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        riddles >> easy >> 5 counters in a bag
        
(Message started by: Noke Lieu on Nov 3^rd, 2014, 2:37am)

Title: 5 counters in a bag
Post by Noke Lieu on Nov 3^rd, 2014, 2:37am

Aw heck, make them marbles if you like...

At least 3 of them are black, any others are white.

What's the probability that the 3rd counter you draw from the bag (without replacement) is black?

Title: Re: 5 counters in a bag
Post by Grimbal on Nov 3^rd, 2014, 3:28am

"At least 3", does it mean the number of black counters is uniformly distributed between 3, 4 and 5?

Either the quesiton is very simple or I am missing something.

Title: Re: 5 counters in a bag
Post by towr on Nov 3^rd, 2014, 9:57am

[hide]at least 60%

In any case, it doesn't matter whether you look at the third or the first or the Kth.[/hide]

Title: Re: 5 counters in a bag
Post by Noke Lieu on Nov 3^rd, 2014, 6:10pm

on 11/03/14 at 03:28:06, Grimbal wrote:

Either the quesiton is very simple or I am missing something.

Just a simple question (it is in easy)... possibly the trap was trying to get you to overthink it... ;)

Yes I was intending there to be either uniformly 3, 4 or 5 black counters.

There's one part of me that wonders about conditional probability- only because I fear that it crops up more than it actually does.

Title: Re: 5 counters in a bag
Post by Noke Lieu on Nov 9^th, 2014, 8:59pm

ah well...

It was something that I might have messed up, bt I was interested in a quirk...

When I was a younge man, my teacher told me that if in trouble or in doubt, write the sample space out. Not terribly practical in quite a few situations, but not the worst adivce either...

bbbww[/b]
bbwwb
bwwbb
wwbbb
bbwbw
bwbbw
wbbbw
bwbwb
wbbwb
wbwbb

bbbbw
bbbwb
bbwbb
bwbbb
wbbbb

bbbbb

which is 11/16 successes.

But 1/3(1+6/10+4/5) = 12/15

and 11/16 =/= 12/15...

so what's going wrong?

Title: Re: 5 counters in a bag
Post by towr on Nov 9^th, 2014, 10:37pm

on 11/09/14 at 20:59:24, Noke Lieu wrote:

so what's going wrong?

The samples aren't equi-probable. It's like saying "what's the chance of throwing a 1 with a die" and describing the sample-space as:
* one
* not-one
and coming up with an answer of 1/2.

At the very least, the three sample-subspaces should be weighed equally (if 0, 1 or 2 white tokens are in fact equally likely), and not weighed by 1, 5 and 10.
The easiest way to fix it so you get the right sample-space, is to give the two extra tokens an identity. So e.g. use permutations of bbb12

Actually, that might still suggests a different sample-space than you might expect. If each of the two extra token has a 50% chance of being black, the cases of 0, 1 or 2 extra black tokens are not equi-probable; having 1 extra black is twice as likely. (But ultimately the answer it gives is the same)

Title: Re: 5 counters in a bag
Post by Noke Lieu on Nov 10^th, 2014, 1:48am

y'know, I know that, but can't see it clearly enough to trust myself... One day, one day... ::)