Only people with an IQ of 115 or higher can get this right

[Mai72]

lol, i got some golden balls for ya.

 

[Thebobo]

The boxes aren't relevant really?

 

[spacejamz]

I'd say 50%. Since you already chose 1 gold ball this eliminates the box with the 2 silver balls. So, it's 50/50.

Fern

Since there are only two boxes with gold balls so you have 4 possible balls to deal with. Since you already pulled one (gold) ball out, there are three balls remaining that can be pulled, two gold and one silver so it is 2 out 3.

 

[Hitman928]

Since there are only two boxes with gold balls so you have 4 possible balls to deal with. Since you already pulled one (gold) ball out, there are three balls remaining that can be pulled, two gold and one silver so it is 2 out 3.

But you have to stick with the same box that you started with, you can't switch boxes, so it changes the math. It's no longer the probability of getting a gold ball (which is 2/3) but rather are you in the box with 2 gold balls, or the one with 1 gold and 1 silver.

 

[sandorski]

But you have to stick with the same box that you started with, you can't switch boxes, so it changes the math. It's no longer the probability of getting a gold ball (which is 2/3) but rather are you in the box with 2 gold balls, or the one with 1 gold and 1 silver.

Moot. You don't know which box you picked. So only the amount of balls left to choose from matters.

 

[Mayne]

ruining people's lives on forums since 1999.

 

[Hitman928]

Moot. You don't know which box you picked. So only the amount of balls left to choose from matters.

Probability is 2/3 but it's not for the reason the poster I quoted was saying, unless I misunderstood his reasoning..

 

[DaveSimmons]

https://en.wikipedia.org/wiki/Bertrand's_box_paradox

^ Read this!

"The correct answer of 2/3 can also be obtained as follows:

• Originally, all six coins were equally likely to be chosen.
• The chosen coin cannot be from drawer S of box GS, or from either drawer of box SS.
• So it must come from the G drawer of box GS, or either drawer of box GG.
• The three remaining possibilities are equally likely, so the probability that the drawer is from box GG is 2/3."

Basically, if you get a gold ball, odds are 2/3 that you picked it from the box with 2 gold balls.

 

[Slew Foot]

50% because once you have already determined that a gold ball is present in a box, you have reset your odds. now there are only 2 possibilities left, the box with 2 golds that will have a second gold, or the box the one of each that will have a silver.

 

[Carson Dyle]

Try visualizing it.

Start with

[GG] [SG] [SS]

You picked a G.

You now have either:

[GG] [S_]

or

[G_][SG]

2 out of 3.

 

[DigDog]

You must draw from the same box. The moment you draw the system ends. The number of balls is irrelevant, its the number of choices and their outcome.
The boxes are 1. 100% s 2. 100% g, and you have not been able to determine which is which.
The double gold can contain infinite golds, the silver infinite silvers, numbers do not matter.

The observer is not deterministic.

In Carsons oh so well designed example, you are only allowed to draw from [s_] or [g_]. Two outcomes, one action.

50%

 

[thebestMAX]

25 or 6 to 4

50% just like every other choice. It will be or it wont.

3 to 2 actually.

 

[Hitman928]

Try visualizing it.

Start with

[GG] [SG] [SS]

You picked a G.

You now have either:

[GG] [S_]

or

[G_][SG]

2 out of 3.

This explanation only works if you can pick from either box on your second turn, but you can't.

 

[Hitman928]

You must draw from the same box. The moment you draw the system ends. The number of balls is irrelevant, its the number of choices and their outcome.
The boxes are 1. 100% s 2. 100% g, and you have not been able to determine which is which.
The double gold can contain infinite golds, the silver infinite silvers, numbers do not matter.

The observer is not deterministic.

In Carsons oh so well designed example, you are only allowed to draw from [s_] or [g_]. Two outcomes, one action.

50%

The bolded is true, the italicized is false. You must consider the likelihood of which box you are in if you pulled a gold to begin with.

 

[snoopy7548]

I immediately thought 50%, because that seems like it's close enough and I really don't know the math to figure it out properly. We're all gonna die at some point so what's the point in trying to figure out the probability of choosing a hypothetical gold ball out of a stupid box?

 

[Humpy]

You must consider the likelihood of which box you are in if you pulled a gold to begin with.

No you don't. You can, but you don't have to.

The question is asking about the probability after the first pick is over and done with, so as someone said the probability resets.

 

[interchange]

The first choice is deceptive because you are inclined to think you chose a box at random when in reality you chose a ball at random. So you chose among one of three gold balls. Two out of the three gold balls are paired with another gold ball, so the correct answer is 2/3.

 

[brianmanahan]

2/3

2/3 is not intuitive but it is correct

 

[brianmanahan]

Basically, if you get a gold ball, odds are 2/3 that you picked it from the box with 2 gold balls.

yep, that's the key thing to think about. the 1-silver/1-gold box is less likely because sometimes when you draw the first ball you'd get the silver ball from that box.

 

[SKORPI0]

Like

 

[local]

Ugh, number of balls is irrelevant. The moment it was determined you drew a gold ball from one of the three boxes and then had to draw from the same box eliminated the third box completely. Thus the odds of drawing a second gold ball is literally just down to which of the two boxes you randomly picked. So the entire exercise is down to a random choice between two boxes. Randomly picking from only two options is 50% odds.

This thing could be rewritten as: You have three numbers 1, 2 and 3. Pick one number at random and 3 is not an option. The odds you get the 1 in that scenario is 50%.

 

[local]

yep, that's the key thing to think about. the 1-silver/1-gold box is less likely because sometimes when you draw the first ball you'd get the silver ball from that box.

Conditions of the test already stated you drew a gold ball, you cannot rewrite the conditions.

 

[Hitman928]

Like

This is not correct.

People, the answer is 2/3. This is a known statistical problem. If you think it's 1/2, you are wrong.

 

[Hitman928]

No you don't. You can, but you don't have to.

The question is asking about the probability after the first pick is over and done with, so as someone said the probability resets.

But the first pick being gold is key to the probability, you can't just ignore it. Did you come to 2/3 or 1/2?

 

[SKORPI0]

This is not correct.

People, the answer is 2/3. This is a known statistical problem. If you think it's 1/2, you are wrong.

You're picking from the same box, where the 1st gold ball came from. You don't know that the other ball is.
1 box had 2 gold balls, 1 box had 1 gold ball and 1 silver ball.
You picked a gold ball from 1 box. Now you're down to 1 box, either with a gold ball or a silver ball.