Probability question - hypergeometric distribution?

[AskingQ]

Hi,

I have never quite worked this type of probability question out, so would like some help please.

Imagine this scenario:
There are 4 people sat around a table, A, B, C and D.
A is sitting opposite C, B is sitting opposite D.
There is a bag with 16 balls numbered 1-16.
The balls are randomly drawn so that each person has 4 balls.

Imagine you are A.
And the balls you have are 1,4,5,9.

Now you want to work out what is the probability that C has a ball that is higher than your highest ball i.e. you want to work out whether C has AT LEAST one of 10,11,12,13,14,15, or 16.

Now the way I look at it is that the:
probability of C having ball 10 is 1/3
probability of C having ball 11 is 1/3
probability of C having ball 12 is 1/3
probability of C having ball 13 is 1/3
probability of C having ball 14 is 1/3
probability of C having ball 15 is 1/3
probability of C having ball 16 is 1/3

But I just don't know what to do next. Any assistance would be appreciated.

Thanks.

 

[Matterwave]

Well I would look at it this way. There are 7 balls that beat your 9, and 5 balls that lose. So if C is picking 4 balls at random from the other 12 balls available, what are his chances of not picking a ball that is greater than 9?

 

[AskingQ]

Ah. Ok. So basically you're saying that of the remaining 12 balls, the probability that C has at least one ball higher than my highest ball which is 9 is (1 - probability that C has all balls lower than mine).

Which comes to

1 - (5/12 * 4/11 * 3/10 * 2/9) = 1 - 120/11880 = 0.98...

That seems rather high to me, am I making a silly mistake somewhere?

Thanks.

 

[pbuk]

Which comes to

1 - (5/12 * 4/11 * 3/10 * 2/9) = 1 - 120/11880 = 0.98...

That seems rather high to me, am I making a silly mistake somewhere?

Thanks.

You are making a silly mistake thinking it is high - look at the balls that are left and think about which combinations C could have that don't have any balls above 9,

 

[Matterwave]

Ah. Ok. So basically you're saying that of the remaining 12 balls, the probability that C has at least one ball higher than my highest ball which is 9 is (1 - probability that C has all balls lower than mine).

Which comes to

1 - (5/12 * 4/11 * 3/10 * 2/9) = 1 - 120/11880 = 0.98...

That seems rather high to me, am I making a silly mistake somewhere?

Thanks.

This is correct. Of the remaining balls, there is only 5 balls that are lower than 9. If C picks 4 random balls out of 12, it's hard for him to always pick out of those 5 which are lower than 9 and get none of the 7 balls that are higher.