Your are viewing a read-only archive of the old DiS boards. Please hit the Community button above to engage with the DiS !
Okay, I've read it, and I think I get it. I think. I'm not convinced I get it.
"If you stick with your first choice, you will end up with the Caddy if and only if you initially picked the door concealing the car. If you switch, you will win that beautiful automobile if and only if you initially picked one of the two doors with goats behind them."
so - because you're twice as likely to have made a mistake first time (compared with switching), you're twice as likely to be correct by switching?
you have a 1/3 chance of having picked the car
2/3 chance you did not
if you open a door with a goat, it's still only a 1/3 chance you picked the car in the first place (2/3 you did not) so switching swings the odds in your favour
you'd keep the goat.
I'll stick with the goat.
you win the same stuff, except one of them isn't a prize, you just switch from 1 door to two doors and it gives you a better chance
if you have 30 random people, there is a 70% chance that two of them will have the same birthday
and she said "yeah, but if you switched and lost you'd be fucking raging"
Best get to it.
You're all missing the main headline news http://www.bbc.co.uk/news/uk-england-lincolnshire-24047706
thanks to evolution it'll probably retain that title for a few years to come
no one has filmed it (or even seen it perhaps) in its natural habitat, so we don't really know what it looks like.
in the first instance you're 33.3% of being right. then what?
in which case you've got a 70% chance of having the same birthday as Monty Hall
and Monty Hall is replaced by Nicholas Parsons
It's not really a maths problem, it's a logic one.
so odds are 1/3 that you'll get the car, and 2/3 that you've got the goat.
Monty opens a door and shows a goat
If you picked the car and switch, you'll get a goat. The odds of this (as above) are 1/3.
If you picked a goat, Monty must have revealed the other goat, so the remaining door must have the car behind it. Again the odds of this are 2/3.
Therefore, logically speaking, you maximise your chances by switching as you've no way of knowing whether you picked a goat or the car up front.
"so odds are 1/3 that you'll get the car, and 2/3 that you've got the goat."
I take that back. It is about maths.
(To be honest, shouldn't have tried to explain something while tired)
and i'm pretty sure it's been done on here before anyway
unless it's on a poker forum
as the elimination of balance
there is no goat
or do you want to change to this one?
Imagine it was a 1 in 100 choice. You pick number one, then he rules out all the other wrong boxes apart from number 48. The odds are clearly in your favour to switch to box 48, unless you magically chose the right box on your first pick. The only reason it is hard to get our heads round is because it is only 1 in 3, so it only very slightly nudge it in your favour by switching. And you would feel a right twat if you smugly changed your mind and then turned out to be wrong. I think Mr Logic did this in a Viz strip actually, went on Deal or No Deal and left with 1p.
The rules are still the same. Behind one of the 1000 doors is car and behind the other 999 doors are chickens. The contestant has to pick one door at random. Then the host has to open 998 doors out of the 999 not chosen by the contestant. The host offers the contestant the choice of sticking with their door or switching to the host's single unopened door. The aim is to win the car.
Now let's look at The rules for the host choosing which doors to open
1) If the contestant has chosen the door with the car, then the host can open any 998 doors leaving a single door with a chicken
2) If the contestant has chosen a door with a chicken, then the host MUST open 998 doors with the chicken leaving the door with the car closed.
So, should the contestant stick or switch?
Well, the probability of the contestant choosing the winning door is 1/1000. Not good odds. The probability of the host having the winning door is 999/1000. However, the host has made it easier for the contestant by opening all his doors except for the one with the car. That single door represents all the other doors and there is a 999/1000 chance of a car behind the host's door.
Therefore, the contestant should switch as the probability of winning the car goes from 1/1000 to 999/1000. That is, the contestant is 999 times more likely to win with switching than sticking. (999/1000 is 999 times bigger than 1/1000)
Let's redo the thought experiment, but with 100 doors. The probability of the contestant choosing the winning door is 1/100. The probability of the host having the winning door is 99/100. If the contestant switches, the probability of winning the car goes from 1/100 to 99/100. That is, the contestant is 99 times more likely to win with switching than sticking. (99/100 is 99 times bigger than 1/100)
Again, but with 10 doors. The probability of the contestant choosing the winning door is 1/10. The probability of the host having the winning door is 9/10. If the contestant switches, the probability of winning the car goes from 1/10 to 9/10. That is, the contestant is 9 times more likely to win with switching than sticking. (9/10 is 9 times bigger than 1/10)
Finally, 3 doors......The probability of the contestant choosing the winning door is 1/3. The probability of the host having the winning door is 2/3. If the contestant switches, the probability of winning the car goes from 1/3 to 2/3. That is, the contestant doubles their chances to win with switching than sticking. (2/3 is twice as large as 1/3).
I don't like Monty Hall problem because it is a simple and exact maths problem deliberately obfuscated by the fuzziness of language. It's only purpose to catch you out because the problem has not been fully explained and you use the ambiguity of language to fill in the gaps and make personal assumptions. The person posing the problem can then show off. The Stephen Fry of maths problems.
a real feature from a gameshow where people are allowed to stick or switch and the idea is to work out which option gives you the highest EV. where is the fuziness of language?
If want to win a goat?
because without the agency of the host knowing what is in the other box, it is not really a Monty Hall problem, right?
there is room for 'good' decision making in deal or no deal though. but all you do whenever offered something is add up the totals in the remaining boxes, divide it by the number of boxes left and compare it to the offer. depending on the situation people might give up some EV for the chance for a bigger prize (which is how i justify occasionally playing the lottery) but there's not any more to it than that. the last switch decision is just 50/50