/sci/ - Science & Math

[Anonymous No.9328877]

Remind me again why the Monty Hall problem isn't 50%?

[Anonymous No.9328886]

This board is so boring... I'm leaving you /sci/

[Anonymous No.9328899]

>>9328877
With reference to that image, you're twice as likely to land in a half-red than a full-black scenario. That's it.

[Anonymous No.9328905]

>>9328899
Except you're not. You clearly see all the outcomes of the problem. 1/3 of those outcomes (full-black scenario) is not a possible outcome with the conditions set by the problem. From the remaining, possible outcomes, 50% of the time you win by switching, 50% you lose.

[Anonymous No.9328962]

>>9328877
Here's one easy explanation:
Assume you're going on the show and have decided in advance you will always choose to switch.
Using this strategy, here are your only possible outcomes:
1) You pick the door with the car, and switching makes you lose.
2) You pick a door with a goat, and switching makes you win (necessarily so because the host is forced to reveal the remaining goat door after you pick a goat door, and between your initial goat pick and his forced goat pick switching here will always leave you one remaining possibility: the car).
3) You pick the other door with a goat, and switching makes you win (for the same reason explained in option 2 above).
So if you always switch you are in fact guarenteed a 2/3 chance of winning each round.

[Anonymous No.9328972]

>>9328877
There are two goats and 1 car.
Chance of picking a door with a goat behind it: 2/3
Chance of picking a door with a car behind it: 1/3

Alright. Now what happens after you pick the door? The host opens one of the other two doors to reveal a goat.

So if the door you picked has a goat behind it, then since the host revealed the other goat, if you switch you get the car.
And if the door you picked has a car behind it, then obviously if you switch you get a goat.

2/3 times you pick a door with a goat behind it, therefore, 2/3 times you should switch.

[Anonymous No.9329108]

>>9328877
Try imagining 666 doors. Only one car is behind them. The rest are goats.

If you choose a door, the host will open 664 doors with goats behind them. Should you switch to the remaining not yet chosen door?

1st choice:
A: 1/666 you chose correctly
B: 665/666 you chose wrong

In case A:
Switch win probability 0a.
NoSwitch win probability 1a.

In case B:
Switch win probability 1b.
NoSwitch win probability 0b.


Strategy always switch (win):
1/666 * 0a + 665/666 * 1b = 665/666

Strategy always stay (win):
1/666 * 1a + 665/666 * 0b = 1/666

[Anonymous No.9329112]

>>9328877
Its not, its just a thought experiment, don't overthink it

If you look at it as instead of "sticking with your choice" you are making a new one, then its 50/50.

[Anonymous No.9329124]

>>9328905
But the remaining possible outcomes are not all equally likely. [Car A / Pick B / Reveal C] is twice as likely as [Car A / Pick A / Reveal C]. That's because [Car A / Pick A] and [Car A / Pick B] are equally likely, and [Car A / Pick A] is split into the [Reveal B] and [Reveal C] branches, whereas [Car A / Pick B] doesn't branch further (for [Reveal A] is impossible).

[Anonymous No.9329126]

Deciding to stay or switch at random, or alternating, over a large enough sample of games, reveals a 50/50 win rate for either staying or switching

If you play that same set but only switching is used, there is a *higher chance of winning than the same det but only staying.

*higher chance is not guaranteed win

[Anonymous No.9329317]

>>9328877
you change = 2/3 car

you stick = 2/3 goat

[Anonymous No.9329327]

>>9328886
WHAT?!?! the science and math board isn't a roller coaster of engaging content?!?!?! how ever could you have known.

fuck off newfag

[Anonymous No.9329332]

>>9329126
>Deciding to stay or switch at random, or alternating, over a large enough sample of games, reveals a 50/50 win rate for either staying or switching
"No."

[Anonymous No.9329423]

>>9328877
because you only initially pick the correct door 1/3rd of the time, should be obvious from your image

[Anonymous No.9329483]

>>9328877
Because the host only ever reveals a door with a goat.

[Anonymous No.9329511]

>>9329317
The problem I see with this is the chart claims that if you choose a door with a goat behind it and change you automatically win a car which is false. You can change from a goat door to a goat door because there are two of them.

[Anonymous No.9329520]

>>9329511
>which is false
nope - go to the wikipedia page

no wonder you're confused

[Anonymous No.9329559]

>>9329332
If its 66% win when switching and 33% win when staying, it goes to thinking in a set of games where half the contestants stayed and half the contestants switched, the overall win rate would be between 1/3 and 2/3 which is 1/2

[Anonymous No.9329563]

>>9329511
there are 3 doors at the start
you pick 1 could have goat (2/3), could have car(1/3)
Host then opens the door (with the goat) that you did not pick
then see >>9329317

[Anonymous No.9329696]

>>9329511
>You can change from a goat door to a goat door because there are two of them.
You're failing to take into account that the host has to reveal a door after your initial pick, and if your initial pick is a goat then the host is forced to reveal the other goat door.
He can't choose to reveal the door with the car behind it. That wouldn't make any sense.
>>9329559
>in a set of games where half the contestants stayed and half the contestants switched, the overall win rate would be between 1/3 and 2/3 which is 1/2
That's true, but it's not the same thing you wrote here:
>>9329126
>Deciding to stay or switch at random, or alternating, over a large enough sample of games, reveals a 50/50 win rate for either staying or switching
The win rate isn't 50/50 for either staying or switching. It's 2/3 for switching and 1/3 for not switching.
Also I don't know what you mean by:
>*higher chance is not guaranteed win
When did anyone try to argue switching would give you a 100% chance of winning?

[Anonymous No.9329746]

>>9328877
If you don't understand the math behind it, you can easily write like 6 lines or code to run 10^6 simulations and get back the statistical results, which end up being ~66% win rate if you always switch iirc.
I did it once, could post the mathematica code later today if you really want.

[Anonymous No.9329962]

Here's a program that runs the montyhall game 20,000 times and records the results, along with the ability to read and mull over the source code to make sure it's working as you might expect. Just click the </> button to see the code

https://www.openprocessing.org/sketch/480963

[Anonymous No.9329985]

>>9328877
If you have 100 000 doors and then you pick one and the host (knowing where the car is) opens 98 000 of them then its pretty obvious you picked a shit door.

[Anonymous No.9330042]

The monty hall solution ought to be stickied at this point

[Anonymous No.9330047]

>>9329985
why the coprophobia?

[Anonymous No.9330109]

>>9329696
I assumed that the aspect of switching not guaranteeing a win would encourage some people to believe there is no specific benefit to switching which would lead to 50/50 mentality.

For the example of being a contestant who got to play the game once and only once in their lifetime, having a 33% chance of losing by switching wouldn't be to encouraging. You could tell them "STATISTICALLY..." but if they ended up losing by switching they'd just hate you forever.

having a higher chance of winning by switching means relatively little when you only get to play the game once and you could still lose even if you switched.

[Anonymous No.9330117]

>>9328877
I kind of wonder why shows like this or Deal or No Deal even ran. Weren't they worried actually intelligent people would make off with a lot of the prizes?

[Anonymous No.9330131]

>>9330109
Good thing they have a choice to hate themselves if they don't switch (and loose in 2 of 3 cases).

[Anonymous No.9330134]

>>9330117
To fill airtime cheaply of course.

[Anonymous No.9330135]

>>9330134
you could just add trivia questions and lose a lot fewer cars that way

[Anonymous No.9330144]

Because with the switching strategy, if you pick a goat you always win since only the other goat can be revealed, and then you switch into a car.
Probability of choosing a goat is 2/3, so this is your probability of winning.

[Anonymous No.9330146]

>>9330117
>actually intelligent people
>america
>tv game show

pick two

[Anonymous No.9330179]

>>9330144
Exactly.

>>9329108
Strategy choose door randomly at end:
1/666 * 1/2 + 665/666 * 1/2 = 666/666 * 1/2 = 50%

Hmm. Maybe that murders Intuition a bit.

Strategy choose door randomly at end (Original Monty-Hall).
1/3 * 1/2 + 2/3 * 1/2 = 50%

How nice of the Game designers to allow for a 2/3rds chance of winning a car!

[Anonymous No.9330193]

>>9330179
I'm not sure about your math, bud.

[Anonymous No.9330196]

>>9330193
Why?

x% is a Postfix operator that is a shorthand for (x * 1/100)

[Anonymous No.9330199]

>>9330193
Also the 2/3rds was in relation to the "Always switch" strategy.

[Anonymous No.9330219]

>>9330193
> 8 - 9 = -1
> 10 - 0 = 10
> add the two together

[Anonymous No.9330224]

>>9330117
This doesn't apply to Deal or No Deal because you can open any box at any time.

[Anonymous No.9330227]

>>9330193
>1511848397337.jpg
I hate bastards who do write "9" in that way.

[Anonymous No.9330295]

>>9329511
>>9329112
>OP ignores correct responses and samefags retarded brainlet posts

Newfaggots let me teach you something interesting
In the "options" field put the word "sage" and it doesn't bump these retarded threads

[Anonymous No.9330872]

The solution of 2/3 win chance when switching is based on the false premise that you have 3 choices at the beginning.

You don't. The rules of the game pre-determine you to picking one of the two remaining doors at the end of the game. No matter what door you choose at the beginning of the game, it will ALWAYS be one of two remaining doors.

[Anonymous No.9330929]

>>9330872
Please leave, this is beyond even the normal levels of retardation seen on this board.

[Anonymous No.9330947]

>>9330929
So what is your counterargument?

You are not choosing between three (or n number of) doors. You are choosing between two sets of doors. One set contains one door that will under no circumstance be revealed, and the other set contains two (or n number of) doors, one of which (or n-1 of which) will be revealed to be a goat door.

[Anonymous No.9330949]

>>9328877
because you're making the revealss after you pick a door which has the car behind it appear twice as likely as they actually are.

let's say the car is behind door A, and you pick door A. This is as likely as you picking door B or C. if you picked B then C would definitely be shown.
if you picked C then B would definitely be shown.
If you picked A then there is a half chance that B would be shown and a half chance that C would be shown.
So each of those outcomes are half as likely as the outcome where you pick a non-car door and the other goat is revealed.

so instead count each of those wins as half a win and your diagram will give you the right answer.

You're welcome, friend :)

[Anonymous No.9330975]

>>9330947
You're conflating "capable of being revealed by the host" with "being an option you can pick."
The host can't reveal a door with a car behind it, but that doesn't mean you can't pick a door with a car behind it.
1) There are 3 options, 1 car option and 2 goat options.
2) Now go into the possibilities with the constraint that you'll always opt to change your first answer after being shown a door by the host and you get the following:
A) Initial pick is a car door. Host opens another door and shows you a goat. You change to the remaining door which is also a goat. Result: Lose.
B) Initial pick is a goat door. Host opens another door and shows you a goat. You change to the remaining door which is the car. Result: Win.
C) Initial pick is a goat door. Host opens another door and shows you a goat. You change to the remaining door which is the car. Result: Win.
2 Wins / 3 Possible Outcomes = 66.67% chance of winning when always opting to switch.

[Anonymous No.9331007]

>>9330872
Did you read this?
>Strategy choose door randomly at end (Original Monty-Hall).
>1/3 * 1/2 + 2/3 * 1/2 = 50%

[Anonymous No.9331016]

>>9330947
Try calculating the Game with 666 doors and 1 car where the host removes 664 goat doors AFTER you chose one AND THEN GIVES YOU THE OPTION TO SWITCH TO THE REMAINING DOOR.

>>9329108
>>9330179
Strategy always switch (win):
1/666 * 0 + 665/666 * 1 = 665/666 (Choose right & switch -> loss)

Strategy always stay (win):
1/666 * 1 + 665/666 * 0 = 1/666 (Choose wrong & stay -> loss)

Strategy choose door randomly at end:
1/666 * 1/2 + 665/666 * 1/2 = 666/666 * 1/2 = 50%

[Anonymous No.9331082]

>>9329962
THANKS THIS HELPED