Right, isn't that the point of the question? What old time things did we do for one reason (cloven hooves) that turned out to be right for completely different reasons (health and safety)
frosty99c
joined 1 year ago
I just call it an urban hike. It helps that one of the neighborhoods in my city is up a huge hill, but I'd still call it a hike if I was out walking all day.
Three games
I'd love to see what he does with 100 years of solitude. There are only like 4 names for 15 different characters
In the original the possibilities for a prize behind the doors 1,2,3 are:
A) YNN B) NYN C) NNY
In (A) - A.1 you choose door 1 and then stay, you win A.2 you choose door 1 and switch, you lose A.3 you choose door 2 and stay, you lose A.4 You choose door 2 and switch, you win A.5 you choose door 3 and stay, you lose A.6 you choose door 3 and switch, you win
By staying, you lose in 2 of 3 cases (A.3 and A.5)
By switching you only lose in 1 case (A.2)
It works out for (B) and (C) the same way. You have a 2/3rds chance of winning if you switch and a 1/3rd chance of winning if you don't.
This isn't a trick or anything, the math is pretty clear and you can actually write out all the scenarios and count it up yourself. It's just a little counterintuitive because we aren't used to thinking in terms of conditional probabilities this way.
Another way to think about it is the probability of losing. If the contestant loses, it means that they picked correctly on their first choice and then swapped. This will happen 1/3rd of the games, because there is a 1 in 3 chance of picking correctly the first time. So, if you have a 1/3rd chance of losing by swapping, then it follows that you have a 2/3rds chance of winning by swapping (choosing incorrectly at the start and then switching to the correct door)
The original Star Wars would never have specifically created a character just to sell merchandise...
Do you know the third door is never correct? Because then the probability doesn't change.
Scenario 1: You chose 1/2 at first with a 50% chance of being correct, I introduce a 3rd door (but it isn't a legit possibility), so the actual choice for you is still 50/50 (between doors 1 and 2)
Scenario 2: If you think it's possible that 3 could be correct (but it actually never is) then, no, you wouldn't want to switch. By staying with your first choice has a 50% chance of winning, by switching it only has a 33% chance. But there's no way to know this ahead of time (because as soon as you know you shouldn't switch bc 3 is the wrong door, then you're back in scenario 1)
Scenario 3: For completeness, let's say the 3rd door can be correct sometimes. Then it doesn't matter if you switch or not. It's a 33% chance of winning either way. If there is a chance it can be correct, then your first choice doesn't matter at all and the second choice is the 'real' choice bc that's the only time you're able to choose from all real possibilities.
The only reason that the Monty Hall problem changes probability in the second choice is because you are provided more information before the switch (that the opened door is absolutely not the one with the prize)
Agreed. And I've never read anything quite like The Savage Detectives. His short stories are great too, and you can find a lot of them online published by the New Yorker.
Yes, it's the same concept. The same math/logic behind it doesn't change. You're choosing 1/3 or you are choosing 2/3 and I'll tell you which of the two is incorrect. It's just easier to visualize with 100 doors instead.
I'm not sure I'm following the other angle..there are 3 correct possibilities at the start but I can only choose 2? Or there are 2 possibilities and then you introduce a 3rd door that is never correct?
The original draft probably said "nearly a 300% increase" and then the editor didn't know the difference between percent increase and basic multiplication.