From Julien Beasley comes a coincidence of dice:
Two people are sitting at a table together, each with their own bag of six “DnD dice”: a d4, a d6, a d8, a d10, a d12, and a d20. Here, “dX” refers to a die with X faces, numbered from 1 to X, each with an equally likely probability of being rolled.
Both people randomly pick one die from their respective bags and then roll them at the same time. For example, suppose the two dice selected are a d4 and a d12. The players roll them, and let’s further suppose that both rolls come up as 3. What luck!
What’s the probability of something like this happening? That is, what is the probability that both players roll the same number, whether or not they happened to pick the same kind of die?
There are six types of numbers, which correspond to appearing on 1-6 of the dice. Their probabilities are summarized in the table below.
| Number | Calculation | Probability |
|---|---|---|
| $1-4$ | $\dfrac{1}{6} \cdot \left(\dfrac{1}{4} + \dfrac{1}{6} + \dfrac{1}{8} + \dfrac{1}{10} + \dfrac{1}{12} + \dfrac{1}{20} \right)$ | $\dfrac{93}{720}$ |
| $5, 6$ | $\dfrac{1}{6} \cdot \left(\dfrac{1}{6} + \dfrac{1}{8} + \dfrac{1}{10} + \dfrac{1}{12} + \dfrac{1}{20} \right)$ | $\dfrac{63}{720}$ |
| $7, 8$ | $\dfrac{1}{6} \cdot \left(\dfrac{1}{8} + \dfrac{1}{10} + \dfrac{1}{12} + \dfrac{1}{20} \right)$ | $\dfrac{43}{720}$ |
| $9, 10$ | $\dfrac{1}{6} \cdot \left(\dfrac{1}{10} + \dfrac{1}{12} + \dfrac{1}{20} \right)$ | $\dfrac{28}{720}$ |
| $11, 12$ | $\dfrac{1}{6} \cdot \left(\dfrac{1}{12} + \dfrac{1}{20} \right)$ | $\dfrac{16}{720}$ |
| $13-20$ | $\dfrac{1}{6} \cdot \left(\dfrac{1}{20} \right)$ | $\dfrac{6}{720}$ |
For the Fiddler and Extra Credit, I thought of each player having a single weighted die with the above corresponding probabilities, as the final answer for either part does not have any condition about which die is rolled for each player.
The answer can be calculated by summing the squares of the probabilities.
$$4\cdot\left(\dfrac{93}{720}\right)^2 + 2\cdot\left(\dfrac{63}{720}\right)^2 + 2\cdot\left(\dfrac{43}{720}\right)^2 + 2\cdot\left(\dfrac{28}{720}\right)^2 + 2\cdot\left(\dfrac{16}{720}\right)^2 + 8\cdot\left(\dfrac{6}{720}\right)^2 = \dfrac{3}{32}$$Julien’s puzzle idea didn’t stop there. Instead of two people sitting at the table, now suppose there are three.
Again, all three randomly pick one die from their respective bags and roll them at the same time. For example, suppose the three dice selected are a d4, a d20, and a d12. The players roll them, and let’s further suppose that the d4 comes out as 4, the d20 comes out as 13, and the d12 comes out as 4. In this case, there are two distinct numbers (4 and 13) among the three rolls.
On average, how many distinct numbers would you expect to see among the three rolls?
Using the weighted die from above, I wrote code:
The expected number of distinct numbers you would expect to see is $\dfrac{8489153}{3110400} \approx 2.7292$