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Lior
Jun 19th, 2002, 06:05 AM
Hi all,
Here is a little "probability paradox" I encountered while surfing on the net:

There are 2 balls inside a jar. we don't know the color of each ball,
but what we do know is that the balls must be black or white.

Now, each ball has the probability of 50% to be either black or white.
Let's mark the situations inside the jar:
If there are 2 white balls inside the jar, then the situation is WW.
If there are 2 black balls inside the jar, then the situation is BB.
if one of the balls if white and the other one is black then the situation is BW.

Now, the probability for WW is 1/4.
the probability for BB is 1/4.
and the probability for BW is 1/2.

Now, let's add a new ball into the jar, and let's make sure the new ball inserted is white.

Now, let's check the probabilities for the different situations and for selecting a white ball.

WWW: the probability for this situation is 1/4 because it happens only if the situation was WW before the white ball was inserted.
if the situation is WWW then the probability for selecting a white ball is 1.
pay attention that this claim works in both directions:
if the probability to select a white ball is 1, then the situation MUST be WWW.
take a minute to be totally convinced at this point, because it will be important for the rest.

WWB: the probability for this situation is 1/2 because it happens only if the situation was BW before the white ball was inserted.
if the situation is WWB, then the probability to select a white ball is 2/3.
Again, on the other direction: if the probability to select a white ball is 2/3, the situation MUST be WWB.

WBB: the probability for this situation is 1/4 because it happens only if the situation was BB before the white ball was inserted.
if the situation is WBB, then the probability to select a white ball is 1/3.
or if the probability to select a white ball is 1/3 then the situation MUST be WBB.

BBB: the probability for this situation is 0, because we know a white ball was inserted into the jar.
if the situation is BBB, then the probability to select a white ball is 0.
or if the probability to select a white ball is 0, then the situation MUST be BBB.

Now, after all of the above, let's see that's the probability to select a white ball.

there is a probability of 1/4 to get WWW which gives us a white ball in a probability of 1.
1/4 multiply by 1 is 1/4.

there is a probability of 1/2 to get WWB which gives us a white ball in a probability of 2/3.
1/2 multiply by 2/3 is 1/3.

there is a probability of 1/4 to get WBB which gives us a white ball in the probability of 1/3.
1/4 multiply by 1/3 is 1/12.

now let's add up all of the sums:
1/4 + 1/3 + 1/12 = 2/3

That's why the probability to select a white ball is 2/3.
Now we must remember that above we concluded that if the probability to select a white ball is 2/3 the situation MUST be WWB.
This stands in contradiction with the fact that we don't know what's the situation inside the jar.

Happy solving...

JPicasso
Jun 19th, 2002, 01:31 PM
What's the question?

Lior
Jun 19th, 2002, 02:13 PM
Settle the contradiction of course.

JPicasso
Jun 19th, 2002, 02:44 PM
I guess I don't see the problem...
(sorry, I'm not trying to spoil things for the sake of spoiling)

Look at it this way, if each ball (when there are two) has an equal
change of being black or white (or both white, or both black),
then it stands to reason that about
1/2 of the time I'll pull a while ball.
(unless we find out that they are both black and I keep drawing from the same bag :D )

If you add a guaranteed white ball, it's logical that those
odd increase. So, 2/3 does not sound unreasonable.

I'm not seeing a contradiction.

Guv
Jun 20th, 2002, 10:36 AM
There is no paradox. There is only confusion due to a complicated analysis.

Basically, the situation is as follows. 1/4 WWW, P(White) = 1/4 * 1 or 3/12. P(Black) = 1/4 * 0 or 0

1/2 BWW, P(White) = 1/2 * 2/3 or 4 /12 P(Black) = 1/2 * 1/3 or 2/12

1/4 BBW, P(White) = 1/4 * 1/3 or 1/12 P(Black) = 1/4 * 2/3 or 2/12

TotalP(White) = 8/12 or 2/3
TotalP(black) = 4/12 or 1/3

The above is a correct analysis, and agrees with your computation of probabilitites.

The paradox is resolved by noting that the following statment is not valid. Again, on the other direction: if the probability to select a white ball is 2/3, the situation MUST be WWB.

galway
Jun 20th, 2002, 10:40 AM
The contradiction he speaks of is in the first part of his post he determines that the only way to have a 2/3 probability of getting a white is if there is WWB. The he goes on to determine that the overall chance of pulling a white marble not knowing what the situation in the jar is 2/3. It is claimed that this is a contradiction because part one says 2/3 probability neccesitates WWB, but part 2 doesn't have that stipulation.

The reason for this "contradiction" is because they are 2 different problems. In part 1 he is calculating the chances of pulling a white for 3 differnet jar configurations. In problem 2 you are calculating the chances of pulling a white with no particular jar configuration. These are 2 different problems and it is just a coincedence that they are both 2/3

galway
Jun 20th, 2002, 10:49 AM
The contradiction he speaks of is in the first part of his post he determines that the only way to have a 2/3 probability of getting a white is if there is WWB. The he goes on to determine that the overall chance of pulling a white marble not knowing what the situation in the jar is 2/3. It is claimed that this is a contradiction because part one says 2/3 probability neccesitates WWB, but part 2 doesn't have that stipulation.

The reason for this "contradiction" is because they are 2 different problems. In part 1 he is calculating the chances of pulling a white for 3 differnet jar configurations. In problem 2 you are calculating the chances of pulling a white with no particular jar configuration. These are 2 different problems and it is just a coincedence that they are both 2/3

smwwilson
Jul 5th, 2002, 03:12 PM
this is not a paradox. its so counter-intuitive if you just think about it for one second. the first case is for a single occurence and the last case is all-encompassing of all the probabilities. dont be tricked