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Earlier quoted context omitted.
I'm willing to forgive pretty much anyone who's fooled by something that also fooled Paul Erdos: https://www.wired.com/2014/11/monty-hall-erdos-limited-minds...
Something tells me Erdős didn't resort immediately to personal insults when faced with the problem, though.
Another statistically unintuitive problem (which I've witnessed a lecture hall enter a state of uproar over): There are 2 red and 2 blue balls in a box. One ball is removed at random, what are the odds that the ball is blue? Now we repeat the problem, but before examining the ball, we remove a second ball. We observe that the second ball is blue. In this case, what are the odds that the first ball is blue?
A neat trick to reason about those cases is to use odds. The prior odds for the first ball being blue vs red is 1:1. The odds for the first ball being blue vs red, given the second ball is blue is 1:2. We can just multiple the odds to get (1*1):(1*2) = 1:2 as the posterior odds.
This doesn't seem impressive because the prior is 1:1, but using this method you can easily calculate the odds in the scenario where there are 4 red and 2 blue balls. The prior odds is 2:4=1:2, the conditional odds is (1/5):(2/5)=1:2, 1:2 * 1:2 = 1:4, i.e. 1/5 chance that the first ball was blue.
I may be missing something (not a mathematician, probabilist) but doesn’t the fact that the host would never reveal the car change the calculation? In the table shown in the article game 3 & 6 are nonsensical so with the remaining 4 games the odds are even between switch & stay.
So this is at least in part about the "Monty Hall Problem" and why it's solution not intuitive. The article missed an important angle: when the host opens a door, he's giving you more information , which explains why it's better to switch. If you're the host, you need to know which door the car is behind to do your job 2/3 of the time, to avoid revealing it. It's this quality of unexpected information exchange that I…
When you choose the first door, you are partitioning the "board" into two parts: the door you chose, and "everything else". By opening a door, Monty lets you cover 100% of the "everything else" partition using only one guess. So now you get to choose between partition 1, which covers 1/3 of the board, and partition 2, which covers 2/3 of the board.
that 71% leaves 29% of _academics_ not getting the elementary math of the problem. I'm baffled this is _that_ hard?
Earlier quoted context omitted.
Something tells me Erdős didn't resort immediately to personal insults when faced with the problem, though.
I'm not sure about that since Erdos had the typical pre-GenZ-wokeness vocabulary of a nerd, where everything they said was worded as an insult to people less smart or with different hobbies than him. To the point where it's in his Wikipedia article: > Women were "bosses" who "captured" men as "slaves" by marrying them. Divorced men were "liberated". > People who stopped doing mathematics had "died", while people who…
Probably, because the husband-based humor was the same.
Earlier quoted context omitted.
I don't understand how this could be. If you start over when the host picks the car, then isn't that the same as the host picking the goat every time, i.e. the same as the host knowing.
Fair warning: I haven't rigorously verified the math on this, and tfa is all about the importance of vigorously verifying the math. In the case where the host revealing the car means you restart, there are three equally likely situations after the host opens a door: - You picked the car, and the host revealed a goat - You picked a goat, and the host revealed the other goat - You picked a goat, and the host reveals th…
Assuming, WLOG, that we choose door 1, that leaves us with 6 equally likely cases just before that final correction (or lack thereof):
A) Car 1, Monty 2
B) Car 1, Monty 3
C) Car 2, Monty 2
D) Car 2, Monty 3
E) Car 3, Monty 2
F) Car 3, Monty 3
An intuitive way to think about the problem is from the perspective of the host. Sometimes the host has 2 goats he can pick from to show, and sometimes he only has 1 goat he can show. 66% of the time the contestant chose a goat, so the host has to reveal the only other goat 66% of the time. So switch.
Earlier quoted context omitted.
Interestingly, if the host picks randomly (and if he reveals the car, you... start over, or you get the car, or you get nothing, or ... it doesn't matter because it happens not to have happened in the time we're considering) then you are faced with a 50/50 chance.
Yeah - and at least for me this was part of the confusion. The problem is often stated incorrectly. Critically the host always removes a goat. If the person presenting the problem just says “the host opens one of the doors”, but doesn’t specify he’s always revealing a bad door then it’s not clear why that matters. The many door example makes it easy to intuit as well.