[ 3 / biz / cgl / ck / diy / fa / ic / jp / lit / sci / vr / vt ] [ index / top / reports ] [ become a patron ] [ status ]
2023-11: Warosu is now out of extended maintenance.

/sci/ - Science & Math

View post   
File: 2.11 MB, 4032x2268, 1598396421311834884237969756109.jpg [View same] [iqdb] [saucenao] [google]
12048834 No.12048834 [Reply] [Original]

Why doesn't the probability work like this? Why couldn't you represent it with a 3 sided die and then later a coin flip?
Perhaps true answer is that the 1 in 3 chance was an illusion, you always had the 1 in 2 chance but didn't know until after door 3 opened. Seems like common sense though.
If you understand, please represent this in conceptual tokens if possible, if not a coin toss, then what tokens are necessary?

>> No.12048837

Jesus, sorry about the pic

>> No.12048844
File: 2.18 MB, 4032x2268, 15983965674012542100246505682179.jpg [View same] [iqdb] [saucenao] [google]
12048844

>> No.12050680

>>12048844
>tokens
what? no.
total probability across all doors that there is a prize = 1.
initial choice door prob = 1/3.
monty knowingly reveals a goat, so prob on that door = 0.
1=1/3 + 0 + x
x is the prob on the remaining door, and is obviously 2/3.

>> No.12050693
File: 50 KB, 374x382, monty.png [View same] [iqdb] [saucenao] [google]
12050693

>> No.12050868

>>12048834
To understand the Monty Hall problem you need to think around the corner. After opening 1 door you remain with only two left. If you now switch it's basically the same as if you have choosen 2 doors right from the beginning: the one that has been opened already and the one you have switched to. If you don't switch it's basically the same as if all other doors have been opened right away after your first choice (think of it like there is just a minor delay between opening the two doors).

Now ask yourself: by what chance is the car behind 2 doors with 1/3 each?

>> No.12051063

>>12048834
The best way to think about it is to consider more doors. Say you have 1000 doors, 999 with nothing behind them, and 1 with the car. You pick a random door of the 1000, and then good ol Monty says "now I'll remove 998 doors that definitely don't have the car behind them." You had a 1/1000 chance to get the car upon initially picking, and now by switching you have a 999/1000 chance of getting the car and should always switch. Now reduce the number of doors to 3 and you have a similar situation that seems much less intuitive, but you should still always switch

>> No.12051160

>>12051063
>he randomly leaves door 731 closed when opening the 999 doors
>gee I wonder where the ferrari is XDDD

>> No.12051192

>>12048834
The best way of thinking about the monty hall problem (that I have found) is this.
There are three doors:
DDD
Behind one door is a Car, behind the other two are Goats. The door which the Car is behind is random. This means there are three configurations:
1) CGG
2) GCG
3) GGC

The contestant picks a door at random. Lets say door 2.
In scenario 1, a goat is revealed on the right. In scenario 2, a goat is revealed on the left or the right. In scenario 3, a goat is revealed on the left.
If you switch from the door you originally picked (door 2), to the remaining closed door, what are the odds you win a car?
Switch in scenario 1, from door 2 to door 1. You win
Switch in scenario 2, from door 2 to whichever closed door remains. You lose
Switch in scenario 3, from door 2 to door 3. You win

Tally it up
Scenario 1 = win
Scenario 2 = loss
Scenario 3 = win
That's 2 wins in 3 scenarios. Or 2/3.

You can repeat the process for whichever door you like. With as many additional doors as you like