Recently whilst playing pool i was reaching into the table to lift the
balls out so as i could set up the rack.
I reached in with both hands at the same time and at the exact same
time i lifted out 4 balls in each hand. When i set the balls on the
table the 4 balls in my left hand were red and the 4 balls in my right
hand were yellow
In total there were 15 balls inside the table. 7 were red, 7 were
yellow and 1 was black.

I want to know what the chances were of this happening, that is the
chances of me reaching into the table and lifting out 4 balls of the
same colour in one hand and 4 balls of the same colour in the other
hand.

I also want to see how you work this out as it has been puzzleing me
for some time now.

Hello!
The probability of the event that you take 4 balls of the same colour
(any colour) in one hand and 4 balls of the other colour in the other
one is about 0.54%

The procedure to compute it is the following. Let's first calculate
the probability of getting 4 red balls in your left hand and 4 yellow
balls in your right hand.

When you pick up the first ball with your left hand, there is a 7/15
probability that you pick up a red ball. If you do pick up a red ball,
then there will be a 6/14 probability (6 red balls, over 14 balls
left) that you pick up another red ball. Again, if you pick a red
ball, then there is a 5/13 probability that the next one will be red
once again, and for the final ball with your left hand there is a 4/12
probability. Therefore, the probability of picking 4 red balls with
your left hand is:

(7/15)*(6/14)*(5/13)*(4/12) = 0.0256...

Now let's continue with the right hand. There are now 11 balls left in
the table. Therefore, the probability of picking a yellow ball is
7/11. If you pick a yellow ball first, then there is a 6/10
probability of picking another yellow one. Following the same logic as
above, the next probabilities are 5/9 and 4/8. Therefore, the
probability of picking 4 yellow balls in your right hand, given that
you picked 4 red ones in your left hand is:

(7/11)*(6/10)*(5/9)*(4/8)

We thus conclude that the probability of picking 4 red balls in your
left hand and 4 right balls in your right hand is:

(7/15)*(6/14)*(5/13)*(4/12)*(7/11)*(6/10)*(5/9)*(4/8) = 0.0027

However, you asked for the probability of picking 4 balls of any one
color in one hand and 4 of the other color in the other one.
Therefore, we must also take in account the probability that you get 4
yellow balls in your left hand and 4 red balls in your right hand.
Obviously, this probability is exactly the same: 0.0027. We thus get
that the probability of picking 4 balls of any one color in one hand
and 4 of the other color in the other one is 0.0027+0.0027 = 0.0054
(or 0.54%)


I hope this helps! If you have any doubt regarding my answer, please
don't hesitate to request clarification before rating it. Otherwise, I
await your rating and final comments.

Best wishes!
elmarto

Thats excellent thank you very much, this has been bugging me for
quite a while now.
Thank you

Thank you for the nice comments and rating!