well known proba problem

perceval · #1

At work last friday i brought up the following proba problem and the open space productivity almost stopped due to the argument, i thought it might be fun to share here.

Its a fairly well know problem so you might already know of it, but almost everybody that hear it for the first time gets it wrong.

Suppose you have 3 Boxes:
- One with 2 Gold coins
- One with 2 Silver coins
- One with 1 Gold and 1 Silver Coin

Suppose you randomly pick a box, then take one of the 2 coins out from the box.
This Coin turns out to be a Gold coin. What is the probability that the second coin in the box is also Gold ?

Kirby · #2

HermanHiddema · #3

EdLee · #4

perceval, I can't believe you didn't convert this for here

Suppose you have 3 Go bowls:
- :white:

One with 2 Shell stones
- :black:

One with 2 Slate stones
- :white:

One with 1 Shell and 1 Slate stone

Suppose you randomly pick a bowl, then take one of the 2 stones out from the bowl.
This stone turns out to be . What is the probability that the second stone in the bowl is also ?

Kirby · #5

perceval · #6

EdLee wrote:

perceval, I can't believe you didn't convert this for here

Suppose you have 3 Go bowls:
- :white:

One with 2 Shell stones
- :black:

One with 2 Slate stones
- :white:

One with 1 Shell and 1 Slate stone

Suppose you randomly pick a bowl, then take one of the 2 stones out from the bowl.
This stone turns out to be . What is the probability that the second stone in the bowl is also ?

shame on me :oops:

tj86430 · #7

I was quite recently involved in a lengthy (10 pages or so) discussion concerning this problem on another forum, and later there was another, even lengthier discussion concerning a similar problem, but where the wording was a bit more vague, on the same forum.

The other problem was like this (I have attempted to remove the vagueness of the wording): There are 150 cards. 40 of them are white on both sides, 50 of them are black on one side and white on the other side and 60 of them are black on both sides. You take a card, and see that one side is white, but you don't see the other side. What is the probability that the other side is white too?

Bonobo · #8

perceval · #9

tj86430 wrote:

I was quite recently involved in a lengthy (10 pages or so) discussion concerning this problem on another forum, and later there was another, even lengthier discussion concerning a similar problem, but where the wording was a bit more vague, on the same forum.

The other problem was like this (I have attempted to remove the vagueness of the wording): There are 150 cards. 40 of them are white on both sides, 50 of them are black on one side and white on the other side and 60 of them are black on both sides. You take a card, and see that one side is white, but you don't see the other side. What is the probability that the other side is white too?

i'll have a go at this one:

tj86430 · **#10**

EdLee wrote:

tj86430 · **#11**

perceval wrote:

tj86430 wrote:

I was quite recently involved in a lengthy (10 pages or so) discussion concerning this problem on another forum, and later there was another, even lengthier discussion concerning a similar problem, but where the wording was a bit more vague, on the same forum.

The other problem was like this (I have attempted to remove the vagueness of the wording): There are 150 cards. 40 of them are white on both sides, 50 of them are black on one side and white on the other side and 60 of them are black on both sides. You take a card, and see that one side is white, but you don't see the other side. What is the probability that the other side is white too?

i'll have a go at this one:

Bonobo · **#12**

I don’t get it.

Choices at first:

- BB
- BW
- WW

I choose a box and take out a B stone, so the WW option has just been eliminated, hasn’t it? How should then a 2/3 chance remain? There is no /3 anymore, so it should be 2/2 -> 1:1 chance, no?

OK, I’m slightly dyscalculic, meaning I have to take of my shoes when counting to 20 ;-)

Please just disregard if I’m simply too stupid.

perceval · **#13**

Bonobo wrote:

I don’t get it.

Choices at first:

- BB
- BW
- WW

I choose a box and take out a B stone, so the WW option has just been eliminated, hasn’t it? How should then a 2/3 chance remain? There is no /3 anymore, so it should be 2/2 -> 1:1 chance, no?

OK, I’m slightly dyscalculic, meaning I have to take of my shoes when counting to 20 ;-)

Please just disregard if I’m simply too stupid.

Definitly not stupid, everybody starts with this argument, that is the beauty of the problem.
Here is some food for thought:

Lets change the game, now each 3 boxes contains 1000 coins:
- one contains 1000 G
- one contains 1000 S
- one contains 999 S and one G

You draw a coin from a box and notice its a gold coin, what is the probability that the box contains 999 other Gold coins ?

If you happen to draw a silver coin, what is the probability that the box contains 999 silver coins ?

tj86430 · **#14**

Bonobo wrote:

I don’t get it.

Choices at first:

- BB
- BW
- WW

I choose a box and take out a B stone, so the WW option has just been eliminated, hasn’t it? How should then a 2/3 chance remain? There is no /3 anymore, so it should be 2/2 -> 1:1 chance, no?

OK, I’m slightly dyscalculic, meaning I have to take of my shoes when counting to 20 ;-)

Please just disregard if I’m simply too stupid.

tj86430 · **#15**

perceval wrote:

everybody starts with this argument

I beg to disagree.

daal · **#16**

Bonobo wrote:

I don’t get it.

Choices at first:

- BB
- BW
- WW

I choose a box and take out a B stone, so the WW option has just been eliminated, hasn’t it? How should then a 2/3 chance remain? There is no /3 anymore, so it should be 2/2 -> 1:1 chance, no?

OK, I’m slightly dyscalculic, meaning I have to take of my shoes when counting to 20 ;-)

Please just disregard if I’m simply too stupid.

Now that I've been told the answer, I think I can explain it. First, take off your shoes.

Bantari · **#17**

tj86430 wrote:

Bonobo wrote:

I don’t get it.

Choices at first:

- BB
- BW
- WW

I choose a box and take out a B stone, so the WW option has just been eliminated, hasn’t it? How should then a 2/3 chance remain? There is no /3 anymore, so it should be 2/2 -> 1:1 chance, no?

OK, I’m slightly dyscalculic, meaning I have to take of my shoes when counting to 20 ;-)

Please just disregard if I’m simply too stupid.

I don't know if I can explain it any more clearly than it has already been explained, but:

1. originally there are three black stones, right? (lets call them A, B and C, and let's also say that A & B are in the same bowl)
2. If 1. is correct, then you can take any one of those three with equal probability, right?
3. If 2. is correct, then there is 1/3 chance that you have taken the stone we decided to call A; 1/3 chance that you have taken B and 1/3 chance that you have taken C, right?
4. If 3. is correct, then there is 2/3 chance that there is another black in the same bowl (1/3 if you had taken A + 1/3 if you had taken B) and 1/3 chance that there is a white in the same bowl (in case you had taken C)

I dunno, its late and I am drunk, but....

The argument (and the result) you give is the same as if (after agreeing that the WW box is eliminated) you had all the stones in the same box. So, after pulling out a B stone, you are left with BBW in the box, and then indeed the chance of pulling another B is 2/3.

However - this is not the case that we have all the remaining stones in the same box. We have only one stone left in the box, and we are set on this particular box, right? Intuitively, it seems to me that this should make a difference in the result, no matter how we twist the math.

Here is what I think.

The chances are:
- either we have initially picked from the box BB - in which case the chance to pick another B stone is 100% or
- we have initially picked from box BW - in which case the chance to pick another B stone is 0%

Since the chances of initial pick from BB and BW are the same (given 3 boxes they are 33% each, but since we KNOW per the assumption we did not picked box WW, the chances are 50% each BW and BB.) Either way - given that BB and BW have the same chance of being picked - it seems to me that the result should be 50% (100% + 0% / 2)

Or am I just too drunk to think?....

perceval · **#18**

tj86430 wrote:

perceval wrote:

everybody starts with this argument

I beg to disagree.

ok

, would you agree that a lot of people starts with this argument, including a lot of people with a mathematical background ?

Bantari · **#19**

daal wrote:

Now that I've been told the answer, I think I can explain it. First, take off your shoes.After having taken out a black stone, you are right in saying that the WW box is eliminated. So, how many stones are left? 3. How many of them are black? 2. Chance of one of them being black? 2/3. Now put your shoes back on.

By this logic, if the initial boxes were WW, BW, and BBBBB - then the result would be 4/5? But it clearly should not matter how many Black stones are in the Black-only box... we have the same chance of picking black Stone from this box regardless of how many there are (provided there are more than 1 and all are Black.) No?

tj86430 · **#20**

perceval wrote:

tj86430 wrote:

perceval wrote:

everybody starts with this argument

I beg to disagree.

ok

, would you agree that a lot of people starts with this argument, including a lot of people with a mathematical background ?

Definitely

well known proba problem

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