sorry i thought someone who is familiar with this kind of thing might be able to do it easily. i certainly didn't expect it to take 5 hours for the right person (although i admit it might take me 5 weeks to work it out).

unfortunately, my spec hasn't really been satisfied yet:


there's no particular real world example yet, although the dice are a good example IF the number is 6, or i suppose you could think of a 9-sided dice if you wanted to work with 9 numbers.

but this knowledge would be handy for many different purposes and i'm a bit sad that i don't remember any of this stuff from school. i'd like to try to get my head around the math again without resorting to buying some old books and studying it for months. combinations and permutations.

i wish i remembered from school what that meant.

thank you bytecoin that is very close to the random data.
to pay the bounty though, i must see how you arrived at your final numbers. did you use a formula or did you write out numbers on a page 46000 times?

i suppose instead of random data i could populate my test table with exactly 1 instance of every combo (46656*46656 records?) and then count the matches, but doing that would assume 'hard-coding' of the number of product variations (6), which i'd rather avoid.

Thank you for spotting this, I couldn't figure out why his numbers were so different.

or your interpretation of my description

it seems extremely unlikely to me given the description, that 0 matches would occur 83% of the time.

here's a few example rows from the data:

[6,3,6,6,3,4] + [6,3,4,3,6,6] = 6 matches
[1,5,2,6,1,5] + [2,3,5,6,5,1] = 5 matches
[4,2,1,1,5,6] + [1,5,3,3,3,1] = 3 matches
[5,6,3,5,5,1] + [1,2,5,4,3,2] = 3 matches
[4,2,5,2,6,6] + [2,1,1,1,1,3] = 1 match
[3,5,5,4,4,5] + [1,1,6,1,6,2] = 0 matches

i can use random.org all day long but i'd much rather find out the actual mathematical formula.

edit: here's some more...

[1,5,2,6,1,5] + [2,3,5,6,5,1] = 5
[4,2,1,1,5,6] + [1,5,3,3,3,1] = 3
[5,6,3,5,5,1] + [1,2,5,4,3,2] = 3
[5,2,6,2,4,1] + [2,5,5,4,3,6] = 4
[1,2,2,5,6,2] + [1,4,3,6,5,6] = 3
[5,6,3,2,2,4] + [1,3,5,3,2,2] = 4
[2,4,3,6,4,1] + [1,5,2,2,2,6] = 3
[3,1,4,3,1,1] + [5,4,1,4,6,2] = 2
[6,3,3,1,3,3] + [4,1,5,6,1,4] = 2
[3,5,1,2,1,1] + [6,4,3,5,2,2] = 3
[5,4,6,3,2,5] + [4,4,3,3,2,1] = 3
[4,6,4,2,4,5] + [3,3,2,1,5,3] = 2
[4,2,5,2,6,6] + [2,1,1,1,1,3] = 1
[1,4,1,1,6,5] + [5,1,1,1,3,3] = 4
[2,2,6,4,4,1] + [1,6,5,3,2,6] = 3
[2,4,5,4,4,3] + [2,4,4,6,2,4] = 4
[3,3,1,5,3,3] + [2,3,2,5,5,6] = 2
[1,1,3,2,2,6] + [6,4,1,3,2,4] = 4
[5,1,2,6,2,1] + [4,6,2,3,3,6] = 2
[6,6,5,2,3,6] + [2,5,5,3,5,2] = 3
[2,1,6,4,2,4] + [4,1,2,6,5,4] = 5
[1,2,5,2,6,1] + [3,5,3,3,2,3] = 2
[3,5,5,5,5,6] + [5,6,6,6,5,1] = 3
[5,2,3,2,4,4] + [4,5,4,1,1,1] = 3
[6,1,5,2,1,4] + [3,3,5,3,3,2] = 2
[1,2,5,6,4,1] + [2,2,5,1,4,3] = 4
[6,2,6,2,3,6] + [3,6,3,1,2,3] = 3
[3,4,2,6,5,1] + [2,6,4,6,6,4] = 3
[6,6,2,2,6,4] + [2,5,5,4,4,3] = 2
[5,4,6,1,4,3] + [6,2,5,3,5,1] = 4
[3,2,4,5,2,2] + [1,5,6,6,3,4] = 3
[1,3,4,1,2,3] + [6,2,6,6,1,4] = 3
[4,1,4,5,4,3] + [5,4,4,6,1,6] = 4
[4,2,4,6,4,5] + [3,2,4,4,6,1] = 4
[6,3,4,6,2,3] + [3,5,1,2,3,4] = 4
[3,5,5,4,4,5] + [1,1,6,1,6,2] = 0
[3,5,1,4,6,1] + [2,3,2,4,3,2] = 2
[6,6,6,4,6,3] + [4,5,6,6,5,6] = 4
[3,2,3,3,6,4] + [3,2,5,6,5,5] = 3
[5,2,6,1,1,1] + [5,6,3,4,3,3] = 2
[3,1,6,6,5,4] + [1,1,1,3,4,5] = 4
[6,5,2,1,2,3] + [6,1,4,1,3,2] = 4
[5,3,4,3,4,4] + [6,6,1,2,5,2] = 1
[5,5,3,4,1,6] + [2,1,1,2,3,2] = 2
[6,5,5,5,1,6] + [4,3,3,6,5,5] = 3
[5,4,3,6,4,2] + [3,2,1,2,1,5] = 3
[2,3,5,1,1,6] + [6,4,5,6,3,2] = 4
[4,6,1,2,6,1] + [2,1,4,4,2,2] = 3
[3,3,1,3,4,1] + [1,4,6,5,4,6] = 2
[4,2,3,5,3,2] + [6,4,4,5,4,6] = 2
[5,2,4,1,6,1] + [6,1,1,2,2,4] = 5
[6,3,4,4,5,5] + [4,2,3,1,4,2] = 3
[5,1,1,1,2,1] + [5,4,6,1,6,5] = 2
[3,4,2,5,4,3] + [4,1,2,3,6,2] = 3
[5,1,5,1,6,3] + [2,1,5,5,5,6] = 4
[6,5,4,3,3,4] + [5,3,4,1,4,1] = 4
[1,3,1,6,6,3] + [1,1,1,4,4,1] = 2
[4,3,4,4,1,2] + [3,6,1,6,1,3] = 2
[5,5,1,1,6,1] + [2,1,1,4,1,2] = 3
[3,6,1,5,1,6] + [2,6,2,6,2,3] = 3
[6,3,5,6,4,5] + [3,1,3,3,1,3] = 1
[6,3,4,3,2,4] + [4,4,4,4,1,3] = 3
[4,2,3,3,5,1] + [5,2,4,6,2,3] = 4
[5,2,1,5,4,2] + [4,3,2,4,6,4] = 2
[6,4,4,3,3,4] + [2,5,4,1,4,5] = 2
[4,2,3,1,4,1] + [1,2,3,5,2,3] = 3
[1,2,4,1,3,3] + [1,6,2,2,3,6] = 3
[1,5,6,5,3,3] + [5,6,5,5,1,4] = 4
[4,5,5,6,1,5] + [4,2,6,4,2,1] = 3
[4,5,3,2,6,6] + [5,4,1,1,6,2] = 4
[1,2,2,5,4,6] + [3,6,2,5,2,1] = 5
[1,2,4,1,5,3] + [3,4,4,1,4,5] = 4
[5,2,5,5,2,1] + [6,2,1,4,3,4] = 2
[4,3,4,1,4,5] + [4,1,3,3,2,1] = 3
[3,1,2,3,2,1] + [1,2,4,1,1,1] = 3
[6,4,5,1,6,4] + [2,6,3,4,5,3] = 3
[4,1,3,5,6,2] + [5,3,1,4,5,4] = 4
[5,5,5,4,4,2] + [5,2,3,3,1,2] = 2
[4,1,5,1,2,5] + [1,5,2,4,4,4] = 4
[5,2,5,2,5,4] + [5,2,5,1,1,5] = 4
[4,3,5,3,4,3] + [1,3,3,4,5,2] = 4
[2,2,4,4,4,4] + [4,2,4,5,6,2] = 4
[4,6,5,3,2,4] + [5,4,5,5,4,6] = 4
[2,1,2,1,2,5] + [4,4,5,1,1,3] = 3
[1,4,6,6,3,6] + [6,5,4,1,4,3] = 4
[5,3,6,2,1,4] + [4,1,6,1,1,6] = 3
[3,2,3,6,3,2] + [6,5,5,4,3,4] = 2
[3,1,5,2,1,1] + [4,3,5,2,2,2] = 3
[4,5,6,2,5,3] + [2,5,4,2,1,3] = 4
[6,6,6,1,4,2] + [2,4,4,5,6,3] = 3

thank you kjj, should i send to your signature address or a different one?

also, for a bonus 0.05, what's the chance that there are no matches?

thanks

edit: i ran the numbers with a lot of data from random.org and after 5831 tries, i got the following ratios, which disagree quite a lot with your figures:

matches   count(matches)   (count(matches)/5831*100)
0   37   0.6345
1   359   6.1567
2   1222   20.9570
3   2099   35.9973
4   1676   28.7429
5   417   7.1514
6   21   0.3601

sorry i don't know where it needs to go

yes
sounds correct, give me a few minutes to think about it


Can you simplify them into a % for me? I don't get how to take that statement and work out the 'chance' that it occurs.

Should P1 be equal to P6 (1 in 46656) or am I just guessing wrong on that (wild guess)?

Fact 1:
Alice chooses 6 items from the following list of products:
A1, A2, A3, A4, A5, A6

Fact 2:
Bob chooses 6 items from the following list of products:
B1, B2, B3, B4, B5, B6

Fact 3:
They are free to choose the same product more than once, so for example, Alice might choose to buy the following 6 items:
A1, A1, A2, A2, A2, A6

Fact 4:
Each 'A' product is used with a corresponding 'B' product, so for example, an A5 needs exactly one B5 to be of any use, and a B5 needs exactly one A5 to be useful (Let's call this match a 'complimentary pair').

Questions:

From Bob and Alice's 12 items, what are the chances that:

• They have exactly 1 complimentary pair
  (eg, Alice chose A1, A2, A3, A4, A5, A6, Bob chose B1, B1, B1, B1, B1, B1)
  
  
• They have exactly 2 complimentary pairs
  (eg, Alice chose A1, A2, A3, A4, A5, A6, Bob chose B1, B2, B1, B1, B1, B1)
  
  
• They have exactly 3 complimentary pairs
  (eg, Alice chose A1, A2, A3, A4, A5, A6, Bob chose B1, B2, B3, B1, B2, B3)
  
  
• They have exactly 4 complimentary pairs
  (eg, Alice chose A1, A2, A3, A4, A5, A6, Bob chose B1, B2, B3, B4, B4, B4)
  
  
• They have exactly 5 complimentary pairs
  (eg, Alice chose A1, A2, A3, A4, A5, A6, Bob chose B1, B2, B3, B4, B5, B2)
  
  
• They have exactly 6 complimentary pairs
  (eg, Alice chose A1, A2, A3, A4, A5, A6, Bob chose B1, B2, B3, B4, B5, B6)
  
  

For each correct answer with detailed workings I will give 0.05 btc (~$1) for a total of 0.30 btc (~$6).