Throwing Dice N Times

Question

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.

Answer 1

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...

Answer 2

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 $(1-\frac 1{36})^N < 0.5$

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