So, here is the next round. According to the new rules, I shall have the answers PMed as well. Deadline is until Wednesday 12:00 pm (d_g's time).
Assume that there is a big sack with an equal number each of several colored snooker balls (there might be more than the traditional 6 colors though- I am just madly trying to relate this thing to snooker). Now, if one adds 20 balls of a new color to this mix, one wouldn't change the probability of picking two balls of the same color from the bag. (This is w.o. returning the balls after the picks).
The question is how many balls were in the sack, before those balls were added?
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can I ask another one meanwhile? I have no more good fractions for the above one...
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Originally Posted by abextraOk, I'll go for 1/5 then.
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Originally Posted by snookersfun1/4
not that you think, I am just guessing (there is some serious thinking behind this all, just the theory evades me....)This question is lasting much longer than I had thought it would!
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1/4
not that you think, I am just guessing (there is some serious thinking behind this all, just the theory evades me....)
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Originally Posted by snookersfunor wait 1/6 sounds good as well.
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Originally Posted by snookersfunI went for something like 1/9^n (for n = number of completed exchanges),
but this is probability and 'ichs'
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I went for something like 1/9^n (for n = number of completed exchanges),
but this is probability and 'ichs'
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Originally Posted by snookersfunin that case is it really small (tiny fraction)?
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No correct answers yet. Not elvaago's 1/9, nor snookersfun's 2/9...
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