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/sci/ - Science & Math

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4025607 No.4025607 [Reply] [Original]

This shit is 2/3 right /sci/?

I feel like a bunch of people are telling me to ignore the fact you had a higher chance of getting the box with 2 gold coins in the first place, and I'm starting to feel retarded because I'm saying you have to take into account the original choice

>> No.4025645
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4025645

Oh boy, here we go.

Yes you're right. The naive answer would be 1/2, since there seem to be "only 2 cases left", but the right answer is >1/2 because the box with two gold coins was more likely to be chosen.
Such questions are controversial however, since you can't reproduce them properly. That is you can't actually do the experiment (if you try one time), because you can't be sure to not draw a silver coin the first time.
Whatever, the basian logic still says >1/2 and this is the right one from this perspective.

Highly related:
http://en.wikipedia.org/wiki/Boy_or_Girl_paradox

Interesting:
http://en.wikipedia.org/wiki/Sleeping_Beauty_problem

>> No.4025719

>>4025607
Six situations:
1: You pick GG and pull out gold
2: you pick GG and pull out the other gold
3: You pick GS and get gold
4: GS but silver
5: SS silver
6: SS silver

If you know you got a gold coin, theres a 1/3 chance you got it from the GG box, but only a 1/6 chance you got it from GS

Correct me if I'm wrong

>> No.4025726

>>4025719
HOLY FUCK, when did /sci/ learn statistics?

>> No.4025732

>>4025726
Is that wrong?

I wasn't giving the answer the problem, I was just saying yes, you're twice as likely to get gold from GG as from GS

>> No.4025739

http://en.wikipedia.org/wiki/Bayes'_theorem

>> No.4025740

>>4025719
If you drew gold, then it must have come from either GG or GS. So *given that the first draw is gold*, the probabilities of having picked GG or GS must add to 1. Before any draws though, there's a 1/3 probability of drawing G from GG and 1/6 of drawing G from GS.

>> No.4025762

>>4025740
Ah yes.

Well I'm retarded, but getting GG is still more likely correct? if not, then well I have Calc 2 next semester, but maybe I can sit Stats in over the summer

>> No.4025772 [DELETED] 

We had a similar thread the other day.
2/3 is correct.
http://archive.installgentoo.net/sci/thread/4011909

>> No.4025798

I'm getting 1/3.
Since picking a gold coin is garaunteed, we only have to consider they original choice of three boxes.

1. You pick the GG box, you have one gold left
2. You pick the GS box, you have one silver left
3. You pick the SS box, you have one silver left

Only 1 is true, so it's 1/3. Correct me if I'm wrong.

>> No.4025822 [DELETED] 

>>4025798
No.
You know for certain that you have not picked the SS box.
You have 2 chances to pick a gold in the GG box, and 1 chance to pick the gold in the GS box.
Therefore the GG box is twice as likely for you to have picked gold from.

>> No.4025847

>>4025822
Not really though. The box is chosen at random, and only requires ONE gold coin to be present, any duplicates doesn't add to the odds of picking that box, since it only needs to hold ONE to be a legitimate choice. I do however agree on the SS case.

Let's say the content of the boxes are unknown, and with only two boxes remaining, the gold coin case is now null. Ergo, it's 1/2.

>> No.4025857

2/3

If you ended up with one gold coin, there is a 1/6 chance of a G/S box and 2/6 chance of a G/G box.

2/(1+2) = 2/3.

>> No.4025863

>>4025798

Wrong. This isn't even a troll question, it's actually answerable and requires some thought.

It's either 1/2 or >1/2

The thing is, first you pick a box. (****that's important)....

Th GIVEN here is that you did not pick the SS box.

Then, among those two boxes left, given you drew a gold, what are the chances it came from the GG box and what are the chances it came from the GS box.

Clearly it is more likely to have some from the GG box.

2/3 sounds right but it actually may be more complicated than that.

Need a pencil up in this bitch.

>> No.4025890

>>4025863
I understand the explanation and argument for 2/3, but I'm arguing that the box selection is SEPARATE from the coin picking. Meaning, a box with 10000 gold coins is as likely as a box with only two (Since one must remain). Or:
(Number of boxes containing at least two gold coins)
divided by
(Number of boxes that do not)
Or in this case 1/3. OR 1/2 if you decide that the SS box is invalid considering the phrasing of the question.

>> No.4025892 [DELETED] 

>>4025847
It is true that the box is chosen at random, but the fact that gold was the coin that was selected from the box, says something about the probability of the items that are in said box.

Suppose each box had 100 coins, one has 99 silvers and a gold, one has 99 golds and a silver.
If you select a box at random and you can not look inside, and you pick out a random coin and it is gold, clearly it is far more likely to have come from the box which has more gold coins in it to start with, so you would say it probably came from box B.
The probability is not 50/50.

>> No.4025896

Given that we didn't pick the SS box, we have the GS and GG box. If we give each gold coin a subscript, we have the box G1S and G2G3. Then, there are three possibilities, since we KNOW we picked a gold coin already.

1. We picked G1. In this case, the other ball is silver.
2. We picked G2. In this case, the other ball is gold.
3. We picked G3. In this case, the other ball is gold.

Given that those three situations, given that we picked a gold, are exhaustive, the probability of the other coin being gold is 2/3.

>> No.4025906

>>4025892
There is one slight problem there, you do NOT pick a coin at random. You just pick a gold coin, meaning that the coin picking/number of coins does not influent the odds, since it is not a random choice. Also:
>>4025890

>> No.4025919

I changed my steam and gmail password

Will change security question latter

#safety

>> No.4025927

/sci/ is learning how to do bayes through repeating the same problem over and over again.

I'm so proud.

>> No.4025928 [DELETED] 

>>4025906
How is it not a random choice? You choose a random coin from a random box. Then you look at what coin you have selected, and you can say something about the probability of coins in the box now that you have this new information.

>> No.4025939

>>4025919
>Will change security question latter
You should change it to OP's question. If anybody tries to get in your account it'll frustrate the ever living fuck out of them.

>> No.4025942

>>4025645
>can't reproduce them properly
Why?
Disregard cases when you draw silver first, when drawing gold add one to total experiment count, when drawing gold again add one to success count. What's wrong with this method?

>> No.4025964

Let <span class="math">B_i[/spoiler] be the event that box <span class="math">i[/spoiler] is chosen and <span class="math">G [/spoiler] be the event that gold is chosen. Then <span class="math">i=1,2,3[/spoiler]. We have <span class="math">P(G| B_1) = 1, P(G| B_2) = 0, P(G| B_3) = 0.5[/spoiler]. We want to find the probability of the event that the second coin is gold, which is the case if and only if we had chosen box 1. By Bayes' rule, we find <span class="math">P(B_1 | G) = \frac{P(G | B_1)P(B_1)}{P(G|B_1)P(B_1) + P(G|B_2)P(B_2) + P(G|B_3)P(B_3)} = \frac{2}{3}[/spoiler].

>> No.4025965

>>4025942
Nothing is wrong with the method.
In the other thread with the heads or tails selection, somebody wrote a program to calculate this. The experiment was done thousands of times, and the percentage calculated at the end. It came to approximately 2/3, allowing a slight margin for error based on random chance.

>> No.4025976

GG
SS
GS

You reach in and pull out a gold. That means, the possible boxes are:
GG
GS

The probability that the other coin is silver: 1/3 (You picked GS)
The probability that the other coin is gold: 2/3
(You either picked GG, the first gold coin first and left the second in the box, or GG, the second gold coin first and the first one is still in the box)

Basic conditional probability.

>> No.4025992

>>4025847
Lol, the duplicates DO add to the probability of it being in that box. The point is that picking the coin gives us information about what box we picked from! We are more likely to have picked from the first box, because there are more gold coins in it. This is not a complex concept, really. Just put the numbers into bayes theorem if you don't understand.

>> No.4026008

>>4025906
The choice being random is implied

>> No.4026097
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4026097

>>4025942
>>4025965
This is why I wrote
>if you try one time

If you do the experiment one, then you can't be sure you get gold and so you have to do it until you get gold and just count these cases. The method works of course if you don't only "try one time".
These question tend to arise because of the fact that most people cut away everything that led to the question, i.e. the questions says "and the first one turns out to be gold" and this is problematic. You manipulated chances. Here you can interpret it properly by baysian propability theory and everything works out but you can also construct problems like the one I linked, the sleeping beauty problem, where some decission about probability theory have to be takes and there are still papers about these kind of stuff atm

related:

http://en.wikipedia.org/wiki/Self-Indication_Assumption

http://en.wikipedia.org/wiki/Self-Sampling_Assumption

>> No.4026103

>>4026097
Yes, but no experiment involving statistics can be done just one if you want accurate results. Pointing that out is kinda redundant.

>> No.4026121

Monte carlo simulation for anyone who disagree to 2/3;

gold = 0;
silver = 0;
for i = 1:10000
box = round(rand(1)*6);
% 1,2 == GG, 3,4 == SS, 5,6 == GS
if box == 1 || box == 2
gold = gold + 1;
elseif box == 3 || box == 4 || box == 6
% Try again.
elseif box == 5
silver = silver + 1;
end
end
gold/(gold+silver)

ans = 0.66692

You know you haven't picked the second box, so it can just be ignored. It is more likely that you picked the first box, since you got 2 gold coins.

>> No.4026128
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4026128

>>4026103
no, it's a different thing If I want to compute the probability of something by using the axiom which is the definition of a propability (limit of relative cases when the number of the experiments go to infinity) or if I ask a question which is arrived on the determination of a former probability (a decided case) where the answer still depends on the value of the probability.
In this case it's clear because one can seperate and compute the outcome in different cases, yes, but in general, one still has to make a decission on how probabilities are viewed irl (see sleeping beauty problem). This is a question of axiomatisation and the choice can be made based on philosophic arguments.