Game One: You are given a choice of three doors. You pick number one. The host opens one of the other two doors, having been given instructions that, if you pick the car, the host is to open one of the other doors, and if you pick a goat, the host opens the other door with a goat. Stalemate. It is a predetermined outcome.
Game Two: The prior game’s outcome stands. The new choice you have is do you keep door number one, or do you switch?
How do you have a 2/3 chance of winning if you switch?
In: 107
It helps me to think in terms of sets, rather than the three individual options.
There are three choices, A,B and C. You choose one, it doesn’t matter which. You can think of the three choices now being divided into two sets: the ones you chose and the ones you didn’t choose. It’s pretty obvious the set of ones you did not choose is twice as likely to contain the prize. It’s twice as big.
Then Monty Hall reveals one of the two in that Did-Not-Choose set. The 2/3 likelihood that the prize is in that second set does not change, it simply collapses down now to the one unrevealed choice in the Did-Not-Choose set. The set still contains 2/3 of the probability…we just have more clarity where the prize must be, if it’s inside that’s set.
Isn’t this simply a second game?, you asked. I get that. It sure feels that way. And if Monty Hall revealed one of the two options in that set at _random_ then yes, that would be the case. But he does NOT choose at random. He always only intentionally reveals “goats.” A game of chance is not a game of chance when one of the players is intentionally crafting the outcome.
This feels more intuitive if we change the numbers. Instead of three choices to pick from, let’s make it 100. You make your choic. Let’s say you pick #1 out of the 100. It’s 99% likely that the prize is in that huge Did-Not-Choose set.
Monty knows which one is the prize, and which of his 98 others are goats.
_So he reveals 98 goats._ There’s only one option left in the Did-Not-Choose set.
That 99% probability still exists. It’s just been concentrated, purified, into that one remaining option in his set. If it was 99% likely before, it’s still 99% likely now.
But your final choice is still between your set and his set. I’d go for his set.
Here’s my ultra-simple explanation. If I told you you could pick two doors would you agree you’d now have a 2/3 chance to win? If that makes sense to you — here’s how you pick two doors. Let’s say your choice is doors 1 and 2 (don’t tell Monty). Start by picking door 3. Monty will now open one of your doors which doesn’t have the prize — just pick the other door.
Imagine you have 100 doors. You pick one, and the host opens up 98 other doors. You now have a choice to switch to the other door the host didn’t open.
Imagine this game, but with a different phrasing.
You pick one door at random. So you have a 1/3 chance of picking the car.
The host points to one of the other two doors, and says “the car is in here”. If you accidentally picked the car (1/3 chance), they are lying to you. If you picked a goat (2/3 chance), they have to tell the truth.
Statistically, this game is identical. Since the host is not allowed pick the car, they are suggesting semi-honestly that the door they keep closed is the one with the car.
Something that really emphasizes the way the math works is to imagine you have 100 doors instead of 3.
Say you pick door 23. The host then opens all the doors the car isn’t behind, going one by one. 1, 2, 3, 4, on and on, all empty. He skips your door then keeps opening empty doors. 24, 25, 26, on and on. But then he skips door 72 and finishes opening doors. Would you switch to door 72? Obviously.
The same logic applies, albeit much less obviously, to 3 doors.
The second game contains additional information left over from the first game.
The important thing is that *Monty knows where the car is* and uses that information when selecting which door to open. Some of that knowledge flows to you.
A chart I made once:
Car | You choose | Monty opens | You switch | Don’t switch
———|———-|———-|———-|———-
1 | 1 | either 2 or 3 | lose | win
1 | 2 | 3 | win | lose
1 | 3 | 2 | win | lose
There are two common erroneous assumptions made about the problem as presented. The first is that the host acts randomly when host has full knowledge. Second, the host does not open one door, the host opens all unselected non-winning (goat) doors.
This is shown more clearly and is more intuitive when the example is expanded:
The contestant is presented with 100 doors, there are 99 goats and 1 car behind them. They pick #3 (1/100 chance). The host shows all doors except #88 is goat. Now, does contestant stay with #3(still 1/100 chance) or switch to #88, effectively picking all other doors (99/100)? There is no chance the host chose a goat behind #88 unless there was no car available (because it was behind #3).
Let’s say there are a hundred doors you pick one, the host opens 98 doors. Do you still trust your one in a hundred guess?
Think of it as rolling a three-sided die (a 2d triangle with the selected “side” being the edge pointing up), since you don’t really have any information at first, so it’s random. When you roll said die, there’s only a 1/3 chance the car will be picked, therefore, it’s more likely that it’s in one of the doors you didn’t choose, or in this case, in one of the sides in the base. The important thing is: The host will NEVER reveal a side (door) with the car in it, always a goat, so, if there was a 2/3 chance the car was on the bottom, that didn’t change, since it is still a three-sided die, no matter how many or which of those are revealed or hidden, therefore the chance you picked wrong is still 2 in 3, so the real question the host is asking, probability-wise, when they ask if you want to switch, is if you want to invert the game and bet the car is in one of the base sides.
Try thinking of the same situation with a 6-sided die, and you’ll see that it starts to become more and more likely that the car will not be on top, so the safest bet is to say it’s not the one you chose.
Imagine 100 doors instead of 3.
You pick one door which means you have 1% chance of picking the correct door. The host decides to even the field by opening 98 doors. Now there are 2 doors left, that other door has a 99% chance of being the correct door. Basically your original % chances don’t change, it’s only through the act of leveling the field that makes changing doors a higher %
Think of it backwards instead. Remember the host knows which door has the price but you don’t. So the host let’s you open a total of 99 doors, so you pretty much have a 99% chance of nailing the right door. Another way to see this is knowing the fact that the host will never open the correct door, so everytime he opens 1 door it adds 1% to the door he didn’t pick. The only way to lose this is if you stick to your original door [1% chance] or if you were extremely un/lucky in having picked the correct door at 1%.
Basically it’s best to switch doors either way as that gives you a higher %
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