Combinatorics, probability, dice

Question

A friend of mine asked: if I have two dice and I throw both of them, what is the most frequent sum (of the two dice' numbers)?

I wrote a small script:

from random import randrange
d = dict((i, 0) for i in range(2, 13))
for i in xrange(100000):
    d[randrange(1, 7) + randrange(1, 7)] += 1
print d

Which prints:

2:  2770,
3:  5547,
4:  8379,
5:  10972,
6:  13911,
7:  16610,
8:  14010,
9:  11138,
10: 8372,
11: 5545,
12: 2746

The question I have, why is 11 more frequent than 12? In both cases there is only one way (or two, if you count reverse too) how to get such sum (5 + 6, 6 + 6), so I expected the same probability..?

Answer 1

In both cases there is only one way (or two, if you count reverse too)

There are two ways. If the dice are named A and B:

12 = one way: A=6, B=6

11 = two ways: A=5, B=6 and A=6, B=5.

Answer 2

For 11 there is 5 + 6 and 6 + 5 for 12 there is only 6 + 6. You can observe the same thing with 2 and 3.

Answer 3

The question I have, why is 11 more frequent than 12?

First of all, this question assumes that your arbitrary try gives an authoritative result. It doesn’t; the result is pure random and only reliable up to a degree. But in this particular case, the numbers actually reflect the real proportions nicely.

That said, there are two ways to get 11: 5 (first die) + 6 (second die) and 6 (first die) + 5 (second die) but only one way to get 12: 6 (first die) + 6 (second die).

Answer 4

The most frequently met sum is 7, as suggested by your empirical test.

Now, to answer your questions specifically:

• 11 is more frequent than 12 because you get 12 by rolling 6,6, but you can get 11 by 5,6 or 6,5, which is double the probability.
• Based on classic probability theory, the probability of an event occurring is equal to (number-of-beneficial-simple-events-that-trigger-it)/(number-of-all-possible-events). So using this simple formula yields that in order to get a 7, you need to roll one of the following combinations: (1,6), (2,5), (3,4), (4,3), (5,2), (6,1), and you have 6x6=36 events all together. The chance of getting a 7 is P = 6/36 = 1/6, which is as high as it gets.

Check out Probability for more info.