How many dice must be thrown so that there is a better than even chance of obtaining a six?

Explanation

We know that,

Probability of getting six `= frac{1}{6}`

Probability of not getting six in `n` drawn `= frac{5}{6}`



By condition

`1- (frac{5}{6})^n ge frac{1}{2}`

`Rightarrow 1- frac{1}{2} ge (frac{5}{6})^n`

`Rightarrow frac{1}{2} ge (frac{5}{6})^n`

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