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Jacob from Fern Avenue Public School
thought carefully about this problem:
To figure out the counter that wins most often, we first need to
figure out the most common dice rolls.
There are $36$ different possible combinations. Here they are, in
the format "First Die Roll - Second Die Roll - Total":
$1 - 1 - 2$
$1 - 2 - 3$
$1 - 3 - 4$
$1 - 4 - 5$
$1 - 5 - 6$
$1 - 6 - 7$
$2 - 1 - 3$
$2 - 2 - 4$
$2 - 3 - 5$
$2 - 4 - 6$
$2 - 5 - 7$
$2 - 6 - 8$
$3 - 1 - 4$
$3 - 2 - 5$
$3 - 3 - 6$
$3 - 4 - 7$
$3 - 5 - 8$
$3 - 6 - 9$
$4 - 1 - 5$
$4 - 2 - 6$
$4 - 3 - 7$
$4 - 4 - 8$
$4 - 5 - 9$
$4 - 6 - 10$
$5 - 1 - 6$
$5 - 2 - 7$
$5 - 3 - 8$
$5 - 4 - 9$
$5 - 5 - 10$
$5 - 6 - 11$
$6 - 1 - 7$
$6 - 2 - 8$
$6 - 3 - 9$
$6 - 4 - 10$
$6 - 5 - 11$
$6 - 6 - 12$
Number of appearances of $2$: $1$
Number of appearances of $3$: $2$
Number of appearances of $4$: $3$
Number of appearances of $5$: $4$
Number of appearances of $6$: $5$
Number of appearances of $7$: $6$
Number of appearances of $8$: $5$
Number of appearances of $9$: $4$
Number of appearances of $10$: $3$
Number of appearances of $11$: $2$
Number of appearances of $12$: $1$
Looking at all of the totals, we can see that $2$ appears once, $3$
twice, $4$ thrice, and so on until we get to $7$. At this point,
the number of appearances starts to decrease. $8$ appears five
times, $9$ four, and $12$ only once. We can see that $7$ appears
most often, and is therefore most likely to win the race.
But why does $7$ appear most often? It is because there is no
result for the first die rolled that excludes it as a result. If
you look back at the table, you can see that no matter what is
rolled on the first die, seven still has a $1$ in $6$ chance of
being the result. This is not true for any other number. And this
is why $7$ is the most common result.
What if the dice had seven faces? $7$ would no longer be the most
common number, as a roll of $7$ on the first die would eliminate it
as a result. Instead, $8$ is the most common, for the reasons
mentioned above. If the dice had eight faces, then $9$ would be the
most common. There is a pattern here: if two dice are being rolled,
the most common result can be determined by adding $1$ to the
number of faces each die has.
Very well done, Jacob. Here is another way
to display the possible totals when rolling two $1-6$
dice:
|
1 |
2 |
3 |
4 |
5 |
6 |
1 |
2 |
3 |
4 |
5 |
6 |
7 |
2 |
3 |
4 |
5 |
6 |
7 |
8 |
3 |
4 |
5 |
6 |
7 |
8 |
9 |
4 |
5 |
6 |
7 |
8 |
9 |
10 |
5 |
6 |
7 |
8 |
9 |
10 |
11 |
6 |
7 |
8 |
9 |
10 |
11 |
12 |