There are two boxes. You have 10 black and 10 white chess pieces. How do you arrange the pieces in the two boxes, so that the probability to pick a white piece is the largest?
Help please :(
does it matter?
Not sure, heh, that's what I am trying to find out
are you picking a box and random, and then picking a piece at random from a box? question is unclear as written
Yes, you pick a box at random and then a piece at random from the box.
I agree that the question was a bit unclear
if you put all white in one box, probability is \(\frac{1}{2}\) see what happens if you change it, i am willing to bet it makes no difference
That was my thought too, but it seemed a little too easy
for example 5 and 5 probability is \(\frac{1}{2}\times \frac{1}{2}+\frac{1}{2}\times \frac{1}{2}=\frac{1}{2}\)
try it with 3 and 7 i gotta run
\[\frac{1}{2}\times \frac{7}{10}+\frac{1}{2}\times \frac{3}{10}=\frac{1}{2}(\frac{3+7}{10})\] the same
should give an easy proof
Thank you :)
try putting one white piece in a box and all the others in the second box
can you explain the reasoning?
would be very helpful
I got the same answer from another person, but I don't understand it, it seems counterlogical to me
why...the fists box there is a 100% chance of getting a white and the second there is a 9/19 change so prob of white is \[\frac{1}{2}\times 1+\frac{1}{2}\frac{9}{19}=\frac{14}{19}\]
which is a lot bigger than 1/2
Thanks
But why is there a 9/19 chance in the second? Sorry, I suck at this
there are only 19 pieces in the second box (one in the first) of those 9 are white
Ohhhhhhh, just realized
I've gone around thinking thinking there had to be 10 in each
Hah, thanks alot :)
would be a boring problem then ;)
Indeed :)
My intuition messed things up for me
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