Section 1.1: The Initial Dilemma

At first glance, it might seem pointless to switch since you have no information about the contents of either envelope. A random choice seems just as good as any.

However, let’s analyze the situation: suppose the first envelope you select has an amount of x dollars. There are two scenarios to consider:

1. You have picked the envelope with the lesser amount, meaning the other envelope contains $2x. If you switch, you gain $2x - $x = $x.
2. Conversely, if you’ve chosen the envelope with the larger sum, the other one would then contain $0.5x. Here, switching leads to a loss: $0.5x - $x = -0.5x.

Assuming both scenarios are equally probable, switching would yield an expected value of 0.5 * $x + 0.5 * $-0.5x = $0.25x. This suggests that switching might be advantageous.

But this reasoning presents a paradox. If switching is always beneficial, you could continue switching indefinitely, acquiring more money with each turn. Clearly, this is impossible, as we cannot generate unlimited wealth from just two envelopes!

Section 1.2: The Symmetrical Argument

Another way to frame this issue is to recognize the inherent symmetry of the situation. Since there are no observable differences between the two envelopes, the logic behind switching becomes questionable.

Indeed, switching is not advantageous; the initial argument is fundamentally flawed. The amounts in the envelopes are fixed, meaning in scenarios (1) and (2), the variable x is not constant, despite the argument assuming it is.

A more accurate way to model this scenario is as follows: the envelope with the lesser amount contains $x, while the other holds $2x. Re-evaluating the scenarios:

1. If you picked the smaller envelope, switching results in $2x - $x = $x.
2. If you picked the larger envelope, switching gives $x - $2x = -$x.

Given both options are equally likely, the expected outcome is 0.5 * $x + 0.5 * $-x = $0. Thus, switching is neither beneficial nor harmful. Paradox resolved!

Section 2.1: Evaluating the New Situation

This depends on the assumption that there’s a 50% chance the $100 represents the smaller amount (implying an equal chance for it to be the larger one). If this is the case, switching appears to be advantageous: there's a 50% likelihood the other envelope contains $50 (resulting in a loss of $50), and an equal chance it holds $200 (yielding a gain of $100). This leads to an expected value of 0.5 * $-50 + 0.5 * $100 = $25 for switching.

Interestingly, the probabilities don’t need to remain at 50/50. Let’s denote the probability of the $100 being the smaller amount as p. Then switching becomes beneficial when p * $-50 + (1 - p) * $100 > $0, which simplifies to p * $-150 + $100 > $0, ultimately leading to p < 2/3. Thus, if you assess the probability of having selected the smaller envelope as less than 2/3, you should opt to switch.

However, this reasoning feels flawed. Even if p < 2/3 seems like a reasonable assumption, we are still basing our decisions on uncertain probabilities.

Section 2.2: The Importance of Context

Ultimately, the true value of p remains elusive and must relate to the $100 you discovered. For instance, if we estimate p = 0.5 after finding $100, we would also apply that same logic to amounts like $10, $50, or $120. This would imply a continuous inclination to switch, which paradoxically would suggest switching was beneficial from the outset.

Is there a strategy that could provide an advantage? Yes, a surprising approach exists.

Video Description: This video elaborates on the Two Envelope Problem, explaining its complexities and offering insights into potential resolutions.

Section 2.3: Discovering a Winning Strategy

To develop an effective strategy, select a random number. This number must be truly random, ensuring that every number has a chance of being selected. For example, a graph depicting the probability of choosing a number might look as follows:

Next, let’s denote the random number you select as r, while the amount in the first envelope is m. The strategy is straightforward: if m < r, switch; if m > r, don’t switch.

Why does this approach work? There are only three possibilities:

1. If r is less than both envelope amounts, then m > r; therefore, you don’t switch.
2. If r is greater than or equal to both amounts, m < r, leading you to switch.
3. If r falls between the two amounts, you will always switch to the larger envelope.

Crucially, the outcome of scenarios (1) and (2) results in neither gain nor loss, while (3) ensures you always end up with the larger envelope. As long as scenario (3) occurs occasionally, you’ll have a winning strategy and can expect to earn more than $0.

The remarkable aspect of this strategy is that scenario (3) has a non-zero probability of occurring, as we selected r from a distribution that is positive at all points.

Video Description: This video explores the resolution of the two envelope fallacy, providing insights into probability and decision-making.

References

• Two-Envelope Paradox | Brilliant Math & Science Wiki
• A special thank you to Remarkl for insightful discussions and clarifications regarding this problem.

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