From Adam Milligan comes a puzzle about dice, dungeons and dragons (but mostly about dice):
I have three dice (d4, d6, d8) on my desk that I fiddle with while working, much to the chagrin of my co-workers. For the uninitiated, the d4 is a tetrahedron that is equally likely to land on any of its four faces (numbered 1 through 4), the d6 is a cube that is equally likely to land on any of its six faces (numbered 1 through 6), and the d8 is an octahedron that is equally likely to land on any of its eight faces (numbered 1 through 8).
I like to play a game in which I roll all three dice in “numerical” order: d4, then d6 and then d8. I win this game when the three rolls form a strictly increasing sequence (such as 2-4-7, but not 2-4-4). What is my probability of winning?
Extra credit: Instead of three dice, I now have six dice: d4, d6, d8, d10, d12 and d20. If I roll all six dice in “numerical” order, what is the probability I’ll get a strictly increasing sequence?
The number of increasing sequences can be expressed as:
$$\sum_{d_4=1}^4 \left(\sum_{d_6=d_4+1}^6 \left(\sum_{d_8=d_6+1}^8 1 \right) \right)$$
Similarily, the number of increasing sequences for extra credit can be expressed as:
$$\sum_{d_4=1}^4 \left(\sum_{d_6=d_4+1}^6 \left(\sum_{d_8=d_6+1}^8 \left(\sum_{d_{10}=d_8+1}^{10} \left(\sum_{d_{12}=d_{10}+1}^{12} \left(\sum_{d_{20}=d_{12}+1}^{20} 1 \right) \right) \right)\right) \right)$$
I used loops to add up the possiblities. The answer is
$$\frac{48}{4\cdot6\cdot8} = \frac{1}{4}$$I used loops to add up the possiblities. The answer is
$$\frac{5434}{4\cdot6\cdot8\cdot10\cdot12\cdot20} = \frac{5434}{460800} \approx 0.0118$$