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Let's say we're talking about rolling a six. Here's what I know...

For one roll ...

(probability of rolling a six) = 1/6

(probability of NOT rolling a six) = 5/6

(probability of rolling a six) + (probability of NOT rolling a six) = 1/6 + 5/6 = 1

For 'x' number of rolls ...

(probability of rolling a six 'x' times) = (1/6)^x

(probability of NOT rolling a six 'x' times) = (5/6)^x

(probability of rolling at least one six in 'x' rolls) = 1 − (probability of NOT rolling a six 'x' times) = 1 − (5/6)^x

My question:

When trying to calculate the probability of rolling at least one six in 'x' rolls (the last line above), is there an alternative to this 1 − p(NOT six) method? Isn't the probability of rolling at least one six in 'x' rolls somehow a combination of the probabilities of rolling one six, two sixes, three sixes, four sixes etc.?

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    $\begingroup$ Sure. You can just sum the probabilities of rolling exactly one, two, etc. It's less efficient, but perfectly correct. $\endgroup$
    – lulu
    Sep 2, 2021 at 14:29
  • $\begingroup$ Nice to know, but what would that look like written out? Say, for at least one six in 5 rolls. $\endgroup$
    – KQUB
    Sep 2, 2021 at 14:33
  • $\begingroup$ “Probability of NOT rolling a $6$ $x$ times” is ambiguous. It could mean “(not rolling a six) $x$ times” or “not (rolling a six $x$ times.)” $\endgroup$ Sep 2, 2021 at 14:34
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    $\begingroup$ Just sum the terms given by the binomial distribution. $\endgroup$
    – lulu
    Sep 2, 2021 at 14:38
  • $\begingroup$ One approach: At heart, the probability is $1-(1-1/6)^x.$ When you expand $(1-1/6)^x$ you get an alternating sum. That hints at an inclusion-exclusion argument. And, indeed, there is an inclusion-exclusion argument. Let $A_i$ be the cases where the $i$th roll is $6.$ Then you want $P(\lnot(A_1lor\cdots\lor A_x))$ and inclusion-exclusion gives the appropriate alternating sum. $\endgroup$ Sep 2, 2021 at 14:44

1 Answer 1

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Probablity to get "at least" on six in $n$ rolls is obviously

$$P(X=1)+P(X=2)+\dots P(X=n)=1-P(X=0)$$

to calculate this probability you can use the binomial distribution that is

$$P(X=k)=\binom{n}{k}p^k\cdot(1-p)^{n-k}$$


Let's have a simple example:

Let's roll 3 times a die (or one time 3 dice) and calcualate the probability to have "at least one six"

First approach (direct approach)

$$P(X\geq 1)=\binom{3}{1}\cdot\left(\frac{1}{6}\right)^1\cdot\left(\frac{5}{6}\right)^2+\binom{3}{2}\cdot\left(\frac{1}{6}\right)^2\cdot\left(\frac{5}{6}\right)^1+\binom{3}{3}\cdot\left(\frac{1}{6}\right)^3\cdot\left(\frac{5}{6}\right)^0=\frac{91}{216}$$

Second (indirect) approach

$$P(X\geq 1)=1-P(X=0)=1-\binom{3}{0}\cdot\left(\frac{1}{6}\right)^0\cdot\left(\frac{5}{6}\right)^3=\frac{91}{216}$$

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