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The problem is as stated: "How many times do we need to roll four fair dice to get a better than even chance of at least one pair of sixes?"

Using this formula for getting a specific value in exactly $m$ out of $n$ rolls:

$$\binom{n}{m}\cdot\left(\frac16\right)^{m}\cdot\left(\frac56\right)^{n-m}$$

I calculated the probability of favorable outcomes - 1 pair of 6 or 2 pair of 6 as being 11.57% and 0.077%, respectively. Together, that gives us a 11.647% chance. Now, following this logic, how do I get the number of throws of 4 dice that would give us more than a 50% chance of a favorable outcome? I can only think of doing: $$x*11.647>50$$ $$x>4.2$$ So then $x$ would be equal to 5. I don't think this is the right answer tho I have no proof. Can anyone help me with this regard?

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    $\begingroup$ Find the probability of not getting a pair in one throw of the four dice. Raising that to the power of $n$ is the probability of not getting a pair in $n$ throws of the four dice, and you want the smallest $n$ at which that is less than or equal to $\frac12$. Logarithms may help $\endgroup$
    – Henry
    Commented Jun 25, 2021 at 15:19
  • $\begingroup$ Much appreciated! $\endgroup$ Commented Jun 25, 2021 at 16:46

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If $X$ is the number of sixes in a throw of four dice,

$P(X = 0) = \left(\dfrac{5}{6}\right)^4$

$ \displaystyle P(X = 1) = {4 \choose 1} \cdot \frac{1}{6} \cdot \left(\dfrac{5}{6}\right)^3$

$P(X \leq 1) = P(X=0) + P(X=1) = \dfrac{125}{144}$

You want to find number of throws $n$ such that,

$ \displaystyle [P(X \leq 1)]^n \leq \dfrac{1}{2}$

i.e, $\displaystyle \ n \ln \left(\dfrac{125}{144}\right) \leq \ln (0.5)$

Can you take it from here?

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  • $\begingroup$ I think I get it, using your logic I get that $n=5$ . But, isn't the sum of those two probabilities is $1125/1296$ or am I miscalculating? I have one question about this though - so we are going the opposite route, using unfavorable outcomes. But since the formulation of the problem states "at least one pair of sixes" then aren't the unfavorable outcomes $X=0, X=1, X=3$ ? Or maybe in the case of getting 3 sixes, this is a favorable outcome since we have one pair plus another 6? Making "at least one pair of sixes" the equivalent of "getting at least 2 sixes"? $\endgroup$ Commented Jun 25, 2021 at 16:37
  • $\begingroup$ @TamaraTodorovic Sorry I had a calculation mistake. Yes $n = 5$ is correct and it is $1125/1296 = 125/144$. $\endgroup$
    – Math Lover
    Commented Jun 25, 2021 at 16:44
  • $\begingroup$ Alright, thank you very much for the help! $\endgroup$ Commented Jun 25, 2021 at 16:45
  • $\begingroup$ @TamaraTodorovic If $X = 3$, we have $3$ sixes and I consider it that there is at least one pair plus one more six. $\endgroup$
    – Math Lover
    Commented Jun 25, 2021 at 16:46
  • $\begingroup$ Got it, "at least one pair of sixes" is the same as "getting at least 2 sixes", thanks for the clarification. $\endgroup$ Commented Jun 25, 2021 at 16:49

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