Whats the answer b?

It doesn't matter.
In the beginning you picked a box with gold in it (which was 2 in 3), thus eliminating the all silver box.
You're still left with only two boxes, so 1 in 2 you picked the all gold one

If you randomly choose the right box, you have a 0% chance of drawing a gold ball and therefore counting it.

If you randomly choose the middle box, you'll have a 50% chance of drawing a gold ball and therefore counting it.

If you randomly choose the right box, you have a 100% chance of drawing a gold ball, and therefore counting it.

Since you count the gold box every time and the mixed box 50% of the time, 2/3rds of counted events will be from the all gold box.

Was supposed to be right, middle, left obviously

You do one attempt, as per the problem description, and each round is independent, as per... math, except you know you eliminated the only-silver box by drawing a gold ball from your chosen box.
Round two has two potential boxes left, 1 in 2
quod erat demonstrandum

Whats the equation with the original question and whats the equation for the one you replied to?

It can be a thousand attempts or one attempt. It doesnt change the answer which isnt 50% by the way

Guy asked if it was more likely to have picked the gold-only box if there was 1000 gold and a 1 gold 999 silver one. The answer then depends on what specific question you're trying to answer.
Is it more likely you drew the gold ball from the 1000 gold box? Yes, the gold box had a 1 in 1 chance, the other had a 1 in 1000 chance.
Does it influence the probability of the overall outcome? No, still 1 in 2.

I really hope you're not asian. Your parents would kill you for you displaying such a lack of basic mathematical knowledge

>Is it more likely you drew the gold ball from the 1000 gold box? Yes
So you do understand this on some level. Youre also more likely to have drawn a gold ball from the first box in the original question

Ligma

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