For a game I'm trying to determine the frequency that a certain # will show up at a given # of dice being rolled. I know... that question seems odd. Let me try to explain it with real numbers.
So, for 1 die, the frequency for each number will be identical. 1-6 will show up equal number of times.
Now for 2 dice, things get different. I imagine 5,6,7 are going to be the most frequently rolled, while numbers at both ends of the spectrum will show up less or not at all (in the case of 1). I'd like to know how to calculate this list and show them in the proper order, from most frequent to less frequent.
Any thoughts?
@duffymo - It would be really nice though to have some sort of an algorithm to come up with it. It seems that the above way is going to require a lot of hand picking and placing of numbers. If my die count is dynamic up to say 10, doing that by hand will be ineffecient and troublesome I think. :)
As you can see, 7 is the most common roll with two six-sided dice. There are six times more likely chances to roll a 7 than a 2 or a 12, which is a huge difference. There are twice as likely chances to roll a 7 than a 4 or a 10. However, it's only 1.2 times more likely chances to roll a 7 than a 6 or an 8.
When two dice are rolled, there are now 36 different and unique ways the dice can come up. This figure is arrived at by multiplying the number of ways the first die can come up (six) by the number of ways the second die can come up (six). 6 x 6 = 36.
Probability = Number of desired outcomes ÷ Number of possible outcomes = 3 ÷ 36 = 0.0833. The percentage comes out to be 8.33 per cent. Also, 7 is the most likely result for two dice.
There are 6*6 = 36 combinations for two dice.
2 = 1+1 can only appear once, so its frequency is 1/36. 3 = 1+2 or 2+1, so its frequency is 2/36 = 1/18. 4 = 1+3, 2+2, or 3+1, so its frequency is 3/36 = 1/12.
You can do the rest out to twelve.
Any backgammon player knows these well.
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