What this calculator does
In the classic three-door version, staying wins a third of the time and switching wins two thirds. The intuition that it becomes an even chance after a door is opened is wrong, and remains one of the most stubbornly disputed results in elementary probability.
The generalisation makes it obvious. With 100 doors, you pick one at 1% and the host opens 98 losing doors, leaving your original pick and one other. Your first choice was 1% and nothing has changed that; all the remaining 99% has concentrated into the single door left. Switching wins 99 times out of 100.
The formula
Your initial pick has a 1/n chance of being correct and that never changes, because the host actions give you no information about the door you already chose. The remaining (n−1)/n is spread across the other doors, and when the host opens all the losing ones among them it concentrates entirely on the single unopened door. The result depends on the host knowing where the prize is and deliberately avoiding it.
| Term | Meaning |
|---|---|
| Stay | Keep your original door. Wins with probability 1/n. |
| Switch | Change to the remaining unopened door. Wins with probability (n−1)/n. |
| Host knowledge | The host knows where the prize is and never opens it. This is the assumption the result depends on. |
| Conditional probability | The formal framework: the host choice is informative about the other doors but not about yours. |
The inputs explained
| Field | What to enter |
|---|---|
| Number of doors | Number of doors. Three is the classic version; larger numbers make the logic much easier to see. |
When to use it
Settling the argument
The generalised version with many doors usually convinces people the three-door result is right.
Teaching conditional probability
It is the standard example of how information changes probabilities unevenly across outcomes.
Recognising the structure elsewhere
Any situation where an informed party eliminates options while deliberately avoiding one has the same shape.
Worked examples
Every figure in the tables below is produced by this page’s own calculator at build time, so the numbers and the tool always agree. Select any row to load that scenario.
How does the advantage change with more doors?
Staying against switching for various numbers of doors.
| Doors | Win probability if you stay | Win probability if you switch | Switching is better by |
|---|---|---|---|
| 3 doors | 33.3% | 66.7% | 33.3% points |
| 4 doors | 25.0% | 75.0% | 50.0% points |
| 10 doors | 10.0% | 90.0% | 80.0% points |
| 100 doors | 1.00% | 99.0% | 98.0% points |
Questions
Why is switching better?
Because your original pick was made when you had a 1/n chance, and the host revealing losing doors tells you nothing about it. All the remaining probability concentrates on the one door left unopened, so switching collects it.
Does it matter that the host knows where the prize is?
Entirely. If the host opened doors at random and happened to avoid the prize, the odds really would be even between the two remaining doors. The advantage comes from the host deliberate avoidance, which is what injects information into the situation.
Why does it feel like 50-50?
Because two doors remain and it is tempting to treat them symmetrically. They are not symmetric: one was chosen blindly from n doors, the other survived a deliberate filtering process. Equal counts do not mean equal probabilities.
Has this been tested?
Extensively, both by simulation and by physical experiment, and switching wins about two thirds of the time in the three-door game every time. The result caused a famous public dispute in 1990 in which many mathematicians initially and incorrectly insisted it was even odds.
For another counterintuitive probability, see the birthday paradox calculator. For probability conditioned on information, see the conditional probability calculator.