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If I throw one die three times then each time the probability of getting a $6$ is $p={1 \over 6}$.

EDIT

If I want to know the probability of getting at least one $6$, then my intuition (which is clearly wrong) is that I observe three experiments and each time I have a chance of ${1 \over 6}$. So I sum up all ${1 \over 6}$ which yields $3 \cdot {1 \over 6}$.

However, it should be $p= 1-\left({5\over 6}\right)^3$.

Why is $3 \cdot {1 \over 6}$ wrong?

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    $\begingroup$ Please explain why you would make that guess. Offhand, I don't see it at all. I could see it as the guess for at least one $6$ ("I get $6$ on the first roll or the second or the third") but not for at least two. $\endgroup$
    – saulspatz
    Commented Oct 6, 2019 at 17:24
  • $\begingroup$ you seem to add the probability for every throw? but in the first throw you have 1/6, the second 1/6 so 1/36 to have 2 consecutive 6. $\endgroup$
    – trula
    Commented Oct 6, 2019 at 17:38
  • $\begingroup$ As a rule of thumb, “and” translates to multiplying probabilities together, “or” leads to adding them (with correction for non-independence). $\endgroup$
    – amd
    Commented Oct 6, 2019 at 18:15
  • $\begingroup$ @saulspatz, after reconsidering the problem I have in my mind, I edited/changed my question a bit. I hope it doesn't cause too much confusion. Any comments are welcome :) $\endgroup$
    – Philipp
    Commented Oct 6, 2019 at 23:18
  • $\begingroup$ This question is clearer than your original question. However, changing the question after it has been answered invalidates those answers. $\endgroup$
    – N. F. Taussig
    Commented Oct 6, 2019 at 23:42

2 Answers 2

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Since you only ask why your intuition is wrong, that is the question I will answer.

When you add When you add $\frac{1}{6} + \frac{1}{6} + \frac{1}{6} $ you are double counting some of the possibilities. For example, the result $(6,6,6)$ was counted three times. And any outcome that has more than one $6$ was counted more than one time.

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  • $\begingroup$ As we are currently discussing in lecture the rule of adding probabilities I wonder if I can alternatively describe my mistake as follows: Let be $A:=\{6, x, y\}$, $B:=\{x, 6, y\}$ and $C:=\{y, x, 6\}$ sets of the sample space $\Omega:= \{(x, y, z)\vert x, y, z \in \{1, 2, 3, 4, 5, 6\}\}$ and $P(A)$, $P(B)$ and $P(C)$ be the respective probabilities that I get a $6$ on the first, second or third throw. I am only allowed to add them if the sets $A, B$ and $C$ are disjoint. As the three sets are clearly not disjoint it contradicts the rule of adding probabilities. $\endgroup$
    – Philipp
    Commented Oct 7, 2019 at 17:38
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The probability of obtaining a 6 is the same for each throw. Therefore, this is a binomial distribution problem. The probability of exactly $k$ successes in $n$ trials, each of which has probability $p$ of success is $$\Pr(X = k) = \binom{n}{k}p^k(1 - p)^{n - k}$$ where $p^k$ is the probability of $k$ successes, $(1 - p)^{n - k}$ is the probability of $n - k$ failures, and $\binom{n}{k}$ is the number of ways exactly $k$ successes can occur in $n$ trials.

When tossing a fair die, the probability of obtaining a six is $1/6$ for each trial. Hence, the probability of at least one success is \begin{align*} \Pr(X \geq 1) & = \sum_{k = 1}^{3} \binom{3}{k}\left(\frac{1}{6}\right)^k\binom{5}{6}^{3 - k}\\ & = \binom{3}{1}\left(\frac{1}{6}\right)^1\left(\frac{5}{6}\right)^2 + \binom{3}{2}\left(\frac{1}{6}\right)^2\binom{5}{6} + \binom{3}{3}\left(\frac{1}{6}\right)^3\left(\frac{5}{6}\right)^0\\ & = \frac{1 \cdot 3 \cdot 1 \cdot 25}{6^3} + \frac{3 \cdot 1 \cdot 5}{6^3} + \frac{1 \cdot 1 \cdot 1}{6^3}\\ & = \frac{75 + 15 + 1}{6^3}\\ & = \frac{91}{216} \end{align*}

M D's answer explains why your intuition was incorrect. What follows supplements M D's answer.

Your intuition led you to count each outcome as many times as a six appears, which is why you obtained \begin{align*} \Pr(X \geq 1) & = \sum_{k = 1}^{3} \color{red}{k}\binom{3}{k}\left(\frac{1}{6}\right)^k\binom{5}{6}^{3 - k}\\ & = 1\binom{3}{1}\left(\frac{1}{6}\right)^1\left(\frac{5}{6}\right)^2 + \color{red}{2}\binom{3}{2}\left(\frac{1}{6}\right)^2\binom{5}{6} + \color{red}{3}\binom{3}{3}\left(\frac{1}{6}\right)^3\left(\frac{5}{6}\right)^0\\ & = \frac{3 \cdot 1 \cdot 25}{6^3} + \frac{\color{red}{2} \cdot 3 \cdot 1 \cdot 5}{6^3} + \frac{\color{red}{3} \cdot 1 \cdot 1 \cdot 1}{6^3}\\ & = \frac{75 + \color{red}{2} \cdot 15 + \color{red}{3} \cdot 1}{6^3}\\ & = \frac{75 +\color{red}{30} + \color{red}{3}}{216}\\ & = \frac{\color{red}{108}}{216}\\ & = \color{red}{\frac{1}{2}} \end{align*}

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  • $\begingroup$ I calculated the probability for at least two times the same number instead of at least two time a $6$. I edited my question. $\endgroup$
    – Philipp
    Commented Oct 6, 2019 at 22:28

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