well known proba problem

[perceval]

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]

This reminds me of the monty hall problem. Intuition says that it's a 50% chance, but intuition didn't serve me well with the monty hall problem, either.

My rationale:
Two possible events: first box or third box.

[HermanHiddema]

2/3?

There are six possible grabs, each with equal probability:

1. Box GG grab G₁
2. Box GG grab G₂
3. Box GS grab G
4. Box GS grab S
5. Box SS grab S₁
6. Box SS grab S₂

The last three options are eliminated by the fact that we grabbed gold, therefore one of the first three must be the one that happened. If it is either 1 or 2, then the other coin will also be gold. If it is 3, then the other will be silver.

Therefore, we have 2/3 chance of another gold.

[EdLee]

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

Suppose you have 3 Go bowls:
- One with 2 Shell stones
- One with 2 Slate stones
- 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 ?

Monty Hall Problem

[Kirby]

I thought about it more, and I think Herman is correct (I read his answer).

Another way to explain it is that there are 3 gold coins and 3 silver coins in the entire problem. Since you didn't draw a silver coin, the two silver coins in the silver-only scenario can't be from the pool of possibilities.

Therefore, there are 3 gold coins and 1 silver coin in the pool you selected from. You picked 1 gold coin, so there's 2 gold coins and 1 silver coin left. So you have 2/3 chance.

[perceval]

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

Suppose you have 3 Go bowls:
- One with 2 Shell stones
- One with 2 Slate stones
- 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

[tj86430]

Herman has it, 2/3 is correct

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]

50/50

[perceval]

Herman has it, 2/3 is correct

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:

Using the same kind of resaonning that the problem above: lets forget about the cards black on both side, we have not drawn one of them.
What we know is that we have drawn a white side. It is reasonnable to say that the proba to get any White side is the same. There are 40*2+50 = 130 white sides, amongst with 80 belong to a card that is white on both sides , so i would say 80/130 = 8/13 chances of the other side being white

[tj86430]

Monty Hall Problem

This is even more related: http://en.wikipedia.org/wiki/Bertrand%27s_box_paradox

[tj86430]

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:

Using the same kind of resaonning that the problem above: lets forget about the black cards on both side, we have not drawn one of them.
What we know is that we have drawn a white side. It reasonnalbe to say that the proba to get any White side is the same. There are 40*2+50 = 130 white sides, amongst with 80 belong to a card that is white on both sides , so i would say 80/130 = 8/13 chances of the other side being white

Correct

[Bonobo]

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]

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]

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)

[tj86430]

everybody starts with this argument

I beg to disagree.