Games, all sorts

The following table gives the probability for getting at least one success given a number of dice:

The formula for determining odds of success is:

def chance_success(dice):
  return 100 * (1-0.6**dice)

A single die will turn up a value between 1 and 10 inclusive. 60% (1-6) of the time, a single die roll will result in no successes. That means 40% (7, 8, 9, 10) of the time, you'll have at least one success.

When you have a success, 3 of the 4 possibilities give a single success (7, 8, 9), while a roll of 10 results in 2 successes. This means that 75% of the time, you'll get 1 success, and 25% of the time, you'll get two. This averages to be 0.75*1 + 0.25*2, or an average of 1.25 successes (whenever a die shows a success).

Multiplying the average success by the probability of giving a success gives 0.40*1.25, or 0.50. This means that we'll get an average of half a success per die rolled.

Perhaps counter-intuitively, as the odds of success go up, so does the probability you'll botch when you fail. Consider the case where you have 10 dice. The odds of getting at least one success, per the table above, is 99.4%. This gives a 0.6% chance of failure.

A botch occurs when you have a failure and you have at least one one. Once we're in the failure state (0.6% of the time in the above example), we have 2 sub-states. The first is a simple failure. This requires that no ones appeared. The chance of this happening on a single die is 83% (2,3,4,5,6 out of 6 possible failure values). More generically, the probability that all dice resulted in no ones is:

(5/6)**N

As N increases, the probability that no ones are seen begins to drop. The following table demonstrates: