/sci/, I very legitimately don't understand the Monty H

>>8007895
there's three possible outcomes: {goat1, goat2, car}

if you picked a goat (2/3) then switching will get you a car whereas if you picked the car (1/3) switching will get you a goat.

So in 2/3 of possibilities switching will get you a car.

Actually, here's another beef I have with probability. Supposedly, with a hypothetically perfectly random coin, every time you flip the coin, there's a 1/2 chance of heads and a 1/2 chance of tails. EVERY time you flip the coin, regardless of what you've flipped previously.

And yet, to calculate the chance of flipping 3 heads in a row, one uses (1/2)*(1/2)*(1/2), which is 1/8. So on the first flip, there's a 1/2 chance of getting heads, then, on the second flip - since the chance of getting two heads in a row is 1/4, surely the chance of getting tails is 3/4?

What am I missing here? A fundamental understanding of probability, I guess.

>>8007895
You can easily google this to get your answer.

The entire point to this thought experiment is that the person that opens a door after your initial choice knows what is behind each door. Given that he knows what is behind each door and won't open the only door you don't want to select, even a grade schooler can draw a simple diagram for each case and conclude that you're better off switching doors. If the person opening a door for you doesn't know what is behind any of the doors, it's a 50/50 chance, given that he doesn't open the "losing" door after your initial choice, in which case you will win 100 % of the time unless you're literally retarded.

>>8007909
>the chance of getting tails is 3/4?
the odds of getting at least _one_ tails in 2 flips is 3/4

>So on the first flip, there's a 1/2 chance of getting heads, then, on the second flip
it's not "on the second flip" it is "two flips in a row".

You're mixing up two flips which has a sample set {HH, HT, TH, TT} with a single flip {H,T}

>>8007895
Either you pick the right door or the wrong door, it was 50/50 the whole time. ;)

>>8007895
So, you pick a door. There's a 1/3 chance you got the car, and a 2/3 chance you got the goat. Kind of sucks. Then the host (who knows what's behind every door, this is crucial) opens a door (not at random, more on this next) to reveal a goat and offers you the chance to change. So, he knows what is behind each door and he is guaranteed to reveal a goat. If you picked the car (1/3 chance) hen he has 2 doors he can open. If you picked a goat, then he has one door he can open that will reveal a goat. No consider what will happen if you change your choice for each scenario.

Scenario 1: you picked the car, so if you switch, you would not get the car.
Scenario 2: you picked a goat, if you switch you get a car. Now, because this is 2x more likely (there's a 2/3 chance you start with a goat), then there is a 2/3 chance that switching will result in you getting the car.

Another way to look at it is to take scenarios 1 and 2 and add scenario 3 (where you pick the other door that has a goat behind it). You see hat 2/3 times switching gets you the car.

>>8007909
In this case, the probability is just the permutations available.

H:=Heads
T:=Tails

One throw:

P(H) = .5
P(T) = .5

Two throws:

P(xx) = .25, because

P(HH) + P(HT) + P(TH) + P(TT) = 1 (all possible permutations), where P(HH) = P(HT) = P(TH) = P(TT) = .25.

If you expand that to three throws, you get 2*2*2 = 8 different outcomes, and since P(H) = P(T) = .5 for any one throw, all 8 outcomes are equally likely.

>since the chance of getting two heads in a row is 1/4, surely the chance of getting tails is 3/4?

Assuming that you meant "surely the chance of getting two tails in a row is 3/4", you forgot that all probabilities must sum to 1. Would you claim that there are only two possible outcomes from throwing a coin twice, the first being HH and the second being TT? I am sure you can imagine HT and HH, which gives a total of 4 outcomes with equal probability, so each probability must be 1/4.

>>8007909
> since the chance of getting two heads in a row is 1/4, surely the chance of getting tails is 3/4?

If past flips influenced the next flip then this would be true. However, they don't. Each time you flip the coin there is a 1/2 chance of heads/tails. If you flip a coin and get heads (this is a given) then there is a 50% chance you get heads on the next flip. Now, yes, there is a 1/4 chance of getting two heads in a row, but if you are already guaranteed heads you don't even consider the probabilities related to the first flip (so 1/4 goes out the window because it considers both flips when you only want to look at the probability of your next flip).

>>8007895
>But anyway, my beef with it is this: When you are asked whether or not to switch doors or stay with your choice, isn't that a fundamentally different probability set than when you were first asked to choose?
What do you mean by set? Event space perhaps? Anyway, yes obviously they are two different situations with different probabilities.

>I get that the idea is when you first choose, you only have a 1/3 chance of getting it correct. By switching your answer after the reveal, you raise your chances to 1/2, since there are now only 2 doors to choose from.
Raise your chance of what? The probability you chose a car is still 1/3. But now that Monty has removed one of the goats, by switching you are guaranteed to get a goat if you chose the car, and you are guaranteed to get a car if you chose the goat. Since it's more likely you chose the goat (2/3 chance), you are more likely to get the car if you switch.

>You're presented with 2 doors, you have to choose one. Choose to switch, or choose to stay, you've chosen among 2 doors all the same - shouldn't your chances be 1/2?
This is a very common mistake that amateurs make. You shouldn't assume all events are equally likely. As I already explained, the chance you originally chose a goat is higher than the chance you chose a car, and that's why switch is better.

File: nyc-subway.jpg (109KB, 479x641px) Image search: [Google]

109KB, 479x641px

>>8007895
That has to be the easiest equation ever. Do people get this wrong?

>>8007907
>>8007909
>>8007911
>>8007915
>>8007927
>>8007928
If you have already picked a door and got a goat then the worst you can end up with is a fucking goat. You pick another door without question because you're already a loser faggot!

This shit is embarrassing.
>>8007931
>>8007944

>>8007895
OP, think about it this way

There are 100 doors. You choose one and the host opens 98 doors all with goats, is it wiser to switch?

It's the same concept

>>8007979
U wot? Yeah, if you've picked a door with a goat behind it switching is beneficial to you. If you picked the door with the car behind it switching is not beneficial. Because there is only a 1/3 chance you picked the car originally (2/3 chance you picked the goat), there is therefore a 1/3 chance of switching hurting you and a 2/3 chance of switching getting you the car.

It's the same concept as the problem where you pick two gold beads.

>>8007927

Excellent explanation!

>>8008053
Thanks bro, did it help?

>>8007895
>I think I don't understand probability in general.
OP I can confirm that failing to understand the Monty Hall problem does directly imply that you don't understand probability in general. I hope that clears things up.

https://www.youtube.com/watch?v=9vRUxbzJZ9Y
i posted it again

>>8007895
Probability(to choose winner out of 3) = 1/3
Probability(to choose goat out of 3): 2/3

P(win if Switch | if you have already chosen the Winning door) = 0
P(win if not Switch | if you have already chosen the Winning door) = 1

P(win if Switch | if you have chosen a goat door and the other goat door has been opened) = 1
P(win if not Switch | if you have chosen a goat door and the other goat door has been opened) = 0

We are interested about our win probability if we always switch:
Path 1: P(to choose winner out of 3) * P(win if Switch | if you have already chosen the Winning door) = 1/3 * 0
Path 2: P(to choose goat out of 3) * P(win if Switch | if you have chosen a goat door and the other goat door has been opened) = 2/3 * 1
Total probability: 1/3 * 0 + 2/3 * 1 = 2/3

Winning probability if we always stay:
Path 1: P(to choose winner out of 3) * P(win if not Switch | if you have already chosen the Winning door) = 1/3 * 1
Path 2: P(to choose goat out of 3) * P(win if not Switch | if you have chosen a goat door and the other goat door has been opened) = 2/3 * 0
Total probability: 1/3 * 1 + 2/3 * 0

>>8008432
Winning probability if no door has been opened AND
Assume you have to switch to a different door (chosen by you):

1/3 * 0 + 2/3 * 1/2 = 2/6 = 1/3
(you have to discard the winning door OR switch to the winning door with 50% chance)

Assume you have to stick with your chosen door:
1/3 * 1 + 2/3 * 0 = 1/3
(you have already chosen right OR you loose)

Ergo it would not matter, which strategy is chosen by you, if the goat door is not eliminated.

>>8007895
Imagine you have 100 doors instead of just three. You'll agree that your initial chance to pick the right one is 1/100. Then 98 losing doors are revealed to you, leaving only your originally picked door as well as just another door. Do you think that the chances for the prize being hidden behind the original door OR the other one are equal?

>>8007970
Almost everybody gets it wrong when first shown, regardless of education level. There is an inbuilt bias that trips you up

>>8008458
Interestingly, even Paul Erdos was opposed to the correct solution. The proper mindset to understand the problem just didn't click for him for some reason. Though, a simulation finally convinced him, showing that a switching tactic would indeed approach a 2/3 success rate.

>>8008476
Its by far the most interesting part of the whole question, just how many people get it wrong