Well, I still cannot understand why the supposed solution would be the case.
I just do not see why, if there is an equal chance of each door containing the car, the intervention of Monty – without moving the car – would affect the probability.
The impression I got was that the problem somehow relies on second-guessing Monty’s psychology and plans, through the application of unspecified conventions, but now the insistence seems to be that it’s somehow mathematically minimal in a way I can’t penetrate.
I’m reminded very greatly of the time when Numberfile claimed high and low, using sophistic mathematical reasoning, that the sum of all positive integers was -1/12 before being soundly disproven and having to issue an apology.
I never pursued mathematics at a high level, so I’m inclined to stop wasting other people’s time in a probably vain pursuit, but I would still not vote switch or pretend that I’ve been convinced.
There is no guessing of psychology, since this is a mathematical problem. You could easily leave Monty Hall out of the problem statement and talk about an algorithm going through his actions.
Moreover the rules are simple: you get to pick a door, after that a goat will be revealed to be behind a different door than you pick, then you get the chance to switch or not. You know all of this when you pick your door as well.
As you don’t know where the car is, you have 1/3 chance of picking the car, let’s call it A), and 2/3 chance of not picking the car, let’s call it B).
A) If you picked the car by accident, a goat is revealed. When you switch, you’ll get the other goat.
B) If you picked a goat by accident, the other goat is revealed. When you switch, you’ll get the car.
Hence switching will give you the car in 2/3 of the times (namely whenever B) happened).
Note that Monty’s influence on the game is that he essentially eliminates one of the goats from your choices. His opening of a door does not change the probability that the car is behind your door: it’s still 1/3. However, he removes one of the other options for you to pick, thus by switching you move from your 1/3 chance of picking the car initially to the remaining chance of 2/3 that the car is not behind your original pick.
I have to admit that the Numberphile video skips over a lot of very important disclaimers, but the message in the video is not “simply wrong”. The problem lies in that they are not using the “normal” way of looking at an infinite series, but using a different way.
First, if we have an infinite sum of numbers a0 + a1 + a2 + a3 + …, then we can make a sequence of its partial sums:
We say that the series is convergent to a value X, if for every ε > 0, however small ε is chosen, there exists some n such that for all m>n we have -ε < X - Pm < ε. In other words, however small we make the area around the value X, there will always be some step n after which all of the partial sum lie in this small area around X.
For example 1/2 + 1/4 + 1/8 + 1/16 + 1/32 + 1/64 + … converges to 1, since the partial sums look like 1/2, 3/4, 7/8, 15/16, 31/32, 63/64, … They get ever closer to 1, never going over it, and for any small ε eventually all the partial sums will lie between 1 - ε and 1.
In this sense, the sum 1+2+3+4+5+… is not convergent, since we cannot find such a value X that fits, thus the sum 1+2+3+4+5+… has no value under this definition of sum.
Then there is another way to define the sum for sums of a specific format, using the Riemann zeta function. It turns out that this Riemann zeta function agrees with the “normal” way of computing an infinite series as long as that series is convergent. As soon as a series is divergent, we get two different things. According to the “normal” way of computing sums the series will have no value, but according to the Riemann zeta function it does have a value.
So, for example, if we were to ask the Riemann zeta function to evaluate a suitable convergent sum (suitable meaning that it’s a sum of the kind that the Riemann zeta function can compute in the first place), then it would just spit out the value of this convergent sum. If we were to ask the Riemann zeta function to evaluate the divergent sum 1+2+3+…, it still spits out a value, namely -1/12.
However, this is not the value of the sum in the “normal” way of defining values of infinite series. Sadly, they don’t make this all too clear in the Numberphile video.
To compare it to something tangible, suppose I have a function f that is defined on the integers, and if I feed f any integer, it will just spit out 0. So f (0) = 0, f (2) = 0, f (-43) = 0, etc.
Now I have a different function, g, defined on all of the real line, and g(x) = 0. Then if we only look at the integers, it turns out that g and f completely agree, since g(0) = 0, g(2) = 0, g(-43) = 0, etc.
However, for any number that is not an integer, we see that g is defined, while f is not, e.g. g(1/2) = 0, while f (1/2) has no value.
Note that it is unreasonable (and this is a little bit what the Numberphile video does) to claim that f (1/2) should be 0. For example, we could have just as easily worked with g’(x) = sin( π * x ), which also agrees with f on the integers, but g’(1/2) = 1. The only reasonable thing to say is that f has no value at 1/2.
Or in the Numberphile case, that 1+2+3+4+… has no value, if we evaluate it using the “normal” way of evaluating infinite series.
I rewatched the video, and I was perhaps a bit too kind above.
The idea that 1 + 2 + 3 + 4 + … can be assigned the value -1/12 in a meaningful way is not at all a fallacy.
However, in the video they try to explain it with “intuition”, which is not at all a healthy attitude with divergent sums. For example, shifting values in infinite divergent sums has a great impact on its partial sums. In fact, even for some convergent sums this is more or less a deadly sin: by rearranging the terms of a sum like 1/2 - 1/3 + 1/4 - 1/5 + 1/6 - 1/7 + … I can get any value I want.
So, the bottom line is, that the video tries to explain a meaningful and astounding mathematical idea in a way that makes absolutely no sense. The steps are correct, the explanation is not.
Here’s a better video about it. Knowing Mathologer it should be easier to follow than the Numberphile video:
Your solution does not explain in step 2 how the option is eliminated, thus you can’t conclude from it in step 3 that your chances have improved. It’s not as simple as “an option is eliminated”, the way it is eliminated matters to the solution.
But fair enough, if we have an evil Monty Hall, who only offers the switch when it would imply switching to a goat, then even my pedantic correction would be false, since the extra information in such version would be that there’s a goat behind the door when you switch.
I like that the experimental chance of finding the car when switching is only about 50%. The reason being that the car is fixed behind door number 1, and here on the forum, it seems we’re not at all choosing our door with a uniform random distribution.
I meant to say a bad option is removed. You got me. Good one.
My son actually did a school science fair project on the Monty Hall Problem a few years ago and I helped him write some python code to run some simulations and plot the data with matplotlib.
The easiest way to become convinced of the correctness of the standard solution is to ask a friend to sit down and literally simulate it with you with a deck of cards, and run it like 100 times, without switching, and 100 times with switching. (Although, I guess in COVID times this may be harder ) Or rather, make it more extreme, with more doors, for example, 13 doors, represented by the 13 cards of a given suit from the deck of cards. Then you can do a lot fewer than 100 runs and get very high statistical confidence.
So the protocol would be:
Separate out all 13 cards of a given suit, say, hearts, and set aside the rest of the deck, which won’t be used. One person is the dealer, one person is the player, and the player’s goal is to get the King.
The dealer shuffles the 13 cards and lays them out face down.
The player picks one of the cards by indicating it.
The dealer then peeks at the card picked without the revealing the card.
If the player’s selected card was the King, the dealer secretly mentally picks a random number X from 1-12. Then one by one, from left to right in order, they peek at each of the remaining 12 unpicked cards, and if the card is the X of hearts (Ace=1, Jack=11, Queen=12), they keep it face down, else they turn it face up.
Otherwise, if the player’s picked card was not the King, the dealer one by one, from left to right in order, peeks at each of the remaining 12 unpicked cards, and if the card is the King of hearts, they keep it face down, else they turn it face up.
Either way, there will now be only two face down cards left.
The player then either chooses to switch to the other face down card or to keep their original picked card, as their final choice.
Reveal and see if the player’s final choice was the King or not!
Repeat steps 2-6 a total of, say, twenty times, first ten times where you will always choose to switch on step 5, and ten times where you will always choose to keep on step 5.
(with 13 cards/doors, twenty total trials should be enough that with extremely high likelihood, you will get the right result even if you get pretty “unlucky” on a few trials).
You’ll likely find your brain very quickly latches on to the correct intuitions once you see the actual results in-person, staring you in the face, over a bunch of trials.