Monty Hall Problem Calculator

Compute the exact win probability for staying versus switching doors. The counterintuitive answer: in the classic 3-door game, switching wins 2/3 of the time and staying only 1/3.

Quick Facts

Method
P(stay) = 1/N; P(switch) = (N−1)/(N · (N−1−K))
The host knows the prize location and always opens losing doors. For the classic N=3, K=1 game this gives stay 1/3, switch 2/3.

Your Results

Calculated
Win probability if you SWITCH
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Switch to a remaining unopened door
Win probability if you STAY
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Keep your original pick
Switching advantage
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How many times better switching is
Better strategy
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Recommendation

Ready

Set the number of doors and how many the host opens, then calculate.

Understanding the Monty Hall Problem

The Monty Hall problem is a famous probability puzzle based on the American TV game show Let's Make a Deal, hosted by Monty Hall. You face three doors. Behind one is a car; behind the other two are goats. You pick a door — say Door 1. The host, who knows where the car is, then opens one of the other two doors to reveal a goat — say Door 3. He now offers you a choice: stick with Door 1, or switch to the remaining closed door, Door 2. Is it to your advantage to switch?

The counterintuitive answer is yes: you should always switch. Switching wins the car 2/3 of the time; staying wins only 1/3 of the time. Most people's intuition says the odds are 50/50 once one door is open, and that intuition is wrong.

Why switching wins two out of three times

When you first choose, you have a 1/3 chance of picking the car and a 2/3 chance of picking a goat. That 2/3 does not disappear when the host opens a door — because the host never opens the car and never opens your door, he is giving you information. If your original pick was a goat (which happens 2/3 of the time), the host is forced to reveal the other goat, and the car sits behind the one remaining closed door. So 2/3 of the time, switching lands you on the car. Only in the 1/3 of cases where you originally picked the car does switching lose.

The formula

For the general game with N doors where the host opens K losing doors after your pick:

  • P(win if you stay) = 1/N — your first pick's odds are fixed and never change.
  • P(win if you switch) = (N−1)/(N · (N−1−K)) — the probability that your first pick was wrong, spread evenly over the closed doors you could switch to.

This requires K ≥ 1 (the host opens at least one door) and K ≤ N−2 (at least one door remains to switch to). In the classic game N = 3 and K = 1, giving stay = 1/3 and switch = 2/3. When the host opens every losing door except one (K = N−2), the switch probability simplifies to (N−1)/N, and stay + switch add up to 1.

Reference points

  • 3 doors, host opens 1: stay 33.33%, switch 66.67% (switching is 2× better).
  • 4 doors, host opens 2: stay 25%, switch 75% (3× better).
  • 10 doors, host opens 8: stay 10%, switch 90% (9× better).
  • 100 doors, host opens 98: stay 1%, switch 99% (99× better).

The 100-door version is the clearest intuition pump: you pick one door out of a hundred, the host throws open 98 goats, and asks whether you want to keep your near-hopeless original guess or take the one door he pointedly left closed. Almost everyone switches.

Frequently Asked Questions

Why is switching better than staying?
Your first pick has a 1/3 chance of being right, so there is a 2/3 chance the car is behind one of the other two doors. The host, who knows the layout, always opens a goat door — never the car and never your door. That 2/3 probability collapses onto the single remaining unopened door, so switching wins 2/3 of the time while staying stays at 1/3.
Isn't it 50/50 once one door is opened?
No. It would be 50/50 only if the host opened a door at random and happened to reveal a goat. But the host is not random — he knows where the car is and deliberately avoids it. That deliberate choice is what makes the two remaining doors unequal: yours stays at 1/3, the other jumps to 2/3.
Does the host's knowledge matter?
Yes, it is essential. The whole 2/3 advantage depends on the host always knowing where the car is and always opening a losing door. If the host opened a door blindly and it just happened to be a goat, switching and staying would each be 50/50. This calculator assumes the standard rules: an informed host who never reveals the prize.
What happens with more doors?
The switching advantage grows. With N doors where the host opens all but one other door, staying wins 1/N and switching wins (N−1)/N. With 100 doors, staying wins just 1% while switching wins 99% — a dramatic illustration of the same logic behind the 3-door case.

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