I can accept the fact that on a Roulette wheel (as long as there are no defects or imbalances in the wheel or ball) that the odds are the same each spin and previous spin outcomes have no influence over the current spin. However, if I see black come up 32 times in a row I am betting on red for the next spin.
After you find out there’s a goat behind door #2, you have a 50% chance whether you stay on 1 or move to three. There are only two possible outcomes at that point (car or goat), so either way it’s a coin flip.
You can test it empirically. It’s clearly not 50%.
There are only two options at that point. It MUST be 50-50.
People will claim that you’re wrong, but you’re 100% correct. It’s always 50-50. You either win, or you don’t.
The important thing is that the host will always show you a goat, meaning the only way the other door has another goat is if you just so happened to pick the car the first time.
Take the situation to the extreme and imagine a hundred doors, and after you pick a door, the host opens 98 doors, all of them with goats behind them. Now which seems more likely, that you chose right the first time, or that the other door has the car?
My comment is clearly sarcastic…
You’re wrong, but you’re in good company. It’s a very counterintuitive effect. One technique that can be helpful for understanding probability problems is to take them to the extreme. Let’s increase the number of doors to 100. One has a car, 99 have goats. You choose a door, with a 1% chance of having picked the car. The host then opens 98 other doors, all of which have goats behind them. You now have a choice: the door you chose originally, with a 1% chance of a car… or the other door, with a 99% chance of a car.
Oh that’s so weird. I get it from a proof perspective but it feels very wrong.
My brain tells me it’s two separate scenarios where the first choice was 99:1 and after eliminating 98 there’s a new equation that makes it 50:50.
Now you have 2 choices: the door you chose, or the only other door left. One has a goat and one has a car. That’s fifty-fifty.
In your explanation, the door originally had a 1% chance, but after showing 98 goats, it has a 50% chance.
No. Taking it to the extreme with 100 doors, your first pick was a 1% chance to get the car. The host then shows you 98 other doors that all have goats.
What’s more likely? That you picked the right door when it was 100:1? Or that the other door is the one with the car?
There are 2 choices so they are equally likely
One of those choices is, itself, one of two choices. So it’s secretly three, but two of them look identical.
They’re really not.
The important thing is that the host will always show you a goat, meaning the only way the other door has another goat is if you just so happened to pick the car the first time.
Take the situation to the extreme and imagine a hundred doors, and after you pick a door, the host opens 98 doors, all of them with goats behind them. Now which seems more likely, that you chose right the first time, or that the other door has the car?
The host’s intentions are irrelevant. Numerically, there are only two choices. That makes it fifty-fifty.
You think that even in the hundred-door case? Test it. Hell, even test it in the 3 door case. It is empirically not 50%.
If the host had an even chance to show you either door, you’d be right, but since the host always shows you a goat, the two events (picking a door and choosing whether to switch) are no longer independent, since if you pick a goat it forces the host to pick the other goat.
The important thing is that the host will always show you a goat, meaning the only way the other door has another goat is if you just so happened to pick the car the first time.
Take the situation to the extreme and imagine a hundred doors, and after you pick a door, the host opens 98 doors, all of them with goats behind them. Now which seems more likely, that you chose right the first time, or that the other door has the car?
Your first paragraph made it click. Thanks!