Lottery Odds
Thread Starter
Joined: Jun 2001
Posts: 3,991
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From: UK
Lottery Odds
Anyone out there know how to calculate these?
On the Thunderball you pick 5 numbers between 1 and 39 and also 1 number (the thunderball) between 1 and 14.
The odds of winning the jackpot (£500,000) are 1 in 8,060,598.
If you do a syndicated entry which covers all the thunderballs (ie 14 entries covering each thunderball from 1 to 14) then what are the odds of winning the jackpot (assume choice of the 5 numbers is the same throughout).
Thanks for any help
On the Thunderball you pick 5 numbers between 1 and 39 and also 1 number (the thunderball) between 1 and 14.
The odds of winning the jackpot (£500,000) are 1 in 8,060,598.
If you do a syndicated entry which covers all the thunderballs (ie 14 entries covering each thunderball from 1 to 14) then what are the odds of winning the jackpot (assume choice of the 5 numbers is the same throughout).
Thanks for any help
Joined: Jul 2006
Posts: 130
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From: MAN
Try here
FFB....
Try here,
Excellent background to the whole gaming probability calculations
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Regards
DaveA
Try here,
Excellent background to the whole gaming probability calculations
Additionally, this one's fun!
Enjoy
Regards
DaveA
Joined: Apr 2009
Posts: 816
Likes: 0
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)
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)
