper prior.