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

[Hitman928]

I get how you are thinking. I took statistics in grad school. However, based on the OP's wording of the question you are still wrong. It doesn't matter how many times you copy and paste your reply.

So you think this is different than the Bertrand's Box Paradox?

 

[Mayne]

So you think this is different than the Bertrand's Box Paradox?

its totally different.

 

[Hitman928]

its totally different.

How so?

 

[brianmanahan]

its totally different.

lol

now i get why they banned you at mmo, you're just trying to cause trouble

 

[Engineer]

50%.

No clue if that's right but seems logical. Did not read past the OP.

 

[Hitman928]

lol

now i get why they banned you at mmo, you're just trying to cause trouble

Yep, it makes sense now.

 

[brianmanahan]

No clue if that's right but seems logical. Did not read past the OP.

don't, it's a trap

i couldn't stop thinking about it last night and was unable to fall asleep for several hours

 

[snoopy7548]

How has anyone not read the wiki about this yet?

There are 3 possibilities once the first gold was chosen:
1) the gold chosen was #1 from the box with 2 gold.
2) the gold chosen was #2 from the box with 2 gold.
3) the gold chosen was from the box with one of each.

2/3 it's the box with 2 golds. Not rocket science. Probability is often counter-intuitive, but there's not point ignoring the maths when it's been explained again and again.

That makes a lot of sense and is simple, I get it now. I think you're the genius here.

That said, it's still 50%.

 

[Tweak155]

I get how you are thinking. I took statistics in grad school. However, based on the OP's wording of the question you are still wrong. It doesn't matter how many times you copy and paste your reply.

You seem to be hung up on the wrong event. Both balls have already been picked once the box was picked, so you need to figure out what the probability is that you picked the box with 2 gold balls in it.

 

[loafbred]

I've had a nagging feeling that I've been overlooking something, so I just reread the first four pages of the thread. Now that I'm fully awake and not watching TV, I see something in Darwin333's post from page 4 that I didn't absorb before:

He isn't rewriting the conditions. If you pull a gold ball you are more likely to have pulled it from the box with only gold balls. If you pulled a silver ball you are more likely to have pulled it from the box only silver balls. This information must be considered in future questions which is why the answer is 2/3.

Now I'm thinking that the paired gold balls could skew the 1/2 conclusion, but would it increase it to 2/3? It seems to me that it wouldn't increase the odds that much.

 

[brianmanahan]

I've had a nagging feeling that I've been overlooking something, so I just reread the first four pages of the thread. Now that I'm fully awake and not watching TV, I see something in Darwin333's post from page 4 that I didn't absorb before:

Now I'm thinking that the paired gold balls could skew the 1/2 conclusion, but would it increase it to 2/3? It seems to me that it wouldn't increase the odds that much.

it approaches 2/3 as the number of simulations increases

http://js.do/code/225139

 

[Hitman928]

it approaches 2/3 as the number of simulations increases

http://js.do/code/225139

Here's what it looks like as you increase the number of trials. Notice the "window" around 67% gets smaller and smaller.

 

[WelshBloke]

Its 50%

You're being asked to choose between two boxes. One with a gold ball in and one with a silver ball in.

The box with the two silver balls is irrelevant and doesn't come into the problem as asked in the op.

 

[loafbred]

Its 50%

You're being asked to choose between two boxes. One with a gold ball in and one with a silver ball in.

The box with the two silver balls is irrelevant and doesn't come into the problem as asked in the op.

I was in your camp for the first ten hours or so, but I'm now in the "~sixty something percent" camp.

 

[Darwin333]

Its 50%

You're being asked to choose between two boxes. One with a gold ball in and one with a silver ball in.

The box with the two silver balls is irrelevant and doesn't come into the problem as asked in the op.

Why would previous information be irrelevant to future probability?

This is very much like the Monty Hall Problem.

 

[paperfist]

How has anyone not read the wiki about this yet?

There are 3 possibilities once the first gold was chosen:
1) the gold chosen was #1 from the box with 2 gold.
2) the gold chosen was #2 from the box with 2 gold.
3) the gold chosen was from the box with one of each.

2/3 it's the box with 2 golds. Not rocket science. Probability is often counter-intuitive, but there's not point ignoring the maths when it's been explained again and again.

Maybe my IQ is only 113, but I still don't see how it can be 2/3 as you can't pick from multiple boxes, only 2 which your hand is already in.

Mayne is somewhere ROFL as grown men continue to argue about balls.

 

[loafbred]

Maybe my IQ is only 113, but I still don't see how it can be 2/3 as you can't pick from multiple boxes, only 2 which your hand is already in.

Mayne is somewhere ROFL as grown men continue to argue about balls.

Keep in mind that I've only recently fallen in line with the 2/3 camp, and was staunchly against them earlier. It finally occurred to me that when you reach for a box, two of the boxes have twins, and only one has both colors. When you select a box and get your ball of either color, you're more likely to have a twin box than not.

I agree that the probability of Mayne laughing at us is greater than 50%

 

[PowerEngineer]

This is very much like the answer to the Monty Hall curtain question which many people just refuse to believe. Put simply, to meet the first criteria of picking a gold ball your pick must be one of the following:

1. Gold ball #1 in the two gold ball box
2. Gold ball #2 in the two gold ball box
3. The gold ball in the one gold ball box

Each of these gold ball picking possibilities are equally likely. For two of these possibilities (#1 and #2), the next ball you pick out of the box will be gold. The next ball you pick out of the box will be gray for only ones possibility (#3). So the answer is 2/3 or 66.6% of the times you try this your second ball will be gold.

If you do not believe this answer then you can prove it to yourself by getting three boxes and some colored balls and actually running a couple dozen trials.

 

[WelshBloke]

Why would previous information be irrelevant to future probability?

Because the previous information was for a different problem than the question the OP is asking.

The question the OP is asking is specifically about the boxes with the gold balls in and doesn't give you a choice of picking the box with the silver balls in.

 

[WelshBloke]

Put simply, to meet the first criteria of picking a gold ball your pick must be one of the following:

1. Gold ball #1 in the two gold ball box
2. Gold ball #2 in the two gold ball box
3. The gold ball in the one gold ball box

Each of these gold ball picking possibilities are equally likely. For two of these possibilities (#1 and #2), the next ball you pick out of the box will be gold. The next ball you pick out of the box will be gray for only ones possibility (#3). So the answer is 2/3 or 66.6% of the times you try this your second ball will be gold.

That's not the choice the OP is giving you or the odds he is asking you to work out.

The question the OP is asking you is to calculate the odds of picking a gold ball when you have two boxes, one with a gold ball in and one with a silver ball in.

 

[Carson Dyle]

This is very much like the answer to the Monty Hall curtain question which many people just refuse to believe. Put simply, to meet the first criteria of picking a gold ball your pick must be one of the following:

1. Gold ball #1 in the two gold ball box
2. Gold ball #2 in the two gold ball box
3. The gold ball in the one gold ball box

Each of these gold ball picking possibilities are equally likely. For two of these possibilities (#1 and #2), the next ball you pick out of the box will be gold. The next ball you pick out of the box will be gray for only ones possibility (#3). So the answer is 2/3 or 66.6% of the times you try this your second ball will be gold.

If you do not believe this answer then you can prove it to yourself by getting three boxes and some colored balls and actually running a couple dozen trials.

Good explanation.

 

[Hitman928]

That's not the choice the OP is giving you or the odds he is asking you to work out.

The question the OP is asking you is to calculate the odds of picking a gold ball when you have two boxes, one with a gold ball in and one with a silver ball in.

PowerEngineer is correct, maybe try to reread the OP.

 

[William Gaatjes]

Well, it is how someone interprets the question.
I looked up the definition of probability in general (after translating it to my native language).
Probability means the ratio between a given outcome versus all possible outcomes.
When reading the question, i read it as all possible outcomes for the golden ball minus the outcome that already happened. Because that is the intuitive outcome.
And then it is indeed a 50% chance.
But probability in general means all given outcomes divided by all possible outcomes.
And then it is 66%.
With that in mind, i am also in the 66% camp now.
I am still curious and want to write a program.

 

[Skyclad1uhm1]

Maybe my IQ is only 113, but I still don't see how it can be 2/3 as you can't pick from multiple boxes, only 2 which your hand is already in.

Mayne is somewhere ROFL as grown men continue to argue about balls.

Say you do the test 200x, where you pick each of the boxes 100x (we ignore the double silver as that doesn't matter anyway).
You pick the double gold box 100x and the G/S box 100x.
If you randomly take a ball out of the gold box you will get gold ball #1 50x and gold ball #2 50x.
From the G/S box you will get the gold ball 50x and the silver ball 50x.

Out of those 150 picks where you drew gold first, how many will give gold on the second try too?
50x for G #1 and 50x for G #2, while the silver one gets picked 50x too. So it's 100/150 or 2/3

The whole reason why it comes down to 2/3rd is because you leave out the option where you pick silver first. If you would make it 'You blindly remove one random ball from the boxes (Including the S/S box), what is the chance that the remaining one is gold?' or 'You select a box, and one gold ball is removed from it by a third party, what is the chance that the remaining one is gold?' then it would be 1/2. But the first random draw takes out half those of the G/S box.

 

[dasherHampton]

It's a question of whether you view the two selections as a unit or as separate events. Similar to the old "100 coin flips" model.

The odds of flipping heads 100 times in a row is billions to one (or higher). If you somehow manage to flip 99 heads in a row you could beat impossible odds with the final flip - IF you view the 100 flips as a set of 100 events.

But taken as an individual event the odds of the 100th flip being heads is 50/50, even after 99 straight "heads" flips. All preceding flips are irrelevant.

This problem embraces the same philosophy. If you see the two selections as a single event the odds are 2/3. If you view the second selection as it's own event the odds are 50/50, the box with two silvers having become irrelevant.