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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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 Sol | Comment 7 of 9 |

Since the sum of dice (call this T) and sum of two cards are identically distributed, one ought to look at the properties of the # on the cards.

(1) The expected value of sum of the dice is 7.
(2) The sum of dice takes integer values over 2, 3, ..., 12.
(3) The number of choices for T is given by min(T-1, 13-T), or 1,2,...,6,5,...,1 for T=2, ..., 12

The second statement assumes that the cards either all takes integer values, or all takes 0.5 + integer values.  The expected value of each card is 3.5.  Now look at the distribution of T:

T=2  (1,1)
T=3  (1,2), (2,1)
T=4  (1,3), (2,2), (3,1)
...
T=7  (1,6), (2,5), ..., (6,1)
...

So for T=2 (12), there is only one choice
for T=3 (11), there are two choices
...
for T=7, there are 6 choices.

With this let's figure out the minimum value of each card.  Naturally, max(minimum value of cards) = 1 as minimum of sum of two cards is 2.  Suppose minimum is 1.  Then there must be two cards with 1 (1+1, 1x) for T=2.  Then there must be one card with 2 (1+2, 2x) for T=3.  Similarly, three cards with 3 (1+3, 3x) for T=4.  For T=5, we are stuck because there are already three choices from 2+3; but with two cards with 1, we cannot add another card with 4 because altogether there will be 5 choices for T=5.  Thus minimum is not 1.

Now suppose the minimum is x<=0.  Then the rest of the cards are one 2-x, two 3-x, three 4-x and at least three 5-x.  This violates that there are only 9 cards total.

Now suppose the minimum is 0.5.  There can't be more than one card with 0.5.  Now since there is only one choice for T=2, and one card is 0.5, there must be only one card with 1.5.  Then for T=3, there must be two cards with 2.5 since there is only one 1.5 card.  For T=4, there are two choices already with 1.5+2.5 (2x).  For an extra choice we need exactly one card with 3.5.  For T=5, there are already two choices 1.5+3.5 (1x), 2.5+2.5 (1x).  So there must be two 4.5 cards.  For T=6, there are 1.5+4.5 (2x), 2.5+3.5 (2x), so there must be one 5.5 card.  For T=7, there are 1.5+5.5 (1x), 2.5+4.5 (4x), so there must be one 6.5 card.

Apply the same logic will verify the rest of the values for T.

So to recap, the values of the cards are:

0.5 (1x)
1.5 (1x)
2.5 (2x)
3.5 (1x)
4.5 (2x)
5.5 (1x)
6.5 (1x)

Edited on August 9, 2004, 11:05 pm
 Posted by Bon on 2004-08-09 19:34:22

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