betting 6 numbers out of 100 on 16 different occassions to get one match
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2 Answers

Bet 6 numbers out of 100 numbers on a single pick would be 6/100 = 0.06.

Odds of losing on that draw would be 94/100 = 0.94.

 

If the question is asking for the odds of winning at least once then:

P(win at least once) = 1 - P(losing 6 times in a row)

P(win at least once) = 1- (0.94^6)

P(win at least once) = 1 - 0.68986978105

P(win at least once) = 0.31013021894

Odds of winning at least once are about 31 out of 100.

(very confident of correct)

 

 

If the question is asking for the odds of winning exactly once (not 2, 3, 4, 5, or 6 times), then:

P(winning exactly once) = P(winning on one draw) * (P(losing on one draw) )^5 * (# ways to arrange 1 win and 5 losses)

P(winning exactly once) = 0.06 * 0.94^5 * 6

P(winning exactly once) = 0.26420544806

Odds of winning exactly once are about 26 out of 100.

(slightly less confident of correct)

by Level 13 User (103k points)
This seems to be accurate with my personal experince over about 1 year.   Ilke the answer very much.

john
by

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