Which of the following three outcomes has the greatest chance of occuring? (Assuming normal six-sided fair dice) - throwing a die 6 times and getting at least one 6? (i.e. one or more sixes) - throwing a die 12 times and getting at least two sixes ? - throwing a die 18 times and getting at least three sixes ?

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Good question.

The probability of NOT getting what you want in 6 throws is

(5/6)^6 = 0.3348979767

The probability of NOT getting what you want in 12 throws is

(5/6)^12 + 12[(5/6)^11]/6

The first term in the above expression is the probability of getting zero 6's in 12 throws. The second term is the probability of getting exactly one 6. The multiplication by 12 in the second term is because there are 12 equally likely ways that exactly one 6 in 12 throws can come up.

= 0.11215665477 + 12*0.13458798573/6

= 0.11215665477 + 0.26917597142

= 0.3813326262

This is the probability of getting less than two 6's in 12 throws.

In general, as the number of trials increases, the probability of getting LESS than the expected proportion of 6's (1/6) also tends to increase. In this particular problem, your likelihood of success is better with 6 throws than with 12 or 18 throws.

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