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 Using Cards as Dice (Posted on 2004-08-09)
A deck of nine cards can be numbered, so that the sum of the numbers on a randomly chosen pair of cards totals to an integer from 2 to 12 with the same frequency as rolling two standard dice. What are the numbers on the nine cards?

 See The Solution Submitted by Brian Smith Rating: 2.7500 (4 votes)

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 Solution | Comment 8 of 9 |

For the Dice we have: (written as Integer: Frequency (specific additions))

2: 1 (1+1)
3: 2 (1+2, 2+1)
4: 3 (1+3, 3+1, 2+2)
5: 4 (1+4, 4+1, 2+3, 3+2)
6: 5 (1+5, 5+1, 2+4, 4+2, 3+3)
7: 6 (1+6, 6+1, 2+5, 5+2, 3+4, 4+3)
8: 5 (2+6, 6+2, 3+5, 5+3, 4+4)
9: 4 (3+6, 6+3, 4+5, 5+4)
10: 3 (4+6, 6+4, 5+5)
11: 2 (5+6, 6+5)
12: 1 (6+6)

So there are 36 possibilities (6 choose 2 permutations with replacement = 6^2).

Now what do we have for the cards? We have 9 choose 2 combinations without replacement, which is 9!/[2!*(9-2)!] = 36. This is convenient. We just have to have the same number of possible ways of getting each number.

Well, I tried putzing around for a while, and tried things like 1, 1, 2, 5, 6, 6 (with 3 more cards to go), 1, 1, 2, 3, 4, 5, 7 (with 2 more cards to go) and a couple others. I kept getting stuck, and wished that I could pick a number between 3 and 4, for example.

Then it hit me. What if I used x.5 numbers? The way I could solve this is by starting a new, but similar, problem statement, and then scale it down to the actual problem. So instead of solving for integers 2-12, I solved for EVEN integers from 4-24 (so I doubled each integer). Then I limited myself to only putting odd numbers on the cards (so when I scaled back down, these would all be x.5 numbers).

So some of the easy card numbers to pick were 1 and 3 to make 4, and 11 and 13 to make 24. I don't want to add another 1, 3, 11, or 13 again because this will either make more of a number that I don't need any more (like another 1+3 to make a second 4 when I only need one), or it will make numbers that I am not looking for (like 1+1=2 or 13+13=26).

So just for kicks I'll add the other odd numbers in between, 5, 7 and 9. Let's see what we have so far with 1, 3, 5, 7, 9, 11, and 13.

4: 1 (1+3)
6: 1 (1+5)
8: 2 (1+7, 3+5)
10: 2 (1+9, 3+7)
12: 3 (1+11, 3+9, 5+7)
14: 3 (1+13, 3+11, 5+9)
16: 3 (3+13, 5+11, 7+9)
18: 2 (5+13, 7+11)
20: 2 (7+13, 9+11)
22: 1 (9+13)
24: 1 (11+13)

Not bad, we haven't broken any rules yet. We need another way of getting 6. We could add another 3 and then have 3+3=6, but then we would also have another 1+3=4 and we don't want any more ways to get 4. We could add another 5 to get another 1+5=6, without obviously messing anything up. Let's see how that works.

4: 1 (1+3)
6: 2 (1+5, 1+5)
8: 3 (1+7, 3+5, 3+5)
10: 3 (1+9, 3+7, 5+5)
12: 4 (1+11, 3+9, 5+7, 5+7)
14: 4 (1+13, 3+11, 5+9, 5+9)
16: 4 (3+13, 5+11, 5+11, 7+9)
18: 3 (5+13, 5+13, 7+11)
20: 2 (7+13, 9+11)
22: 1 (9+13)
24: 1 (11+13)

We're still doing ok. Now we need another way of getting 22. We could add another 11 and then have 11+11=22, but then we would also have another 11+13=24 and we don't want any more ways to get 24. We could add another 9 to get another 9+13=24, without obviously messing anything up. Let's see how that works. 1, 3, 5, 5, 7, 9, 9, 11, and 13.

4: 1 (1+3)
6: 2 (1+5, 1+5)
8: 3 (1+7, 3+5, 3+5)
10: 4 (1+9, 1+9, 3+7, 5+5)
12: 5 (1+11, 3+9, 3+9, 5+7, 5+7)
14: 6 (1+13, 3+11, 5+9, 5+9, 5+9, 5+9)
16: 5 (3+13, 5+11, 5+11, 7+9, 7+9)
18: 4 (5+13, 5+13, 7+11, 9+9)
20: 3 (7+13, 9+11, 9+11)
22: 2 (9+13, 9+13)
24: 1 (11+13)

There you go. So the answer, based on this method, is: 0.5, 1.5, 2.5, 2.5, 3.5, 4.5, 4.5, 5.5, 6.5

Just cleaned up some spacing

Edited on August 11, 2004, 3:07 pm
 Posted by nikki on 2004-08-11 14:56:20

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