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I know how to calculate the probability of throwing at least one die of a given face with a set of dice, but can someone tell me how to calculate more than one (e.g., at least two)?

For example, I know that the probability of throwing at least one 4 with two 6-sided dice is 27/216, or 1 - (3/6 x 3/6 x 3/6). How do I calculate throwing at least two 4s with four 6-sided dice?

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The probability of at least one four with two dice is $1-(\frac 56)^2=\frac {11}{36}$ –  Ross Millikan Sep 9 '12 at 17:19
    
@RossMillikan Why is that? Could you please explain. –  ihatetoregister Apr 3 '13 at 11:14
    
@ihatetoregister: The chance you don't get a $4$ with the first die is $\frac 56$. Of that, the chance that you don't get a $4$ the second time is again $\frac 56$. So the chance you get no $4$ is $(\frac 56)^2$. The chance you get at least $1 \ \ 4$ is the rest of the time. Many of these probability questions are easier to answer if you count the other probability and subtract from $1$. –  Ross Millikan Apr 3 '13 at 13:29
    
@RossMillikan Thank you, now I understand. But I still find it weird that the "same" logic does not get me the same answer if you count the chances to get four. Eg $ \frac 1 6 \frac 1 6 \ne (11/36) $. I'm obviously missing something but I'm not sure of what. –  ihatetoregister Apr 3 '13 at 14:55
    
Think I got it: The change to get four on first throw + not getting four on first throw but on the second, right? $ \frac 16 + \frac 56 \frac 16 = \frac{11}{36} $ –  ihatetoregister Apr 3 '13 at 15:15

2 Answers 2

You are asking for the distribution of the number $X_n$ of successes in $n$ independent trials, where each trial is a success with probability $p$. Almost by definition, this distribution is binomial with parameters $(n,p)$, that is, for every $0\leqslant k\leqslant n$, $$ \mathrm P(X_n=k)={n\choose k}\cdot p^k\cdot(1-p)^{n-k}. $$ The probability of throwing at least two 4s with four 6-sided dice is $\mathrm P(X_4\geqslant2)$ with $p=\frac16$. Using the identity $\mathrm P(X_4\geqslant2)=1-\mathrm P(X_4=0)-\mathrm P(X_4=1)$, one gets $$ \mathrm P(X_4\geqslant2)=1-1\cdot\left(\frac16\right)^0\cdot\left(\frac56\right)^4-4\cdot\left(\frac16\right)^1\cdot\left(\frac56\right)^3=\frac{19}{144}. $$

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The probability of no 4 with four 6-sided dice$(p_1)=(\frac{5}{6})^4$

The probability of exactly one 4 with four 6-sided dice$(p_2)$ $=4\frac{1}{6}(\frac{5}{6})^3$ as here the combinations are $4XXX$ or $X4XX$ or $XX4X$ or $XXX4$ where $X$ is some other face$≠4$

So, the probability of at least two 4s with four 6-sided dice$=1-p_1-p_2$ $=1-((\frac{5}{6})^4+4\frac{1}{6}(\frac{5}{6})^3)$ $=1-(\frac{5}{6})^3(\frac{5}{6}+\frac{4}{6})=1-\frac{125}{144}=\frac{19}{144}$

The probability of throwing at least 4 by one 6-sided dice $=\frac{3}{6}=\frac{1}{2}$

The possible combinations are $XXYY$, $XYXY$, $XYYX$, $YXXY$, $YXYX$, $YYXX$ where $1≤Y≤3,4≤X≤6$

So, the required probability of throwing exactly two occurrences of at least 4 is $^4C_2\frac{1}{2}\frac{1}{2}(1-\frac{1}{2})(1-\frac{1}{2})=\frac{3}{8}$ using Binomial Distribution.

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I'm not looking for the probability of exactly 4, I'm looking for the probability of at least 4, twice in four throws. I'm sorry that wasn't clear. –  Drew Miller Sep 9 '12 at 6:32
    
So, "throwing at least two 4s with four 6-sided dice" implies two at least 4, right? –  lab bhattacharjee Sep 9 '12 at 6:40
    
Correct. I think it's ((3/6)^2)*((6/6)^2), but I have no confidence or really understan why. –  Drew Miller Sep 9 '12 at 6:45
    
Could you please explain your approach? By the way, I've modified the answer. –  lab bhattacharjee Sep 9 '12 at 6:48

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