Marilyn vos Savant and the Monte Hall Problem
Posted: Tue Jun 18, 2024 11:49 pmFor anyone who enjoys probability, there's an interesting puzzle widely known as the "Monte Hall Problem". It gets its name from (and is often described by analogy to) the television show "Let's Make a Deal".
In the problem, there are three closed doors. Behind each of the doors is a prize. One of the prizes is highly valued and desired (such as a car). The other two prizes are far less valuable gag prizes (such as goats). The player first chooses a door, but is not shown what is behind it. Instead, the host (Monte Hall, for example) shows the player one of the other doors which will always have a gag prize (since the host knows what's behind each door). The player is then given the opportunity to keep their original unopened choice or to switch to the remaining unopened door. The game may be spiced up with some wheeling and dealing along the way. The question arises: What is the best strategy for the player to obtain the valuable prize?
At first glance, it appears that there can be no "strategy" because with only 2 doors left, there appears to be an equal chance of choosing the gag gift or the valuable prize. But that turns out not to be the case, which was quite a surprise to many "experts" in probability. The unexpected solution was originally published by Steve Selvin (1975) and then more widely published by Marilyn vos Savant (1990) as described in this video:
I came across this problem (and video) somewhat randomly, and I had a hard time believing it myself. Even after watching the video, it was still a bit fuzzy and counterintuitive to me. So I wrote a small Python simulation program to test it out. Here's my Python code:
Sure enough, when I ran that program, it showed that a player who stays with their original choice (whatever it was) will only win the prize about 1/3 of the time. However, a player who switches their choice will win the prize about 2/3 of the time. That's a big difference, and it's very counter-intuitive. The Wikipedia page and the video give very good explanations for why this happens. I'll paraphrase it myself here:
When the player makes their first choice, all three doors hold equal probability of containing the prize. So the player can truly pick at random. But once that first choice is made, and one of the other non-prize doors is opened, there are only two remaining cases:
At first glance, those cases seem useless because if the player knew which case they were in, then they would also know which door held the prize. But that's not quite true because the player knows one more bit of information that they might not have considered.
Remember that the player knows that their original pick (whatever it was) had a 2 out of 3 chance of being a goat, and only a 1 out of 3 chance of being the prize. This means that the player (with only two unopened doors remaining) knows that they have a 2 out of 3 chance of being in Case A (where they initially chose a door with a goat), and only a 1 out of 3 chance of being in Case B (where they initially chose a door with the prize). Given the greater chance of being in Case A, the player should make their decision as if they were in Case A, and switch their choice. This will result in winning 2 out of 3 times rather than 1 out of 3 times.
In the problem, there are three closed doors. Behind each of the doors is a prize. One of the prizes is highly valued and desired (such as a car). The other two prizes are far less valuable gag prizes (such as goats). The player first chooses a door, but is not shown what is behind it. Instead, the host (Monte Hall, for example) shows the player one of the other doors which will always have a gag prize (since the host knows what's behind each door). The player is then given the opportunity to keep their original unopened choice or to switch to the remaining unopened door. The game may be spiced up with some wheeling and dealing along the way. The question arises: What is the best strategy for the player to obtain the valuable prize?
At first glance, it appears that there can be no "strategy" because with only 2 doors left, there appears to be an equal chance of choosing the gag gift or the valuable prize. But that turns out not to be the case, which was quite a surprise to many "experts" in probability. The unexpected solution was originally published by Steve Selvin (1975) and then more widely published by Marilyn vos Savant (1990) as described in this video:
I came across this problem (and video) somewhat randomly, and I had a hard time believing it myself. Even after watching the video, it was still a bit fuzzy and counterintuitive to me. So I wrote a small Python simulation program to test it out. Here's my Python code:
- Code: Select all
import random
total_trials = 100000
stay_wins = 0
switch_wins = 0
random.seed()
for i in range(total_trials):
# Place prize behind a random door
prize = random.randint(1,3)
print ( "The prize is behind door " + str(prize) )
# Randomly pick an initial guess
first_pick = random.randint(1,3)
print ( "Player's first random door pick is " + str(first_pick) )
# Randomly open one of the non-prize doors
first_show = random.randint(1,3)
while (first_show == prize) or (first_show == first_pick):
first_show = random.randint(1,3)
print ( "Show that door " + str(first_show) + " does not have the prize" )
# Show which choice wins and update that strategy
if first_pick == prize:
print ( "Stay wins" )
stay_wins += 1
else:
print ( "Switch wins" )
switch_wins += 1
# Show current wins for each strategy
print ( "===================" )
print ( "Stay and win: " + str(stay_wins) + ", Switch and win: " + str(switch_wins) )
print ( "===================" )
# Uncomment for an interactive Python shell:
# __import__('code').interact(local={k: v for ns in (globals(), locals()) for k, v in ns.items()})
Sure enough, when I ran that program, it showed that a player who stays with their original choice (whatever it was) will only win the prize about 1/3 of the time. However, a player who switches their choice will win the prize about 2/3 of the time. That's a big difference, and it's very counter-intuitive. The Wikipedia page and the video give very good explanations for why this happens. I'll paraphrase it myself here:
When the player makes their first choice, all three doors hold equal probability of containing the prize. So the player can truly pick at random. But once that first choice is made, and one of the other non-prize doors is opened, there are only two remaining cases:
Case A: The player had initially chosen a door with a hidden goat. If this happens to be the case (which is unknown to the player), then the player should choose to switch to the other remaining door in order to win the prize.
Case B: The player had initially chosen the door with the hidden prize. If this happens to be the case (which is again unknown to the player), then the player should choose to stay with that door in order to win the prize.
Case B: The player had initially chosen the door with the hidden prize. If this happens to be the case (which is again unknown to the player), then the player should choose to stay with that door in order to win the prize.
At first glance, those cases seem useless because if the player knew which case they were in, then they would also know which door held the prize. But that's not quite true because the player knows one more bit of information that they might not have considered.
Remember that the player knows that their original pick (whatever it was) had a 2 out of 3 chance of being a goat, and only a 1 out of 3 chance of being the prize. This means that the player (with only two unopened doors remaining) knows that they have a 2 out of 3 chance of being in Case A (where they initially chose a door with a goat), and only a 1 out of 3 chance of being in Case B (where they initially chose a door with the prize). Given the greater chance of being in Case A, the player should make their decision as if they were in Case A, and switch their choice. This will result in winning 2 out of 3 times rather than 1 out of 3 times.