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.
I would have agreed with you a couple of weeks ago, but this video explains it well. It wouldn’t be such a well known fallacy if it wasn’t so counterintuitive.
https://youtu.be/ytfCdqWhmdg?si=bNplB3ftYAfvnLYO
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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.
Yeah it’s very counterintuitive
You’re incorrect. It is indeed a higher chance to switch from #1 to #3. You should look up Monty Hall paradox. It’s in the link that you replied to that explains it.
You can test it empirically. It’s clearly not 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.