Monty Hall Calculator
See why switching doors wins more often, for the classic three-door game or any door count.
What it does
The Monty Hall problem is a probability puzzle built on a simple game. A prize sits behind one of several doors. You pick a door but do not open it. The host, who knows where the prize is, opens some of the other doors to reveal goats, then offers you the chance to switch to the one remaining closed door. The question is whether switching helps.
It does, and the reason is that the host's choice is not random. Your first pick wins with probability 1/n. That probability does not change when the host opens doors, because nothing that happened after your pick can move the prize. The remaining closed door carries all of the probability that was spread across the doors you did not pick, which is (n-1)/n.
With the classic three doors, your first choice is right one third of the time and wrong two thirds of the time. When you are wrong, the host is forced to leave the prize door closed, so switching wins. That happens two thirds of the time. Staying wins only in the one third of cases where your first pick was already correct.
The general case follows the same logic. With n doors, the host opens n - 2 losing doors and leaves exactly one alternative. Staying wins 1/n of the time. Switching wins (n - 1)/n of the time. With ten doors, staying wins 10 percent and switching wins 90 percent.
The result surprises people because the two remaining doors look symmetric. They are not, because the host treated them differently: one was chosen by you before any information arrived, and the other survived a deliberate process that avoided the prize. The calculator reports both probabilities and the number of doors the host opens, so the pattern is easy to see as you change the door count.
How to use
- Enter the number of doors in the game, from 3 to 100.
- Read the stay probability, which is always 1 divided by the door count.
- Read the switch probability, which is the rest of the probability.
- Check the number of doors the host opens, which is the door count minus 2.
- Compare the two probabilities to see the advantage of switching.
- Try a larger door count to see how quickly the advantage approaches certainty.
Formula
- n
- Total number of doors at the start of the game.
- P(stay)
- Probability of winning if you keep your first choice.
- P(switch)
- Probability of winning if you move to the only other closed door.
- n - 2
- Losing doors the host reveals before the switch offer.
The result assumes the host always opens n - 2 losing doors, always avoids the prize, and always offers the switch. If the host opens a door at random, or only sometimes offers the switch, the probabilities change and the simple 1/n versus (n-1)/n split no longer applies.
Worked example
Example: the classic three-door game
Three doors, one prize, and a host who knows where it is. You pick one door, the host opens one losing door, and you may switch.
- Start with three doors: n = 3.
- Stay probability: 1 / 3 = 0.3333, or 33.33 percent.
- Switch probability: (3 - 1) / 3 = 2 / 3 = 0.6667, or 66.67 percent.
- Doors the host opens: 3 - 2 = 1.
- The two probabilities add to 100 percent, so every game ends in a win or a loss.
- Switching wins in two of the three equally likely starting positions.
Reference table
| Doors | Host opens | Stay wins | Switch wins |
|---|---|---|---|
| 3 | 1 | 33.33 % | 66.67 % |
| 4 | 2 | 25.00 % | 75.00 % |
| 5 | 3 | 20.00 % | 80.00 % |
| 10 | 8 | 10.00 % | 90.00 % |
| 100 | 98 | 1.00 % | 99.00 % |
Limits
- Assumes the host knows the prize location and always opens only losing doors.
- Assumes the host always makes the switch offer, with no strategic behavior.
- Requires at least three doors; with two doors there is no losing door to open.
- Treats all doors as equally likely at the start, with exactly one prize.
Common mistakes
- Assuming the two remaining doors are equally likely after the host opens one.
- Forgetting that the host's knowledge is what makes switching favorable.
- Believing that a previous loss makes the prize more likely behind your door, which is the gambler's fallacy.
- Applying the 1/n and (n-1)/n split when the host opens doors at random instead of deliberately.
- Confusing the probability before the host acts with the probability after, which are not the same question.
Try it yourself
Quick checks. Answers sit tucked under each question.
With 4 doors, what is the win probability if you switch?
75.00 percent, and the host opens 2 doors.
With 10 doors, what is the win probability if you stay?
10.00 percent, while switching wins 90.00 percent.
With 100 doors, how many doors does the host open?
98 doors, leaving your pick and one other door closed.
Questions
Why is switching better in the Monty Hall problem?
Your first pick is right only 1/n of the time. The host then concentrates the remaining probability on one other door, so switching wins the other (n-1)/n of the time.
Does the host's knowledge matter?
Yes. If the host opened a door at random and happened to reveal a goat, the advantage of switching would shrink or disappear depending on the rules.
What if there are four doors?
Staying wins 25 percent and switching wins 75 percent, because the host opens two losing doors and leaves one alternative.
Is the Monty Hall problem the same as the birthday problem?
No. Both involve probability, but the birthday problem asks about shared dates in a group, while Monty Hall asks how new information changes a choice.
What if I prefer my lucky door?
Preference does not change the probabilities. The math favors switching under the stated rules, though a single game can still go either way.
Why does the probability not split evenly between two doors?
The host's choice depends on where the prize is, so the two closed doors are not symmetric. One was picked before the host acted, and the other survived a process that avoided the prize.