The ‘Monty Hall’ problem is best described by the wikipedia entry:
The Monty Hall problem is a probability puzzle, loosely based on the American television game show Let’s Make a Deal and named after its original host, Monty Hall. The problem was originally posed (and solved) in a letter by Steve Selvin to the American Statistician in 1975. It became famous as a question from reader Craig F. Whitaker’s letter quoted in Marilyn vos Savant‘s “Ask Marilyn” column in Parade magazine in 1990.
Suppose you’re on a game show, and you’re given the choice of three doors A, B and C.
Behind one door is a car; behind the other two are goats. You pick a door, say door A, and the host, who knows what’s behind the doors, opens another door, say door B, which has a goat. He then says to you, “Do you want to change your choice to door C?”
Is it to your advantage to switch? (It is!)
Many readers of vos Savant’s column refused to believe switching is beneficial and rejected her explanation. After the problem appeared in Parade, approximately 10,000 readers, including nearly 1,000 with PhDs, wrote to the magazine, most of them calling vos Savant wrong.[4] Even when given explanations, simulations, and formal mathematical proofs, many people still did not accept that switching is the best strategy.[5]Paul Erdős, one of the most prolific mathematicians in history, remained unconvinced until he was shown a computer simulation demonstrating vos Savant’s predicted result.[6]
First we need to define some notation:
A, B – events
P(A) – the probability of event A occurring
P(B) – the probability of event B occurring
P(A | B) – the probability of event A occurring, given that event B has already occurred
P(B | A) – the probability of event B occurring, given that event A has already occurred
Bayes Theorem is defined as:
The following reasoning is from Julian Havil’s book “Impossible?”
Assign symbols to the events:
A – the event “car is behind door A”
B – the event “car is behind door B”
C – the event “car is behind door C”
MA – the event “Monty opens door A” … similarly for MB , MC
Assume door A is chosen initially by the player, so Monty can open door B or C:
P(MB | A) = ½ , P(MB | B) = 0, P(MB | C) = 1
So, since A, B and C are mutually exclusive events:
P(MB) = P(MB | A)P(A) + P(MB | B)P(B) + P(MB | C)P(C) = ½ x ⅓ + 0 x ⅓ + 1 x ⅓ = ½
Now, the player can stick or change. If they stick with door A, their probability of winning the car is:
P(A | MB) = P(MB | A)P(A) / P(MB) = (½ x ⅓) / ½ = ⅓
If they switch to door C, their probability of winning the car is:
P(C | MB) = P(MB | C)P(C) / P(MB) = (1 x ⅓) / ½ = ⅔