Calcul Marylen S Fallacy

Probability Calculator

This calculator analyzes the famous Marilyn-style Monty Hall reasoning problem, often misspelled online as “marylen’s fallacy.” Enter the number of doors, how many non-winning doors the host reveals, and your decision strategy to see the exact probabilities of winning by staying or switching.

Expert Guide to Calcul Marylen’s Fallacy

The phrase calcul marylen’s fallacy usually refers to a calculator or explanation for the probability puzzle made famous by Marilyn vos Savant’s discussion of the Monty Hall problem. Although the spelling varies online, the underlying issue is the same: many people instinctively believe that after one losing door is revealed, the remaining unopened doors should each have a 50% chance of hiding the prize. That intuition feels natural, but it is wrong under the standard assumptions of the problem.

In the classic game-show setup, you first choose one door out of three. One door hides a prize and the other two hide goats. The host knows where the prize is. After you pick, the host deliberately opens one of the non-winning doors that you did not choose. You are then offered a choice: stick with your original door or switch to the other unopened door. The correct answer is that switching doubles your probability of winning. Staying wins only one-third of the time, while switching wins two-thirds of the time.

Why the mistake happens

The fallacy comes from ignoring the importance of conditional probability. The host’s action is not random. The host does not open just any door. Instead, the host uses information about the prize location and intentionally reveals a losing door. That means the reveal gives you information about the original choice.

Your first pick had a 1 in 3 chance of being correct. Nothing that happens afterward changes the fact that your original door was initially correct only one-third of the time. The remaining probability, two-thirds, must go somewhere. Because the host removes a losing option from the unchosen set, the full two-thirds probability becomes concentrated in the remaining unopened alternative door. That is why switching works.

• Your original pick starts with probability 1/3 of being correct.
• The set of all other doors starts with probability 2/3 of containing the prize.
• The host reveals a goat from that unchosen set without ever exposing the prize.
• The surviving unopened alternative inherits the full remaining probability under the classic 3-door version.

The exact calculation behind the calculator

This calculator generalizes the Marilyn-style logic beyond just three doors. Suppose there are n total doors. You choose one door. The host then opens k losing doors among the doors you did not choose. After that reveal, there are n – k unopened doors left in total, including your original selection.

Under the standard assumptions, the probability of winning by staying remains:

Stay probability = 1 / n

If you switch to one randomly selected unopened alternative door, your probability becomes:

Switch probability = ((n – 1) / n) / (n – k – 1)

Here is the intuition. The chance that your first pick was wrong is (n – 1) / n. If your first pick was wrong, then the prize sits somewhere among the other doors. After the host opens k guaranteed losing doors, there are n – k – 1 unopened alternatives left besides your own. If you choose randomly among those remaining alternatives, your chance of landing on the prize is the wrong-initial-pick probability distributed over those surviving options.

In the classic 3-door problem, n = 3 and k = 1. That gives:

• Stay probability = 1/3 = 33.33%
• Switch probability = (2/3) / 1 = 2/3 = 66.67%

Comparison table: exact probabilities by number of doors

Total doors	Doors opened by host	Stay win rate	Switch win rate	Interpretation
3	1	33.33%	66.67%	Classic Monty Hall result. Switching doubles your chance.
4	2	25.00%	75.00%	With more initial uncertainty, the switch advantage grows.
10	8	10.00%	90.00%	The host’s reveal concentrates almost all value into the remaining alternative.
100	98	1.00%	99.00%	This extreme case makes the logic much easier to feel intuitively.

Notice how the switch advantage becomes more dramatic as the number of doors rises. If there are 100 doors and the host opens 98 goats, your first pick is still only correct 1% of the time. The remaining unopened alternative therefore captures the other 99% probability. This thought experiment is one of the clearest ways to defeat the “it must be 50-50 now” intuition.

Extended case: what if more than one unopened alternative remains?

Many calculators stop at the standard setup with exactly one other closed door left. However, this page lets you analyze broader versions. For example, imagine 10 doors, you pick one, and the host opens only 5 losing doors. That leaves 4 unopened alternatives besides your own door. Your original choice still wins only 10% of the time. The unchosen set still owns 90% of the probability mass, but that 90% is now spread across four surviving unopened alternatives.

In that case:

• Stay probability = 1/10 = 10%
• Switch to one random remaining alternative = 90% / 4 = 22.5%

Even here, switching to a randomly chosen alternative is still better than staying. But it is not as overwhelmingly strong as in the version where the host narrows the field all the way down to one alternative.

Comparison table: exact theory vs practical interpretation

Scenario	Theoretical stay rate	Theoretical switch rate	Odds ratio in favor of switching	Practical takeaway
3 doors, host opens 1	33.33%	66.67%	2.0x	Switching is the clearly dominant strategy.
10 doors, host opens 5	10.00%	22.50%	2.25x	Switching remains superior even when several alternatives remain.
20 doors, host opens 18	5.00%	95.00%	19.0x	Near-total elimination by the host creates a huge switching edge.
50 doors, host opens 20	2.00%	3.27%	1.63x	Switching still helps, but the benefit depends on how much the host narrows the field.

These figures show a subtle but important lesson: the advantage of switching is not merely about the number of doors. It also depends on how many losing doors the host removes while obeying the standard information rule. The more effectively the host filters out bad choices for you, the stronger the remaining alternative becomes.

Common misconceptions about Marylen’s or Marilyn’s fallacy

1. “There are two doors left, so it must be 50-50.”
   This ignores the host’s knowledge. If the host reveals a goat intentionally, the reveal is informative and the probabilities do not reset.
2. “My first choice gets better after the host opens a door.”
   It does not. Your original door had a fixed probability when selected. The host’s reveal does not magically improve your initial guess.
3. “Switching only matters in the 3-door version.”
   False. The same logic scales to many-door versions and often becomes even more obvious.
4. “The puzzle is a trick question.”
   It is not a trick. It is a clean lesson in conditional probability and information structure.

Why statisticians care about this puzzle

The broader importance of this puzzle goes far beyond game shows. It teaches how people can make systematic errors when they fail to account for how information is generated. In statistics, medicine, machine learning, and risk analysis, the same issue appears whenever evidence is filtered or selected rather than drawn randomly.

For example, screening tests, fraud detection systems, legal evidence, and forecasting models all rely on conditional updates. The Marilyn-style fallacy is really a warning against naive probability reasoning. If you overlook the mechanism producing the observed evidence, your final conclusion can be badly distorted.

If you want to deepen your understanding of the underlying ideas, the following resources are particularly useful:

• Penn State Stat 414 on conditional probability
• NIST Engineering Statistics Handbook
• Harvard Stat 110 probability resources

How to use this calculator correctly

To get valid results, you should apply the calculator only when the host follows the standard rule set:

• The host knows where the prize is.
• The host never opens your chosen door.
• The host never reveals the prize by mistake.
• The host always opens the specified number of losing doors.

If those rules change, the probabilities may also change. For instance, if the host opens doors randomly and could accidentally expose the prize, then the classic Marilyn conclusion does not automatically hold. In that altered problem, you must include the host’s behavior in the probability model. That is why careful problem definition matters so much in probability theory.

Final takeaway

The best way to think about calcul marylen’s fallacy is this: your first pick is a low-confidence guess, and the host’s informed reveal transfers useful information to the remaining closed options. In the classic case, that transfer makes switching the rational decision. In generalized versions, the exact strength of the switch advantage depends on both the number of doors and the number of losing doors the host removes, but the central principle remains the same.

If you remember one line, remember this: the host’s action is not neutral information. Once you understand that, the paradox disappears and the calculation becomes clear.

This calculator is educational and assumes the standard Marilyn-style Monty Hall rules. It is not intended for gambling advice or for scenarios where the host behavior is unknown, inconsistent, or random.