re: Solve this OT

In re-reading the problem, it never actually specifies that the balls are chosen from the box without replacement; that is, given the wording you could also do the experiment by:
1. Choose a box
2. Remove one ball at random from the box and look at it.
2a. Put the ball back
3. Remove one ball at random from the same box and look at it.

With those rules, it turns out to be 5/6 picking gold again.
GIVEN YOU ALREADY PICKED A GOLD - it's 2/3 you're in the first box, 1/3 you're in the second. That means you would pick a second gold 2/3 * 1 + 1/3 * 1/2 = 5/6
and simulation shows:
0.833012737925
0.833501195015
0.833995202532
0.83356376496
0.83395551501
0.833647576793
0.832830676433
0.833842267662
0.833897803564
0.832886330521

I think it can be assumed the ball isn’t returned.

Your first post shows the definitive answer and proof.

quote:

---

Once you pull the gold ball, you've eliminated the all silver (AS) box from your options. So in your hand you either have the all gold box (AG), or the mixed box (M). Leaving you only 2 options: 1) The chance of drawing a gold from the M is 0% 2) The chance of drawing the gold ball from the AG is 100% The only way you change those odds are if you place your drawn gold ball back into your box. And then draw again from that same box. But I did not interpret the problem that way. I read it that I now have 1 gold ball in my hand, and there is only 1 ball remaining for me to choose from, in THE SAME BOX. Its 50/50 at that point. I either have the AG box or the M box in my hand. There are simply no other options

---

Same here. Or if it was one box with three balls remaining (G, G, S). But it’s not, it’s one box with one ball remaining.

From what I can gather from the little bit of this argument I've read, I think most of you are misunderstanding the intent of the riddle.

You COULD interpret it as:

These things are true. Now you know stuff and reach into a box. What is the probability of getting a gold ball from this point foward?

But, what I'm pretty sure it ACTUALLY means is:

What is the probability you will get a gold ball in the end if each of these steps are followed.

And the answer is 40%.

quote:

---

And the answer is 40%.

---

Jesus H.

quote:

---

So you DO know.

---

No. You only know you picked from one of the gold containing boxes, you don't know which one.

There is a greater probability, given you've picked a gold ball, that this ball came from the box with two golds rather than one. Thus, your probability for drawing another gold from the same box is greater than not.

quote:

---

With those rules, it turns out to be 5/6 picking gold again.
GIVEN YOU ALREADY PICKED A GOLD - it's 2/3 you're in the first box, 1/3 you're in the second. That means you would pick a second gold 2/3 * 1 + 1/3 * 1/2 = 5/6

---

we have a winner!

bonus on the snake script.

quote:

---

No. You only know you picked from one of the gold containing boxes, you don't know which one. There is a greater probability, given you've picked a gold ball, that this ball came from the box with two golds rather than one. Thus, your probability for drawing another gold from the same box is greater than not

---

Okay, I’m pretty sure I got it based on the two different interpretations or scenarios. Yes assuming not the same/known box being picked from I can see 2/3. But Same box being picked from, 1/2.

quote:

---

Jesus H

---

What?

It doesn't say "NOW" what are the odds of drawing a gold ball. That's "simple" math (for math people).

It says "what are the odds while following all these steps." The trick is to not fall for all the eye candy.

The question is actually the same as if all the balls were in a single box.

Given that you have already picked a gold ball (without replacement), these are the possible outcomes for the second draw:

1. Pick the first gold ball in box A
2. Pick the second gold ball in box A
3. Pick the silver ball in box B

2 out of 3 outcomes are favorable, so without reading any responses in this thread, my guess is 66.67%.