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/sci/ - Science & Math

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6695987 No.6695987 [Reply] [Original]

well, /sci?/

>> No.6695990

Stop

>> No.6695997

>>6695990
no really I'm confused as to how it turns out the way that it does

>> No.6695998

>>6695987
Excuse my ignorance, but am i right in assuming it wouldn't matter at that point? Since it's a 50/50 chance, you might as well flip a coin.

>> No.6696002

>>6695997
Just fucking google it you lazy shit.

>> No.6696010

>Genetics class
>Downtown campus so it's full of pre-meds
>Prof fucked up his flash drive with the lecture on it so he's going to run to his office to get another one
>Since we're doing stuff involving pedigrees and chance TA brings up Monty Hall problem to kill the time
>3/4 of the class doesn't understand it, we spend half the class arguing with them

>Meanwhile during an internship at a science museum I was getting kids understanding it in a couple of minutes

>> No.6696022

>>6695987
The easiest way to understand why it's 33% for this problem is a similar problem.
There are 100 doors. After you choose one all the other doors are opened besides the one you choose and the correct one. What are the chances you choose the correct door on the first guess? 1%, so your chances of the other door being correct is 99% so switching doors is the best idea.

>> No.6696029

>>6695987
Always switch. 2/3 chance of getting the car by switching.

>> No.6696038

>>6695987
Probability the car is behind your choice : 1/3
Probability it's behind the two other doors : 2/3
It's literally like monty the hosts opens two doors instead of one if you decide to change

>> No.6696043

>>6695998
Flipping a coin is not exactly 50/50

>> No.6696048

>>6696043
What's the probability that it lands on the circumference of the coin? Does anybody know?

>> No.6696053

>>6695998
no. it is a conditional problem. you are now endowed with knowing one of the doors has a goat, based on this condition you will want to switch. The opportunity is a 66% now that one of the goats is gone.

>> No.6696066

Easy way to think about it:

You are given 3 doors. 1 contains a car, other 2 contain goats. You may pick a door.
>You pick #1
Now you may switch and pick both #2 and #3 or you may keep #1
>You're not an idiot, so you decide to switch and get #2 and #3.
There's a 2/3 chance you'll get a car.

In the original problem, the host non-randomly showing you a goat from the set of doors you DIDN'T pick makes you effectively get to pick both doors.

>> No.6696077

If you don't get to see what's behind any doors after you pick one and get the chance to change, it's still 2 to 1 against.

>> No.6696082
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6696082

>>6696048

>> No.6696100

>>6696038
>>6696066

Assume we're picking door 1 and door 3 is opened. Why can't we say:
Probability car is behind either door 1 or door 3: 2/3
Probability car is behind door 2: 1/3
Since we know the car isn't behind door 3, the probability it's behind door 1 goes up to 2/3, so you shouldn't switch.

This effectively asks: why is the grouping of doors 2 and 3 NOT arbitrary, and why can't you group doors 1 and 3 together?

>> No.6696106

>>6696077
I may have misunderstood what you said, but:

If you aren't shown a goat on one of the doors, changing will not help your chances. If you pick door #1 and are shown that #2 contains a goat, you can pick #3 with 2/3 chance. However, if you don't know #2 is a goat then you could change to #2 or #3,making your probability of winning 1/3 for each door.

>> No.6696131

>>6696100
If #2 is the car, the host HAS to open #3. If #1's the car, the host has a choice. So, #2 can affects #3 being shown but #1 can't.

>> No.6696136

>>66961
a good question but you assumed that door 3 was opened which makes the probability drop to 1/3 like you said . If you consider the case door 2 is opened it gives you another 1/3 and 1/3+1/3=2/3.

(Draw a tree if you need to convince yourself)

>> No.6696141

your chances of losing before choosing a door is 66%

which means the chances of you picking a goat is 2/3

Assuming you pick the goat since the odds are in that favor, the host will now show one of the doors that is definitely a goat.

Holding onto the fact that initially your odds of picking a got were 66%, we have to assume you've picked the goat, but now the 2nd goat has also been exposed.

so you now switch doors not so much because your odds of picking the car are higher (which is true) but because you're now leaving behind what was probably the other goat.
didnt this shit always used to be posted as troll threads?

>> No.6696145

>>6696136 was meant to quote>>6696100

>> No.6696167

I still don't understand this after watching videos about it and reading about it for an hour now.

I guess this is why I'm a CS major.

>> No.6696191

>>6696167
initially select one door: 1/3 odds of having the prize behind it.

Monty opens a door you didn't select: that choice is now out of the picture.

Total probability that the prize is behind any door = 1.

Prob_total = Prob_initial + Prob_remainingdoor
1 = (1/3) + Prob_remainingdoor
Prob_remainingdoor = 2/3.

>> No.6696196

>>6696191
>Total probability that the prize is behind any door = 1.

Maybe better stated as total probability that there is a prize behind one of the doors. There is a prize in there somewhere.

>> No.6696205

>>6695987
>there are people on /sci/ who haven't encountered this problem

>> No.6696210

>>6696167
What's not to understand? At the very beginning, the probability of choosing the right door is 1/3 and wrong door is 2/3. The fact the host opens a door doesn't impact the fact that you had 1/3 chance of choosing the right door.

Imagine you had 1000 doors and there is only one car behind the 1000, all the rest are goats. The probability you choose the right door on the first try is 1/1000. The host then opens 998 doors that have goats behind them. The fact he opens the doors doesn't change the fact that the probability of you being right is 1/1000. The chance that the other door contains the car is 999/1000. Switch.

This make any sense?

>> No.6696270
File: 16 KB, 555x295, 555px-Monty_tree_door1.svg.png [View same] [iqdb] [saucenao] [google]
6696270

>>6696167
>>6695987
just make a tree diagram the way they taught you in sixth grade
i'm sure that's fresh in your mind

>> No.6696360

>>6696270
Three ( or N ) eggs are in the bowl.
Only one of them is hard boiled.
You select one egg, put it on the table.
The chance that the table egg is hard boiled is 1/3 ( or 1/N ).
The chance that the bowl has the hard boiled egg is 2/3 ( or (N-1)/N ).
The show host breaks one egg ( or N-2 eggs ) from the bowl.
The chance that the bowl has the hard boiled egg is STILL 2/3 ( or (N-1)/N ).

note:

1)
The host does NOT open a random door, the opened door always reveals a goat.

2)
If the problem had 10 doors instead of 3,
after you pick a door, the host would open 8 doors instead of only 1 door.

>> No.6696546

>>6696167
Perhaps it helps if you realized this is the same problem as a situation where you buy some product and later it is revoked.

For example you buy a carton of milk. Later, before you drink it, you find out a large portion of the milk in the store you bought your carton was poisonous. The store owner has tossed out most bad milk and asks if you want to change your carton. Would you say 'nah man, the probabilities stay the same?'