In the monty hall problem 100 doors version, you choose one door from 100, and a host opens 98 goat doors. If the host knows the prize location, always avoids it, and leaves one other door closed, switching wins 99 percent of the time. The result is not based on the two final doors being equal; it comes from how those doors were produced.
The rules that make the answer precise
Use these rules:
- One of 100 doors hides a prize; the other 99 hide goats.
- You choose one door.
- The host knows where the prize is.
- The host opens 98 doors you did not choose, always revealing goats.
- The host leaves your chosen door and one other door closed.
- If your original choice has the prize, the host chooses uniformly at random among the 99 eligible goat doors to leave closed.
That final rule matters when discussing the probability of one specific remaining door. It removes any hidden host preference. Under these rules, switching has a 99/100 success rate before the game begins, and, after seeing which alternative door remains, that specific door has a 99/100 chance of holding the prize.
The basic probability walkthrough
Suppose you select Door 1.
Before the host opens anything:
| Where the prize is | Probability |
|---|---|
| Behind Door 1 | 1/100 |
| Behind one of the other 99 doors | 99/100 |
Your first pick was a blind choice, so it has only a 1/100 chance of being right. The group of 99 doors you did not choose has a combined 99/100 chance.
Now the host opens 98 goats and leaves, say, Door 73 closed.
There are two possible explanations:
- The prize is behind Door 1. This began with probability 1/100. In that case, every other door contains a goat, and the host leaves Door 73 with probability 1/99 because of the uniform-random rule.
- The prize is behind Door 73. This began with probability 1/100. In that case, the host must leave Door 73 closed.
The host is 99 times more likely to leave Door 73 when it contains the prize than when your original Door 1 contains the prize. So, after Door 73 is left closed, the probabilities are:
| Final choice | Chance of winning |
|---|---|
| Stay with Door 1 | 1/100 |
| Switch to Door 73 | 99/100 |
A shorter way to state the strategy is even simpler: you stay only when your first guess was correct, which happens once in 100 games. You switch whenever it was wrong, which happens 99 times in 100 games.
Why 100 doors makes the answer feel obvious
Picture yourself making the first choice with almost no information. Picking the one prize door from 100 is a long shot.
Then the host effectively says, “Your original guess may be right. But I have removed 98 losing choices from the large group you did not pick, and this one surviving door is the meaningful alternative.”
The surviving door is not merely “the other closed door.” It is the result of a constrained process: the host knows where the prize is and cannot reveal it.
That is why the 100-door version is more intuitive than the classic three-door setup. Keeping a 1 percent first guess after the host narrows 99 alternatives to one feels much less defensible.
A concrete round of the game
Assume the prize is behind Door 73.
- You choose Door 1.
- The host cannot open Door 1 because it is your pick.
- The host cannot open Door 73 because it contains the prize.
- The host opens every other door, revealing 98 goats.
- Door 73 is the only available alternative, so switching wins.
Now take the rare case in which the prize is behind Door 1.
- You choose Door 1.
- The host opens 98 goat doors.
- Of the 99 goat doors available to leave closed, the host selects one uniformly at random.
- Switching loses.
Across all 100 equally likely prize locations, staying wins in one case. Always switching wins in the other 99.
Why “two doors remain, so it is 50/50” is wrong
The final count of doors does not erase the earlier information.
Your door was selected by you without knowing the prize location. The other remaining door was selected through the host’s informed, constrained action. Those are not equivalent paths to the final two-door situation.
A useful question for probability and test-prep problems is:
> What process generated the evidence I am looking at?
If a person who knows the answer filters out losing options, their choices carry information. If doors were opened by someone who did not know the prize location, the probability calculation would be different because avoiding the prize would be a lucky event rather than a guaranteed part of the rules.
Run a quick simulation
You can test the always-switch strategy without writing advanced code. For each round, randomly choose a prize door and an initial pick.
```text stay_wins = 0 switch_wins = 0
repeat 10000 times: prize = random integer from 1 to 100 first_pick = random integer from 1 to 100
if first_pick == prize: stay_wins += 1 else: switch_wins += 1
report stay_wins and switch_wins ```
You do not need to model all 98 host openings for this comparison. If the first pick is correct, staying wins. If it is wrong, the standard host rules guarantee that the one remaining alternative is the prize, so switching wins.
Your totals will not be exactly 100 and 9,900 every time. Random samples vary. Over many rounds, though, the stay rate should settle near 1 percent and the switch rate near 99 percent.
For a faster hands-on check, try a probability or cognitive-bias exercise in PurrLearn’s logic quizzes, then explain each result in terms of the information available before and after the reveal.
A worked LSAT-style flaw question
Question
A contestant argues:
> “After the host opens 98 doors, two doors remain. Since one has the prize, each door must have an equal chance. Switching cannot improve my odds.”
Which response most directly identifies the flaw?
A. It overlooks that the contestant might prefer to stay. B. It treats the host’s informed elimination of goat doors as though it were a random elimination. C. It assumes that the prize cannot be behind the contestant’s original door. D. It fails to establish the monetary value of the prize. E. It confuses the probability of a win with the probability that the host opens a door.
Answer: B
The argument uses the final picture—two closed doors—to claim equal probabilities. But it ignores the selection process that created that picture. The host knows the prize location, avoids it, and reveals only goats. Under the stated uniform rule, the remaining alternative door is 99 times as likely to have been left closed because it hides the prize than because the contestant’s original door hides it.
Why the other choices fail:
- A is about preference, not probability.
- C overstates the issue. The original door can still contain the prize; it is simply unlikely.
- D is irrelevant to the reasoning.
- E mentions the host’s action but does not identify the argument’s mistaken assumption.
This is a common flaw-question pattern: an argument treats selective evidence as if it were neutral evidence. Ask what was excluded, who did the excluding, and what they knew.
The one-sentence explanation
Use this version:
> Under the standard uniform host rule, your first door keeps its 1/100 chance, while the specific door the host leaves has a 99/100 chance because the host was forced to preserve the prize door when your first choice was wrong.
FAQ
Does switching always win in the 100-door Monty Hall problem?
No. Switching loses when your initial pick contains the prize, which happens 1/100 of the time. It wins when your first pick is wrong, which happens 99/100 of the time.
Why does switching have a 99 percent success rate?
Your initial choice is correct in only one of the 100 possible prize locations. In every other case, the host must leave the prize door closed as the alternative, so an always-switch strategy wins.
Does the specific remaining door really have a 99/100 chance?
Yes, if the host follows the stated uniform rule when your initial pick is correct. If the host used an unexplained preference for which goat door to leave closed, you could still say that always switching wins 99/100 of the time, but you could not automatically assign 99/100 to every named remaining door without knowing that preference.
What if the host opens doors randomly?
Then this is not the standard Monty Hall problem. A host who does not know where the prize is might accidentally reveal it, and conditioning on the event that they did not changes the calculation.
How does this help with LSAT or GRE reasoning?
It trains you to separate an outcome from the process that produced it. That habit helps with conditional probability, evidence selection, sampling errors, and arguments that quietly treat a non-random process as random.
Conclusion
The 100-door Monty Hall problem is not a 50/50 choice after the reveal. Under a host rule that always reveals goats and chooses uniformly when your first pick is right, staying wins 1/100 of the time and switching wins 99/100. The key lesson is broader than game shows: when evidence has been filtered by someone with information, the filtering process belongs in the reasoning.