The Monty Hall Problem
If you’ve never heard of the Monty Hall Problem before, here is a brief summary.
You are in a game show where you are faced with three closed doors. Behind one of the doors is a brand new car and the other doors hide a goat each. You then choose a door in an attempt to win the car but before the door is opened the host (Monty Hall) opens one of the remaining doors he knows contains a goat. You are then given the opportunity to choose whether to stay with your original selection or switch to the other, remaining closed door.
Intuitively most of us tend to think that it doesn’t really matter because we think there is a 50/50 chance either way but the reality is that statistically you have a greater chance of winning the car if you switch. In fact you stand a 2/3 chance of winning with the switch strategy as opposed to a 1/3 chance with the stay strategy.
If you remain unconvinced feel free to try it out for yourself with a friend or a die and a piece of paper.
Anyway, the point of this post is to say that in the two years or so that I’ve known about this I’ve never heard a concise and easy-to-understand explanation for why this is so. But now I have one…
You have a 2/3 chance of selecting a goat instead of a car. This means that two thirds of the time Monty Hall will reveal the last remaining goat which means that two thirds of the time if you use a switch strategy you will win.
To highlight how this works, imagine the same scenario but with 1000 doors and only 1 car. You have a 1/1000 chance of selecting the car or a 999/1000 chance of selecting a goat. If Monty Hall opens 998 other doors with goats behind them it is extremely likely (999/1000) that the last remaining door will have the car. (Thanks A3!)
You have two strategies: switch or stay. Using the stay strategy, if you accidentally choose a car first off (1/3 chance) you will win but if you choose a goat (2/3) you’ll lose. Using the switch strategy you’ll reverse these odds giving you statistically a 66.6% chance of winning every time.
Simple huh?
Tags: explanation, monty hall, probability, statistics
I’m unconvinced
Regardless of your odds before the first goat is revealed, the fact is that your second choice is between 2 (not 3) equally likely events. Therefore the chance that a goat is behind either 1 of the 2 remaining doors is 50%. No?
That’s what I instinctively thought too. The best way is to actually go ahead and try the two strategies a large number of times and you’ll find that the switch strategy will trend towards 66%.
You have a 66% chance of choosing a goat. Using the switch strategy you’ll end up on a car 66% of the time.
I have an excel spreadsheet that tries this 1000 times and confirms the unintuitive but logical result if anyone wants to see it
Perfect! I have both of your email addresses if you would like me to introduce you to each other.
I just did it using a die where if I roll a:
1 or 2 = 1
3 or 4 = 2
5 or 6 = 3
(Sounds complex but just a way of randomly getting a 1, 2 or 3.)
The first time I roll gives me the location of the car. The second roll gives me the door I blindly choose.
So, if I roll a 5 then the car is behind door 3. If my second roll is a 6 then I have chosen door 3 and, using the switch strategy where the host would open one of the other doors I would lose. But had I rolled a 1, 2, 3 or 4 (which would be a 1 or 2) I would have successfully switched to door 3 with the car.
I did it 10 times and got 7 wins but this sample is too small really. The more you roll the closer you’ll approach 2/3 success.
Or, you could do it the easy way and use a spreadsheet! (But I like dice and pencils)
This is a long-ish explanation, but stay with me – it’s only long because I spelled it out – not because the explanation is difficult.
Here goes…
The problem is in the formula for calculating the probability. Here’s what I mean:
The probability that you will initially select a goat out of the 3 choices is 2/3 (since there are 2 goats and 1 car). Likewise, the probability that you will initially select a car is 1/3. However, the probability that you will end up with a goat is only 1/2 – and the probability that you will end up with a car is also 1/2. The reason is that even if your initial choice is random, Monty will eliminate a goat from the mix. This is true whether your initial selection is a goat or a car.
Let’s look at all possible scenarios:
A) you first choose a goat, you hold
B) you first choose a goat, you switch
C) you first choose the car, you hold
D) you first choose the car, you switch
Note that each scenario involves 2 separate choices, both of which are random. The 1st choice is your initial selection between 3 doors, and the 2nd choice is whether to switch or hold. Then we have to consider each of the 2 choices separately for the calculation.
First, let’s look at the 2 choices:
Choice 1: You select 1 of 3 doors. The probability that your initial choice is a goat is 2/3.
Whether you initial selection is a goat or a car, there is at least one goat that is not selected – Monty eliminates one goat. Then regardless of your initial selection, the remaining 2 doors always have one goat and one car. (pretty neat!). Because of the elimination of one of the goats, the chance that you now hold a car is 1/2 (since there is now one goat and one car).
Choice2: Do you hold or switch? Let’s calculate:
Scenario A: The probability that you first selected a goat is 2/3, the probability that your decision to hold results in a car is 1/2. So the probability that you first selected a goat and your decision to hold results in a car is:
2/3 * 1/2 = 2/6 (or ~66%).
Scenario B:
The probability that you first selected a goat is 2/3, the probability that your decision to switch results in a car is 1/2. So the probability that you first selected a goat and your decision to switch results in a car is:
2/3 * 1/2 = 2/6 (or ~66%).
Scenario C:
The probability that you first selected a car is 2/3, the probability that your decision to hold results in a car is 1/2. So the probability that you first selected a car and your decision to hold results in a car is:
2/3 * 1/2 = 2/6 (or ~66%).
Scenario D:
The probability that you first selected a car is 2/3, the probability that your decision to switch results in a car is 1/2. So the probability that you first selected a car and your decision to switch results in a car is:
2/3 * 1/2 = 2/6 (or ~66%).
So now we see clearly that we have a problem with the calculation – all of the probabilities don’t add up to 1 (100%, that is)! The problem is that the calculation for probability is not formulated correctly. You can verify by random dice rolls and spreadsheet calculations that each of these scenarios yield ~66% (given a large enough sample size).
In fact, your choice is not actually between 2 goats and a car. If we are guaranteed that Monty will always eliminate one goat and allow us to choose again, then the choice is really between 2 doors: one concealing a goat and one concealing a car (the choice is not really between 2 goats and a car). This hold true whether your first choice was a car or a goat.
Damian – not sure if the offer for an introduction was to me or not, but in general, feel free to give out the webmaster@askanatheist.org address to anyone who wants it.
Firstly, 2/6 equals ~33% (not 66%).
Secondly, I don’t think it’s correct to try to get all four scenarios to add up to 100%. I think that for each strategy there are two scenarios and that they should add up to 100%:
Strategy 1 (hold)
A. Goat 2/3 (~66% chance of losing)
B. Car 1/3 (~33% chance of winning)
equals 100%
Strategy 2 (switch)
A. Goat 2/3 (~66% chance of winning)
B. Car 1/3 (~33% chance of losing)
equals 100%
As to your reduction of the chance to 50% at the final hurdle this doesn’t work because we are measuring the chance overall. Not just the final stage.
The simplest way to confirm this is with a pencil and a die using one of the given strategies multiple times. If you use the ‘hold’ strategy I guarantee that you will trend toward a 33% success rate and toward 66% with the ‘switch’ strategy. (A perfect example of the scientific method in action!)
At the time of my post, 2/6 was equal to 66%.
Just kidding – good catch!!
But all of the permutations do have to add up to 100% (or it’s an indication that the probabilities are not expressed correctly). The reason is that the probability that exactly one of any of the possible permutations will occur is 100%.
First, take the example of a bag containing 2 blue beads. If you reach into the back and select one bead, the probability that the bead will be blue is 100% (obviously). Now take a bag with one blue bead and one red bead. If you select a bead this time (and presuming you are equally likely to pick either the red or the blue bead), the probability that the bead will be blue is 50%. The reason is that there are 2 equally likely outcomes so the probability of either one of the outcomes is 100%/2 or 50%. By definition, the probabilities of either outcome must add up to 100%. In other words, the probability that you will either pick a red bead or you will pick a blue bead is 100%/2 for the red + 100%/2 for the blue = 100% for either a red or a blue.
Now let’s put 3 beads in the bag: 2 reds and 1 blue. The probability that you will pick a red bead is 100%/3 for one of the red beads, 100%/3 for the other red bead, and 100%/3 for the blue bead. If we cared about which red bead you pick, then the chance of picking a particular red bead is 100%/3. But since either red bead will count as a “red”, then the chance of picking a red is 100%/3 (for one) + 100%/3 (for the other) = (100%/3) x 2 = 2/3 x 100% ~ 66%. The chance that you pick a blue is still 100%/3 or ~ 33%. Notice that the reason the probabilities add up to 100% is that it is certain that exactly one of the 3 equally possible outcomes will occur (100% = certainty).
So the trick of the Monty Hall puzzle is expressing it correctly:
There are exactly 4 possible outcomes:
1) (a) you pick a goat
(b) Monty removes the other goat from the mix
(c) you keep your selection
2) (a) you pick a goat
(b) Monty removes the other goat from the mix
(c) you switch
3) (a) you pick a car
(b) Monty removes a goat from the mix
(c) you keep your selection
4) (a) you pick a car
(b) Monty removes a goat from the mix
(c) you switch
(notice that there would be 8 possible outcomes if Monty sometimes removes a goat and sometimes removes a car)
Now calculate the probability of each outcome:
1) [probability you picked a goat]
x [probability that Monty removes a goat]
x [probability you keep your selection]
= 2/3 * 1 * 1/2 = 1/3 = probability that you get a goat
2) [probability you picked a goat]
x [probability that Monty removes a goat]
x [probability you switch]
= 2/3 * 1 * 1/2 = 1/3 = probability that you get a car
3) [probability you picked a car]
x [probability that Monty removes a goat]
x [probability you keep your selection]
= 1/3 * 1 * 1/2 = 1/6 = probability that you get a car
4) [probability you picked a car]
x [probability that Monty removes a goat]
x [probability you switch]
= 1/3 * 1 * 1/2 = 1/6 = probability that you get a goat
All of these add up to 100%: 1/3 + 1/3 + 1/6 + 1/6 = 2/6 + 2/6 + 1/6 + 1/6 = 6/6 = 1 = 100%
Here’s the kicker: add up the probability that you get a goat (scenarios 1 & 4):
1/3 (scenario 1) + 1/6 (scenario 4) = 2/6 + 1/6 = 3/6 = .5 = 50%
Add up the probability that you get a car (scenarios 2 & 3):
1/3 (scenario 2) + 1/6 (scenario 3) = 2/6 + 1/6 = 3/6 = .5 = 50%
The probability that you end up with a car or a goat = 100%
Now let’s see what happens if you always switch (scenario 2 & 4):
First scenario:
[2/3 = chance you first pick a goat]
* [1 = chance Monty removes the other goat]
* [1/2 = chance your switch results in a car (because there is now 1 goat and 1 car)]
= 2/6
Second scenario:
[1/3 = chance you first pick a car]
* [1 = chance Monty removes the other goat]
* [1/2 = chance your switch results in a car (because there is now 1 goat and 1 car)]
= 1/6
Your chance of getting a car? 2/6 + 1/6 = 50%
This is consistent with the assertion that the 2nd choice is really the only one that counts. By the time you make the 2nd choice, there is 1 car and 1 goat. Regardless of you previous choice, choosing between 1 car and 1 goat gives you a 50/50 chance of winning the car.
I realise you’re putting a lot of thought into this and I apologise for not fully addressing the formula you are using.
I’ll try expressing this problem another way.
Let’s increase the numbers in this scenario and see what happens. Imagine now that there are 1000 doors and all but one have goats behind them. You are allowed to pick one door, after which Monty Hall opens 998 other doors all revealing goats. You are left with the option to stick with your initial pick or to switch to the last remaining door. Should you switch or stay?
Here we have the initial odds of selecting the door with the car at 1/1000 so it’s very likely that you will have selected a goat. My claim is that by always switching you will win 999/1000 times. Do you agree with this or do you think it’s still a 50/50 chance?
What I’m illustrating here is that stacking the odds at the start can make a huge difference to what intuitively feels like a 50/50 final choice.
Do you see where I’m coming from?
(Also, have you yet sat down with a die and a pencil and experimentally tested the original Monty Hall Problem? If you haven’t, please do, and behold the power of scientific experimentation!)
Ah!! Your scenario where we start with 1000 doors makes sense!! Now I (finally!!) see how it works. You are right: starting with 3 (or more doors), you are more likely to have selected a goat on your first pick. So when Monty removes the remaining goats, switching gives you a better chance at the car.
I also see now why my calculations were off: I calculated the “probability you keep your selection” as if it were a random event with a probability of 1/2. But it isn’t a random event!
The 1000-goat scenario was great explanation – and this has been a really fun thread!
Cheers mate.
Ahhhhh, it’s soooo refreshing to have had an ‘argument’ with someone who wants the truth rather than to prove a point with no regard for truth. (I’ve been having a few of those kinds of encounters lately, as you may be aware).
Thanks, I love you man.
Yes, I should have thought to exaggerate the numbers to highlight how stacking the odds has an effect on the deceptive 50/50 call at the end. I’ll update my easy-to-understand explanation in the original post methinks.
XOXO
Yea, I’ve been following the conversations and jumping in when I get time – great fun!
I always found that playing the scenario out with this one is intellectually unsatisfying; it still leaves one grasping for a handle on exactly WHY this works this way.
MOST of the time (when you choose the wrong door) Monty has no choice but to increase your informtion about the door he doesn’t open, but he can never give you information about the door you originally chose.
The door you originally chose is stuck, then, at a probability of 1/3, while the probability that it was one of the other doors was 2/3. When he reveals a goat behind one of those other doors, the 2/3 probability falls upon the door you didn’t choose. So 2/3 of the time, the door with the car will be the door you didn’t originally choose.