Drawing balls from a pot.

Pigman II

Say you have a pot with eight balls
4x red
1x blue
1x green
1x black
1x white.

Now if you draw the balls out in pairs, what is the probaility that at least one of the pairs will be two reds?

Crumbs

This is obviously to do with the Champions League draw, right?

You need to first work out the probability that none of the reds will be paired and then subtract that from 1. I'll give it a quick bash but I can't guarantee it's correct.

P(all reds drawn in separate pairs) = (4/8)(4/7)(2) * (3/6)(3/5)(2) * (2/4)(2/3)(2) * (1) = 8/35.

P(at least one pair is two reds) = 1 - (8/35) = 27/35.

Just over 77%.

plodder

I worked it out a different way using combinations.

The total number of combinations of 2 balls taken from a set of 8 is 8C2, which is 28. The number of red only combinations is 4C2, which is 6. So, the probability of two pairs of red balls is 6/28 = 21%

To get the probability of at least one red only pair, you need to add the numbers of combinations of one red + one other colour pairs, which is [EDITING MISTAKE] 4*4

So, the probability of at least one red-only pair is (6+16)/28 = 78%

MathsManiac

The concept is correct, but I don't think those calculations are right.

If I understand the question correctly, the 8 teams are being used to create an unordered list of four unordered pairs, and the selection is performed in a manner that makes all these equally likely. A case-by-case analysis yields the following, I think:
(I'm pretending black is really yellow, so that I won't have two colours starting with B!)

case 1: no reds paired: only one possible combination: {RB,RG,RY,RW}
case 2: two red pairs: three possible combinations: {RR,RR,BG,YW},{RR,RR,BY,GW},{RR,RR,BW,GY}.
case 3: one red pair: six possible combinations: {RR,RB,RG,YW},{RR,RB,RY,GW},{RR,RB,RW,GY},{RR,RG,RY,BW},{RR,RG,RW,BY},{RR,RY,RW,BG}.

Assuming all of the above are equally likely, the answer to your question is 9/10.


Does that make sense?

[Edit, I was responding to Crumbs' post]

MathsManiac

plodder wrote: »

I worked it out a different way using combinations.

The total number of combinations of 2 balls taken from a set of 8 is 8C2, which is 28. The number of red only combinations is 4C2, which is 6. So, the probability of two pairs of red balls is 6/28 = 21%

To get the probability of at least one red only pair, you need to add the numbers of combinations of one red + one other colour pairs, which is 4*3 = 12.

So, the probability of at least one red-only pair is (6+12)/28 = 64%

I don't think this is correct either (perhaps obviously, since it's different from mine!!!) The 8C2 only applies if you're only creating one pair from the 8, not if you're dividing the 8 into 4 pairs of 2. 8 distinguishable objects can be divided into 4 distinguishable sets of two objects in C(8;2,2,2,2) ways = 8!/(2!2!2!2!) = 2520 ways. If the four sets are indistinguishable, this reduces by a factor of 4! to 105 ways. But to proceed further, (i.e., trying to count which of these has the combinations of reds you're interested in) I think you'd have to start down a case-by-case road anyway, to avoid double-counting.

(But now I'm worried by the fact that a sample space of 105 can't give an answer of exactly 9/10...)

plodder

MathsManiac wrote: »

I don't think this is correct either (perhaps obviously, since it's different from mine!!!) The 8C2 only applies if you're only creating one pair from the 8, not if you're dividing the 8 into 4 pairs of 2. 8 distinguishable objects can be divided into 4 distinguishable sets of two objects in C(8;2,2,2,2) ways = 8!/(2!2!2!2!) = 2520 ways. If the four sets are indistinguishable, this reduces by a factor of 4! to 105 ways. But to proceed further, (i.e., trying to count which of these has the combinations of reds you're interested in) I think you'd have to start down a case-by-case road anyway, to avoid double-counting.

(But now I'm worried by the fact that a sample space of 105 can't give an answer of exactly 9/10...)

Actually, I fixed a mistake in my original post, so I get the same answer as crumbs now. I think it is correct.

I think the reason why you are getting fewer combinations is that you are treating each of the four red balls as one "R". If you rewrite the problem naming the red balls, R1, R2, R3, and R4, then you will get more combinations.

MathsManiac

I was right to be worried. The combinations I identified in my first post are not equally likely. Crumbs was correct in his first post. 24 of the 105 combinations don't pair any reds, and the remaining 81 do, yielding the same answer as Crumbs gave.

Sincerest apologies.

[Edit: tempted to delete all my posts from this thread, but I won't!]

plodder

So, there was one all-English quarter final drawn (Arsenal v Liverpool) , as predicted. :cool:

Pherekydes

Hehe:

An Citeog wrote:

Thread: Champions league draw

BobbyD10 wrote:

I think the odds of an english quarter final is 43%, but I wonder what are the odds of them all avoiding each other.

Eh, how'd you work that one out?

4 English teams, 8 teams in total ==) odds of an English quarter final = 50%

Odds of them not being drawn against each other also = 50%

Pherekydes

MathsManiac wrote: »

case 1: no reds paired: only one possible combination: {RB,RG,RY,RW}

Assuming this problem is about the Champions' league QF draw, then this statement is incorrect.

The problem is in the original wording. There are not four identical red balls. Each ball is distinct.

Taking this statement again:

case 1: no reds paired: only one possible combination: {RB,RG,RY,RW}

If we see each red as a separate entity, we can see that there will be 16 such combinations. Manu can play each one of four different teams, and for each combo Arsenal can play four, etc., giving 16 pairings with no English pairings.

Carrying on to 2 English pairings, we get 3 combinations and each of these can have 3 different all-continental pairings, giving 9.

And finally, one English pairing: four teams gives six pairings and each of these can have 15 different pairings with one English team, giving 90.

Add them all up and I get 115. 99 of these gives at least one all-English pairing, or just over 86%.

Pherekydes

Slow coach wrote: »

If we see each red as a separate entity, we can see that there will be 16 such combinations. Manu can play each one of four different teams, and for each combo Arsenal can play four, etc., giving 16 pairings with no English pairings.

OK, this is starting to piss me off! I'm still wrong. Obviously, there can be four such combos, not 16.

The total number of distinct draws is what we're looking for.

How many distinct ways can we choose 4 pairs of objects from 8 distinct objects?

Pherekydes

I think the correct formula is:

(2m)!/(m!2^m)

where 2m = n (the number of teams)

This gives 105, which is what MathsManiac gave earlier.

That is, 105 is the number of distinct draws.

Only four of these (3.8%) give no English pairings. 96.2% do give at least one English pairing.

Is that it, finally?