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Old 18th March 2011 | 12:07
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Torque Tonight
 
Joined: Apr 2009
Posts: 816
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From: UK
1 in 575,575. 14 times greater than the original odds because all 14 possibilities for the last number have been covered. One entry will definitely have the correct thunderball.

Original probability is:

(5/39)*(4/38)*(3/37)*(2/36)*(1/35)*(1/14)=(1/8060598)

where the first 5 brackets refer to picking 5 numbers out of a set of 39 where the order in which they are picked does not matter. The sixth term is picking one number out of an independant set of 14 (the thunderball). This is the term that drops out if you cover all 14 thunderball possibilities:

(5/39)*(4/38)*(3/37)*(2/36)*(1/35)=(1/575757)
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