Percentage Probability of Products when Multiplying 2d10

Question

This is my first post to any Stack Exchange, so I only think I know what I'm doing. In short, my objective is to design a math game for my elementary students to practice multiplication facts (math facts 1-10) in a way that's fun and engaging. My idea is to have them roll two ten-sided dice (2d10) and multiply the result. If the product is over a certain threshold (36 for example), then they do 1 damage to a monster (it's basically a monster-fighting game). However, I don't want to set the thresholds too low or high.

Here's what I might know: the average of rolling 1d10 is 5.5, so I would expect the most common results to fall around 25 (5x5) to 36 (6x6). However, I'm not confident that this makes actually checks out from a probability standpoint. Also, I don't know what the probability is exactly of rolling, say, a 54.

So, my question is this: what would be the percent probability of rolling 2d10 and attaining products that pass certain thresholds such as 12, 20, 36, 42, 54, 63, 72, and 81. I know I can guess more accurately on some than others and probably get somewhere, but I don't want students to try and just get frustrated at first when I give them something that they can't beat or if I make it way too easy and therefore not sufficiently rigorous for students to learn from. I hope this poses to be an interesting thought experiment for those who are more naturally gifted in math. I just don't feel confident forging ahead without getting at least some feedback from people whose math skills are more proficient than mine. Let me know if you need more details and thank you!

Answer 1

There probably is not a much easier way to do this than producing a multiplication table and counting the values, or getting a computer to do it for you. You might get something like this:

  *   [,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8] [,9] [,10]
 [1,]    1    2    3    4    5    6    7    8    9    10
 [2,]    2    4    6    8   10   12   14   16   18    20
 [3,]    3    6    9   12   15   18   21   24   27    30
 [4,]    4    8   12   16   20   24   28   32   36    40
 [5,]    5   10   15   20   25   30   35   40   45    50
 [6,]    6   12   18   24   30   36   42   48   54    60
 [7,]    7   14   21   28   35   42   49   56   63    70
 [8,]    8   16   24   32   40   48   56   64   72    80
 [9,]    9   18   27   36   45   54   63   72   81    90
[10,]   10   20   30   40   50   60   70   80   90   100 

The average product is $5.5^2=30.25$, while the median is $24$. The most common values are $6, 8, 10, 12, 18, 20, 24, 30, 40$ as they each appear four times in the table.

If you want the probaility of hitting, or of hitting and exceeding, particular values, then you might use this to just where to put your threshold.

  prob_hit prob_hit_or_exceed
1     0.01 1.00
2     0.02 0.99
3     0.02 0.97
4     0.03 0.95
5     0.02 0.92
6     0.04 0.90
7     0.02 0.86
8     0.04 0.84
9     0.03 0.80
10    0.04 0.77
12    0.04 0.73
14    0.02 0.69
15    0.02 0.67
16    0.03 0.65
18    0.04 0.62
20    0.04 0.58
21    0.02 0.54
24    0.04 0.52
25    0.01 0.48
27    0.02 0.47
28    0.02 0.45
30    0.04 0.43
32    0.02 0.39
35    0.02 0.37
36    0.03 0.35
40    0.04 0.32
42    0.02 0.28
45    0.02 0.26
48    0.02 0.24
49    0.01 0.22
50    0.02 0.21
54    0.02 0.19
56    0.02 0.17
60    0.02 0.15
63    0.02 0.13
64    0.01 0.11
70    0.02 0.10
72    0.02 0.08
80    0.02 0.06
81    0.01 0.04
90    0.02 0.03
100   0.01 0.01

Answer 2

There isn't a nice general formula, but you can simply make a table of all possibilities by sorting the products obtained for the 100 possible rolls of 2d10. This yields the following list (organized as a 10-by-10 grid for convenience): \begin{array}{rrrrrrrrrr} 1 & 2 & 2 & 3 & 3 & 4 & 4 & 4 & 5 & 5\\ 6 & 6 & 6 & 6 & 7 & 7 & 8 & 8 & 8 & 8\\ 9 & 9 & 9 & 10 & 10 & 10 & 10 & 12 & 12 & 12\\ 12 & 14 & 14 & 15 & 15 & 16 & 16 & 16 & 18 & 18\\ 18 & 18 & 20 & 20 & 20 & 20 & 21 & 21 & 24 & 24\\ 24 & 24 & 25 & 27 & 27 & 28 & 28 & 30 & 30 & 30\\ 30 & 32 & 32 & 35 & 35 & 36 & 36 & 36 & 40 & 40\\ 40 & 40 & 42 & 42 & 45 & 45 & 48 & 48 & 49 & 50\\ 50 & 54 & 54 & 56 & 56 & 60 & 60 & 63 & 63 & 64\\ 70 & 70 & 72 & 72 & 80 & 80 & 81 & 90 & 90 & 100 \end{array} Note, for example, that the last $14$ occurs in the $43^\text{rd}$ position (fourth row, third column) so the probability of rolling $14$ or less is $43\%$.

From above list we observe that $24$ is the median roll. The mean roll is $5.5^2 = 30.25$, and there is a 9-way tie for the mode (i.e., 9 distinct numbers that occur 4 times each). What are those 9 ways? Hey, that's a good problem for your students!

An interesting observation is that as the number of sides $n$ of the die grows, the median product grows as $k n^2$, where $k$ is the unique solution in (0,1) of $k \log k = k - 1/2$. This value is $k \approx 0.18668$. Showing this would be a suitable problem only for your more advanced students.