# Calculate the probability of a "good" dice roll

As part of a bigger game-theory problem, I've been trying to solve a rather simple probability question, and I seem to be getting the wrong answers. Here's the problem:

Dice are rolled to determine a pass or fail, and the probability of either is equal. If the roll is a pass, the entire condition passes, and if it is a failure it is re-rolled until either it passes, or has failed 3 times. What is the probability of the entire condition passing?

The way I solved this is by listing all possible states, of which there are four:

fail fail fail = failure
fail fail pass = success
fail pass      = success
pass           = success


based on this, the probability of overall success is $$\frac{3}{4}$$, or 0.75.

To test it, I wrote a simple simulation in python as follows:

def test():
fails = 0
while fails < 3:
if random.choice([True, False]):
return True
else:
fails += 1
return False

passes = 0
for i in range(1000000):
if test():
passes += 1

print(passes / 1000000)


It shows that the probability is 0.875. I'm sure that the simulation is corrent - I rewrote it in several different ways and came up with the same answer each time. So where am I going wrong with the mathematics?

The followup is to generalise this slightly where the probability of a pass or fail roll is not equal (e.g. only a 6 of a 6-sided die means pass). I have no idea how to solve that though.

• You've made the classic probability mistake of using a sample space where not all of the elements are equally likely: (fail, pass) is actually twice as likely as (fail, fail, pass) or (fail, fail, fail). May 5, 2021 at 6:45
• "based on this, the probability of overall success is $\frac{3}{4}$, or 0.75." only if each of the four outcomes is equally likely. May 5, 2021 at 6:45

To see why the probability is $$\frac78,$$ let's slightly rephrase the question: let's say instead of simulating the flips until we get a success (let's say heads) let's just flip the coin three times straight away and then look at the results. A set of three flips is successful if and only if there is at least one heads in the three flips, so the probability of a success must be equal to the probability of getting at least one heads.
So, letting $$X \sim B(3, 0.5)$$ be the number of heads, we want $$\text{P}(X \geq 1).$$ By the law of complements, this is equal to $$1 - \text{P}(X = 0) = 1 - \left(\frac12\right)^3 = \frac78.$$
Alternatively, using a sample space, there are $$2^3 = 8$$ possible sequences of three flips, and only $$1$$ has all three tails, so the other $$7$$ have at least one heads for a probability of $$\frac78.$$
Say $$1, 2, 3$$ are considered success and $$4, 5, 6$$ are considered failure. So we have equal probability of success or failure in each roll of the dice. As we have up to $$3$$ rolls, the probability of success is,
$$P(S) = 0.5 + 0.5 \times 0.5 + 0.5^2 \times 0.5 = 0.875$$