I calculate 1 in 2 since the silver-silver box can now be ignored. You either have the gold-silver box or the gold-gold box.

Am I correct? I'm looking for a trick, like in the now famous Monte Hall problem, but I don't see one.

marmot84
Level 7
Feb 12, 2018

Having drawn a gold marble first, it's more likely that you selected the gold-gold box.

@marmot84 On the initial draw there are six possibilities (six marbles). You had two chances at drawing gold from the gold-gold box, two chances at drawing silver from the silver-silver box, one chance of drawing gold from the mixed box, and one chance of drawing silver from the mixed box.

Once you make your first draw and get gold, you've eliminated three of those possibilities and have three left. You have either drawn gold from the gold-gold box (two chances) or gold from the mixed box (one chance). So there is a 2/3 chance you'll get gold on the second draw.

Edited

@cmadler Oh! I see your line of reasoning. Have to pause to think about it though because I think I have an objection...

Ok... yes, you are correct. Wow! Probabilities are often counter-intuitive and being an intuitive thinker I tend to try that first. Definitely, I'm wrong in this case! Cool!

2/3.

On your initial draw, you have pulled one of three gold marbles, call them b1m1 (box 1 marble 1), b1m2, and b3m1. You know you're drawn one of them, but you don't know which one. Of the three possibilities, in two cases the second marble in the box is also gold, so it's 2/3. This is somewhat related to the Monty Hall problem.

cmadler
Level 6
Feb 12, 2018

@Rugglesby That's correct. But think of it this way.

On the initial draw there are six possibilities (six marbles). You had two chances at drawing gold from the gold-gold box, two chances at drawing silver from the silver-silver box, one chance of drawing gold from the mixed box, and one chance of drawing silver from the mixed box.

One you make your first draw and get gold, you've eliminated three of those possibilities and have three left. You have either drawn gold from the gold-gold box (two chances) or gold from the mixed box (one chance). So there is a 2/3 chance you'll get gold on the second draw.

@Naeem You can't ignore the first marble, though. I've attached an image that should illustrate this. On the first draw you have a 50/50 chance of gold or silver. If you pull gold on the first draw, you have gotten one of three gold marbles, but you don't know which one. There is a 1/3 chance you drew the gold marble from the mixed box (because that's 1 of 3 gold marbles you might have drawn) and a 2/3 chance you drew the gold marble from the gold-only box (because those are 2 of the 3 gold marbles you might have drawn).

Here's another way to think about why this can't be 1/2. The first draw is 50/50 on getting gold or silver, so if you refers the problem, the probabilities must be the same. If the first draw is gold, and then you have a 50/50 chance of getting gold again, that means that if the first draw is silver you'd have a 50/50 chance of getting silver again. When you put all those numbers together, it comes to 1/4 gold-only (1/2 times 1/2), 1/4 silver-only, and 1/2 mixed. Obviously those numbers should really each be 1/3, so 1/2 can't be the right answer.

JimG:

@cmadler look at it this way: you have drawn from one of two possible boxes because only two contained at least one gold marble. You now have two possible outcomes; either you have drawn the gold marble from the box containing one gold and one silver marble, and you will now draw the silver one; or you've pulled one gold marble from a box containing two and will now draw the remaining gold marble. Your chances are 1/2 because you narrowed your options down to two when you chose the box for your first pick.

Edited

@JimG You have drawn from one of two boxes, but they are not 50/50.you have drawn one of three gold marbles. Two of the gold marbles were in the gold-gold box, and one was in the mixed box. So there is a 2/3 chance you drew gold-gold and a 1/3 chance you drew mixed.

Another way to think of it is that in the initial draw you have a 1/3 chance of choosing each box, and then a 1/2 chance for each marble within the box you've chosen. Overall on the first draw you have a 1/6 chance of drawing gold from the mixed box, a 1/6 chance of drawing silver from the mixed box, a 1/3 chance (2/6) of drawing gold from the gold only box, and a 1/3 chance (2/6) of drawing silver from the silver only box. Eliminating the three silver possibilities leaves three gold possibilities.

The answer, counterintuitively enough, is 2/3rds.

Druvius
Level 7
Feb 12, 2018

How so? I'm actually interested to know what I forgot.

Doesn’t matter, gold and silver? I’m taking all three boxes to a pawn shop and cashing in.

PeppermintDreads
Level 7
Feb 13, 2018

Oh forget that. I'm gonna take off with all 3 boxes, and unless you can catch me AND overpower me, they're all mine.

Farfrompukin
Level 3
Feb 13, 2018

Screw these probability questions: I am with you, gangster!

50%. Since you didn't pick a silver marble, the box with two silver marbles is out of play. that leaves 2 boxes left with a 50/50 chance at the gold.

ollieberry
Level 6
Feb 13, 2018

Just about 47% if my mental math is right. Nevermind, I forgot to factor in the third box. All three boxes is closer to 30% for a back of the envelope calculation.

jayneonacobb
Level 7
Feb 13, 2018

It's counter intuitive. The way to think about it is this. Two of the three boxes contain two of the same colour marbles. So when you select the first box at random, there is a 2/3rds chance it contains two like coloured marbles. Looking at one marble doesn't change that, whether it is gold or silver, 2/3rds of the time the second marble will be the same as the first. Since 2/3rds of the time the box you randomly selected contains two like coloured marbles.

Druvius
Level 7
Apr 3, 2018

1/3 the odds that you picked the 2 gold box.

SocraticAddict
Level 6
Feb 13, 2018

Damnit it's 50/50 because you've eliminated the 2 silver

Forgive my flawed brain, but I have to disagree with Bertrand. I grant that prior to picking a box or marble the chances are 2/3 that you will pull 2 marbles of the same colour from the one box. But only a 1/3 chance that is will be 2 gold or 2 silver. By pulling a gold out, you have eliminated the box with the 2 silvers. So you are choosing from either a box with a remaining gold or a silver marble in it. That is only 50% at that point in time. Yet I can see, if you read the question differently, that the 2/3 option is there. I have trouble with Schrodinger's Cat also.

Rugglesby
Level 8
Feb 13, 2018

I don't know that I can explain it, but I think I get it now.

It's related not just to which box is picked (1 in 3) but to the FACT that you drew a gold marble as your first draw.

Think of the chances of drawing gold from either of the two boxes that have gold and thus the chances that you drew gold AND got the box with one gold and one silver... since you drew gold as your first draw, you are more likely to have picked up the box with two gold in it.

The part I can't explain is how the math rules that this means the chancesa re 2 in three. I probably could if I spent more ergs on this... but I am feeling too lazy for that.

Gnarloc
Level 7
Feb 13, 2018

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