How this calculator works
With an informed host who always reveals goats and offers a switch, the original door retains its initial 1/n chance. The unchosen doors share the remaining probability. A uniformly selected remaining alternative has probability (n - 1)/(n times the number of alternatives). This averages over the reveal and does not infer information from which door labels were opened. If the host opens uniformly without knowing, conditioning on all reveals being goats makes the remaining closed doors equally likely.
Formula
Informed host: P(stay) = 1/n; P(switch) = (n - 1)/[n(n - opened - 1)]. Uninformed uniform host, conditional on goat reveals: both probabilities = 1/(n - opened).
Worked example
With three doors and one deliberate goat reveal, staying wins with probability 1/3 and switching with 2/3. If an uninformed host happened to reveal a goat, the two closed doors instead each have probability 1/2.
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Frequently asked questions
Must the host always offer a switch?
The informed model assumes so. A host who selectively offers a switch can change the conditional probabilities.
What if several alternatives remain?
The switch output assumes you choose one of those doors uniformly. You do not receive every remaining door.
Does the host opening goats always help switching?
The reveal protocol matters. A knowledgeable host who avoids the prize differs from an uninformed host conditioned on a lucky reveal.