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#1 2020-10-31 02:48:37

blob123
Member
Registered: 2020-10-31
Posts: 4

Sports draw problem

Hi,
I'm stuck with this football draw problem: could you help me out ?
I have 20 football teams: 11 are from Canada, 9 are from Mexico.
I will draw randomly 10 pairings out of those 20 teams so as to create 10 matches...
What are the probabilities to have :
- only 1 pairing comprised of 2 canadian teams ?
- only 1 pairing comprised of 2 mexican teams ?
- exactly 2 pairings each comprised of 2 canadian teams ?
- exactly 2 pairings each comprised of 2 mexican teams ?
- exactly 3 pairings each comprised of 2 canadian teams ?
- exactly 3 pairings each comprised of 2 mexican teams ?
- exactly 4 pairings each comprised of 2 canadian teams ?
- exactly 4 pairings each comprised of 2 mexican teams ?
- at least 4 pairings each comprised of 2 canadian teams ?
- at least 4 pairings each comprised of 2 mexican teams ?
Many thanks for your help !

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#2 2020-10-31 23:13:08

Bob
Administrator
Registered: 2010-06-20
Posts: 10,053

Re: Sports draw problem

hi blob123

Welcome to the forum.

One caution.  I'm a bit shaky on questions like this so I may be wrong but here's how I'd tackle these.

Make the question easier by simplifying the numbers.

Let's say you have just 6 teams and you're trying to make up three matches.  I'll call them C1 C2 C3 C4 M1 and M2

How many ways can we select the matches?

Pick any team.  There are 5 opponents to choose from.  Pick another team.  There are 3 opponents to choose from.  Pick a final team.  There is just 1 opponent to 'choose' from.

So that makes 5 x 3 x 1 = 15 possible selections.

EDIT:  I've had a further think and the next bit isn't correct yet.  I still think it's ok to try and simplify the problem. I'm working on a modification that will yield the correct version.  Hope to get there later today. LATER EDIT. This is now corrected.

Now require that just two Cs are together.  There are so few possibilities I can easily list them:

{C1C2 C3M1 C4M2} {C1C2 C3M2 C4M1} {C1C3 C2M1 C4M2} {C1C3 C2M2 C4M1} {C1C4 C2M1 C3M2} {C1C4 C2M2 C3M1} {C1M1 C2C3 C4M2} {C1M1 C2C4 C3M2} {C1M1 C3C4 C3M2} {C1M2 C2C3 C4M1} {C1M1 C2C4 C3M1} {C1M2 C3C4 C2M1}

So I have 12 possibilities making a probability of having just2 Cs of 12/15 = 4/5

Is there a way to get this without listing all the possibilities?

Start with a C (4 choices) Choose another (3 choices) Divide by 2 as I'm counting eg C1C2 and C2C1.

The remaining Cs must each be paired with an M; there's just two ways to do this. (eg C3M1 or C3M2 ... last pair is fixed).

So we have 4 x 3 / 2 x 2 = 12.

That approach can now be scaled up for the 20 teams.

Bob


Children are not defined by school ...........The Fonz
You cannot teach a man anything;  you can only help him find it within himself..........Galileo Galilei
Sometimes I deliberately make mistakes, just to test you!  …………….Bob smile

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#3 2020-11-01 01:56:35

blob123
Member
Registered: 2020-10-31
Posts: 4

Re: Sports draw problem

Many thanks Bob for your reply.
I understand your method, even though I'm not sure we have to divide par 2 to account for the same pairings (e.g. C1C2 and C2C1)...
I tried to generalize your method for the 20 teams, but I end up with a huge number of possible combinations of only two Cs drawn together.
So I don't think this is working..

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#4 2020-11-02 03:12:11

Bob
Administrator
Registered: 2010-06-20
Posts: 10,053

Re: Sports draw problem

hi

I haven't forgotten you.  I'm working on a couple of approaches which may help.  Is it possible for you to post the working you tried ... don't worry if it's muddly ... I'll try to work it through.

Bob


Children are not defined by school ...........The Fonz
You cannot teach a man anything;  you can only help him find it within himself..........Galileo Galilei
Sometimes I deliberately make mistakes, just to test you!  …………….Bob smile

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#5 2020-11-02 07:08:48

blob123
Member
Registered: 2020-10-31
Posts: 4

Re: Sports draw problem

It seemed to me that COMBIN(11,i)/COMBIN(20,i) is the probability to end up with i pairing from Canada...
But this doesn't seem to work.

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#6 2020-11-02 15:47:55

Denominator
Member
Registered: 2009-11-23
Posts: 219

Re: Sports draw problem

Hi, here's my two cents but do take it with a grain of salt!
My notation: denote C1 as first match Canada vs Canada; M2 as second match Mexico vs Mexico; CM3 as third match Canada vs Mexico; and so on...

Dealing with the first sub-problem: 'only 1 pairing comprised of 2 Canadian team'
One such arrangement is C1,CM2,CM3,...,CM10.
Let's compute the probability of this arrangement:
P(C1,CM2,CM3,...,CM10) = (11/20*10/19)*(9/18*9/17*2)*(8/16*8/15*2)*(7/14*7/13*2)*...*(2/4*2/3*2)*(1/2*1/1*2) = 0.003048...
This is only one possible arrangement, another arrangement is CM1,C2,CM3,CM4,...,CM10 for example.
There are 10 of these arrangements, therefore the answer to the first subproblem is 0.03048.

So one possible strategy for the first 8 subproblems can be summarised as:
1. Compute the probability of one such arrangement satisfying the conditions
2. Find the total number of arrangements
3. Multiply them both for the overall probability (you can do this because the probability of each arrangement should be the same).

The solution to the last two subproblems is different.
Hope this helps and good luck!

Last edited by Denominator (2020-11-02 15:48:47)


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#7 2020-11-14 07:34:47

blob123
Member
Registered: 2020-10-31
Posts: 4

Re: Sports draw problem

Many thanks !

For 2 pairings with Canada vs Canada, is it correct to write:

P(C1,C2,CM3,...,CM10) = (11/20*10/19)*(9/18*8/17)*(7/16*7/15*2)*...*(2/4*2/3*2)*(1/2*1/1*2) ?

And the number of arrangements of this is combin(10,2)=45 ?

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