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You Throw Two Dice. One of the possible many outcomes that may occur is that you get a six on each die (this outcome is called a double six). How many times must you throw the two dice in order for the probability of getting a double six (on one of your throws) to be at least .50?

I completed this problem by setting up the equation .5 = (1/6)(1/6)N N= 18 times. Is this correct? Can someone please provide me with the proper equation. Thanks so much.

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    Hint: If you throw $n$ times then what is the probability that **no** double six was thrown? – 2017-02-21
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    Further hint: the [binomial distribution](https://en.wikipedia.org/wiki/Binomial_distribution) should be useful here. $Pr(X=k)=\binom{n}{k}p^k(1-p)^{n-k}$ – 2017-02-21
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    So is it (35/36)^N = .5? Do I need to solve using logs? – 2017-02-21
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    Yes (or just substitute some values for positive integer $n$), and actually you are not solving an equation but an inequality: $1-(\frac{35}{36})^n\geq0.5$ for positive integer $n$. – 2017-02-21
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    So is the answer 24.6 times or 25 times? – 2017-02-21
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    24.6 times is not possible. 25 is correct. – 2017-02-21
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    I solved for (35/36)^N = .5 to get 25 times. If you solve for 1-(35/36)^N = .5 don't you 70 times? – 2017-02-21

2 Answers 2

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The probability of getting a double 6 on any throw is $\frac {1}{36}$

Your probability of getting not getting any 1 double 6s on N throws is $(1-\frac 1{36})^N$

Solve for N such that $(1-\frac 1{36})^N < 0.5$

$N \log (\frac {35}{36}) < \log {\frac 12}\\ N > \frac {\log 2}{\log 36-\log 35}$

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Ask yourself the following:

$p(double \space six)=$

That means $p(no \space double \space six)=$

Chaining this: $p(no \space double \space six \space in \space n \space tries)=$

Again negating: $p(at \space least \space one \space double \space six \space in \space n \space tries)=$

Now just solve for what $n$ the last probability is $>0.5$

I can't quite tell what you mean by your equation...

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    The return key just doesn't do anything if pressed only once. His attempt was to set $0.5=\frac{1}{36}N$ and thinks this implies $N=18$ – 2017-02-21