Interesting Probability question that I can't figure out.

[MathBuggy]

Question solved.

 

[Unco]

Here's one approach (Soroban probably has a more obvious method).

There are seven pairs to put our socks in:

_ _ | _ _ | _ _ | _ _ | _ _ | _ _ | _ _

We need to put 6 blues in there, ensuring they're in seperate pairs.

e.g.
B _ | B _ | B _ | B _ | B _ | B _ | _ _

There are in fact 7 places to put the 6 blues, so the number of ways to distribute them would be 7C6 = 7. You can also think of this as there being a pair without a blue everytime, and this can occur 7 times.

If we now place the 4 reds. One red has to go in the emtpy pair (the one with no blues) because otherwise we'd have to stick two whites in there.

e.g
B _ | B _ | B _ | B R | B R | B R | R _
There 6 places (not 7 because of the empty one having to be filled) to distribute those 3 reds, so 6C3 = 20 ways to distrubute the reds.

There are 4 places left for the 4 whites. 4C4 = 1.

Thus the total ways of getting no pairs to match is 7*20*1.

The probability of getting no matching is this divided by the total arrangements. For total arrangements you're placing 6 blues in 14, then 4 reds in 8 etc).

If I have it got it wrong, Soroban, could you please help me out a bit? Thanks.

 

[pka]

You have tackled a very difficult counting problem.
I think that you have greatly over counted the possible pairings.
You did this by considering the pairs as if they were different.
It is as if you were putting two socks each into seven different drawers.

Consider the problem with fewer pairs.
Suppose we had only two blue pair and one red pair.
The total paring would be: (B,B)(B,B)(R,R) or (B,B)(B,R)(B,R).
That is, there would only be two different parings.

Again, take two blue pair, a red pair and a white pair.
(B,B)(B,B)(R,R)(W,W)
(B,B)(B,B)(R,W)(R,W)
(B,B)(B,R)(B,R)(W,W)
(B,B)(B,W)(B,W) (R,R)
(B,B)(B,R) (B,W) (R,W)
(B,R)(B,R)(B,W)(B,W)
Thus, now we have exactly only six possible pairings.

 

[Unco]

Thanks pKa, that makes a lot of sense.

 

[MathBuggy]

I see your logic, but how do you actually go about solving the question...you can't just write down the sets when there are so many socks required...Is there some sort of pattern or formula that relates?

 

[Denis]

Haven't tried some "scientific approach" yet;
but ran simulation and get approximately 17/200;
so in 183 weeks out of 200, he'll get a matching pair at least one morning!

My simulation effectively shuffles a deck of 14 cards:
4 aces, 4 2's and 6 3's;
checks 1st against 2nd (Monday), 3rd against 4th (Tuesday), ...
Anybody see anything wrong with that?

31 13 31 23 32 22 31 : one matching pair only, on Saturday :idea:

31 13 31 23 32 23 21 : a really bad week!

 

[pka]

“you can't just write down the sets when there are so many socks required”
Oh yes, sometimes that is all one can do.
How are you with working with generating functions?
As I said this is a very, very difficult counting problem.
Sometimes a simple looking problem is the most difficult.

 

[Denis]

More on that sock-it-to-me problem...

A simulation using 10 million weeks:
(as soon as a match is found, I jump to next week)

a match found : 9 147 168
1 is the match pair: 1 823 987
2 is the match pair: 1 819 131 (these 2 are really equal, of course)
3 is the match pair: 5 504 050 (3 times one of above, really)

Match pair on:
Mon : 2 967 698
Tue : 2 078 888
Wed :1 460 452
Thu : 1 029 063
Fri : 728 654
Sat : 517 501
Sun: 364 912

Did a search for all days of week matched: 1/1000

 

[happy]

I'll notify the people over here, http://www.mathgoodies.com/forums/topic ... C_ID=26337, so you don't waste anyone's time. :wink:

 

[pka]

I did a MathCad simulation on 5000 random pairing.
I ran it 50 times and got a average of 05.8% of NO MATCHES

 

[Denis]

I did a MathCad simulation on 5000 random pairing.
I ran it 50 times and got a average of 05.8% of NO MATCHES

Could that be a typo, and should be 8.5%, pka?

 

[Denis]

Here's one approach (Soroban probably has a more obvious method).
There are seven pairs to put our socks in:
_ _ | _ _ | _ _ | _ _ | _ _ | _ _ | _ _
We need to put 6 blues in there, ensuring they're in seperate pairs.
e.g.
B _ | B _ | B _ | B _ | B _ | B _ | _ _

There are in fact 7 places to put the 6 blues, so the number of ways to distribute them would be 7C6 = 7. You can also think of this as there being a pair without a blue everytime, and this can occur 7 times.
If we now place the 4 reds. One red has to go in the emtpy pair (the one with no blues) because otherwise we'd have to stick two whites in there.

Well, why place the 4 reds now?
Simply put a red and a white in last section and you're finished,
since it then don't matter where you put the last 3 reds and blues.