# Two Dice = Nine Cards

I have nine cards, each labeled with a real number (possibly negative or fractional) in a special way. When I randomly pick two of these cards without replacement and add their labels together, the probability of each sum is the same as when you roll two fair six-sided dice.

This means there is a $\frac16$ chance the two cards add to 7, a $\frac5{36}$ chance they add to 8, a $\frac19$ chance they add to 5, etc.

How have I labeled my cards?

The cards are numbered:

$0.5,1.5,2.5,2.5,3.5,4.5,4.5,5.5,6.5$.

Reasoning:

There are 36 possible ways to choose two cards out of nine, so the results must be $2,3,3,4,4,4,\ldots,11,11,12$. If we sum all of these, we find that the total is $252$, and that each card is represented eight times. This makes the sum of all nine cards $\frac{252}{8}=31.5$.

For any card with value $x$, the other eight cards sum to $31.5-x$. However, these cards can be divided into four pairs, each of which sums to an integer. Therefore $31.5-x$ is an integer, and $x$ is a half-integer.

Now, consider the eight pairs that contain the card $x$. These eight pairs include $x$ eight times, and each other card once, so they sum to $31.5+7x$. The smallest eight results $(2,3,3,4,4,4,5,5)$ sum to $30$, so $31.5+7x\ge30$. The smallest half-integer satisfying this inequality is $0.5$.

The smallest card must be at most $0.5$, because a sum of $2$ would not be possible if it was $1.5$ or greater. Therefore the smallest card is exactly $0.5$. The rest of the cards follow easily from this.