Without Google (cheating), what do you think the answer is?

[Hugo Boss]

[soon]

without looking at google i'd say that is similar to the monty hall problem.

[Lochdubh]

The box on the far right has 0% possibility of producing a gold ball. The fact that you found a gold ball on your first dip eliminates this box from consideration...it simply was not chosen as THE box.

The box on the far left offers 100% possibility of producing a gold ball

If on the first draw you chose the box in the centre, that box now has a 0% possibility of producing a second gold ball; if you chose the box on the left for the initial draw it still has a 100% possibility of producing a second gold ball.

So for the second dip you are either rummaging around in the centre box or the one on the left. Ergo you have a 50% chance of drawing a gold ball on the second try.

50%.

[Lochdubh]

This took me two tries, and I'm still not dead certain I'm right.

This sort of thing is why I'm a poor card player and failed to gain entrance into the Royal Artillery. My entrance interview ended rather like this:
"Mr. Harris, if we used YOUR computation for the firing table at Larkhill (Royal Skool of Artillery in Wiltshire) instead of shelling Salisbury Plain your battery would be raining rounds on the city centre. Do take a biscuit on your way out."

[Blurt]

Bayes' Theorem.

Mel has become a Reddit savant!

[J0E]

1 in 5? Or 20%

Probably wrong but worth a try.

[asal]

50% because there are only two eligible boxes.

[Lochdubh]

#@asal NOT Joe

^great minds thinking alike, toots.

[asal]

#@asal NOT Joe

^great minds thinking alike, toots.

There'll be some trick in the answer. But I think it's right, toots.

[J0E]

#@asal NOT Joe

^great minds thinking alike, toots.

Actually I changed my mind.

When I read the question over again.

I'd say 1 out of 3 or 33%.

Because the process of elimination would indicate that he's not going to choose the 3rd box and he's going to pick the same box he initially drew from

[Hugo Boss]

The correct answer is 2/3 chance of picking another gold ball. Many will guess 50% initially, until they examine closer (Including me).
It's called the Bertrand's Box Paradox.

Remember, you have to pick the second ball from the same box you picked the first ball from.

By picking the first gold ball, you've now eliminated the box containing two silver balls from the equation. So you're left with two boxes now from which the odds are calculated, and that works out to 2/3.

[soon]

it's always a bummer when they don't win the car.

[Lochdubh]

#@asal NOT Joe

^great minds thinking alike, toots.

Actually I changed my mind.

When I read the question over again.

I'd say 1 out of 3 or 33%.

Because the process of elimination would indicate that he's not going to choose the 3rd box and he's going to pick the same box he initially drew from

That was my first solution. It took me between 5 and 10 minutes to come up with it. Then I reread the problem, and still got it wrong. Well, the world needed ditch-diggers and infantry officers too.

[Hugo Boss]

They asked 53 Psychology freshmen this question who were taking an introductory probabilities course, and 35 incorrectly guessed 1/2 odds.

Only 3 correctly guessed 2/3.

The Bertrand's Box Paradox!

[soon]

i do believe we managed to screw in the light bulb then.