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# odds of winning

Topic closed. 4 replies. Last post 7 years ago by savagegoose.

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New Member

United States
Member #94445
July 20, 2010
13 Posts
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 Posted: July 21, 2010, 10:29 am - IP Logged

As a new lottery player, I'd like to calculate the odds of winning.  Each NYLottery game has posted
odds, however, my calculations give different odds.  Perhaps I am not doing it right, or
they use a different method, but here are my results compared to theirs:

If we use #'s 0-9, and pick one number, then odds of winning are 1 in 10.  Thus, I just have to
buy 10 tickets, play all numbers, and will get a winner 100%.
If 3 numbers are picked, then it's 10x10x10, or 000-999, thus 1:1000 correct?

But, in a single Take5 game, numbers cannot repeat.  Key space is 39.
thus 39x38x37x36x35 = 69,090,840
To win top prize(all 5 #'s matched), their odds are 1:575,757.

How is that calculated?

Pick10 has better odds, eventhough the key space is 80.  Since 20 numbers are picked(1/4),
and you pick 10(1/8), top prize matches 10, thus you have 2x chances to win.  But, that's the same odds as matching 5 to 10 if key space was 40.

Winning Numbers Matched Per Game: 10
Prize: \$500,000
Chances of Winning
on one Game Panel
1:8,911,711

Isnt it 80^10?

In any case, how is that any better than playing slots in Vegas or A.C.?  1c, 5c, 25c, gives more games than \$1 on above.  I played 1c at a casino in Vegas, and won \$10; that's 1000%.

Chief Bottle Washer
New Jersey
United States
Member #1
May 31, 2000
24039 Posts
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 Posted: July 21, 2010, 1:56 pm - IP Logged

<Moved to Mathematics forum>

Please post in the appropriate forum ... thank you.

New Jersey
United States
Member #17843
June 28, 2005
71119 Posts
Offline
 Posted: July 21, 2010, 3:58 pm - IP Logged

As a new lottery player, I'd like to calculate the odds of winning.  Each NYLottery game has posted
odds, however, my calculations give different odds.  Perhaps I am not doing it right, or
they use a different method, but here are my results compared to theirs:

If we use #'s 0-9, and pick one number, then odds of winning are 1 in 10.  Thus, I just have to
buy 10 tickets, play all numbers, and will get a winner 100%.
If 3 numbers are picked, then it's 10x10x10, or 000-999, thus 1:1000 correct?

But, in a single Take5 game, numbers cannot repeat.  Key space is 39.
thus 39x38x37x36x35 = 69,090,840
To win top prize(all 5 #'s matched), their odds are 1:575,757.

How is that calculated?

Pick10 has better odds, eventhough the key space is 80.  Since 20 numbers are picked(1/4),
and you pick 10(1/8), top prize matches 10, thus you have 2x chances to win.  But, that's the same odds as matching 5 to 10 if key space was 40.

Winning Numbers Matched Per Game: 10
Prize: \$500,000
Chances of Winning
on one Game Panel
1:8,911,711

Isnt it 80^10?

In any case, how is that any better than playing slots in Vegas or A.C.?  1c, 5c, 25c, gives more games than \$1 on above.  I played 1c at a casino in Vegas, and won \$10; that's 1000%.

How To Calculate Odds

Lottery Post Forums Search is Your Friend:

http://www.lotterypost.com/search/forums?q='calculate+odds'&t=all

A mind once stretched by a new idea never returns to its original dimensions!

CA
United States
Member #2987
December 10, 2003
832 Posts
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 Posted: July 21, 2010, 6:25 pm - IP Logged

The reasoning behind this is, when one picks 5 numbers out of a pool, there are 120 different ways in which those 5 numbers can be picked, ranging from 1-2-3-4-5 to 5-4-3-2-1, hence the game odds are calculated as (39x38x37x36x35)/(1x2x3x4x5). That formula yields 575,757.

Pick 10 is far more complicated, since 10 different games are being played from one draw.

Blessed Saint Leibowitz, keep 'em dreamin' down there.....

Next week's convention for Psychics and Prognosticators has been cancelled due to unforeseen circumstances.

=^.^=

adelaide sa
Australia
Member #37136
April 11, 2006
3329 Posts
Offline
 Posted: July 22, 2010, 4:30 am - IP Logged

As a new lottery player, I'd like to calculate the odds of winning.  Each NYLottery game has posted
odds, however, my calculations give different odds.  Perhaps I am not doing it right, or
they use a different method, but here are my results compared to theirs:

If we use #'s 0-9, and pick one number, then odds of winning are 1 in 10.  Thus, I just have to
buy 10 tickets, play all numbers, and will get a winner 100%.
If 3 numbers are picked, then it's 10x10x10, or 000-999, thus 1:1000 correct?

But, in a single Take5 game, numbers cannot repeat.  Key space is 39.
thus 39x38x37x36x35 = 69,090,840
To win top prize(all 5 #'s matched), their odds are 1:575,757.

How is that calculated?

Pick10 has better odds, eventhough the key space is 80.  Since 20 numbers are picked(1/4),
and you pick 10(1/8), top prize matches 10, thus you have 2x chances to win.  But, that's the same odds as matching 5 to 10 if key space was 40.

Winning Numbers Matched Per Game: 10
Prize: \$500,000
Chances of Winning
on one Game Panel
1:8,911,711

Isnt it 80^10?

In any case, how is that any better than playing slots in Vegas or A.C.?  1c, 5c, 25c, gives more games than \$1 on above.  I played 1c at a casino in Vegas, and won \$10; that's 1000%.

where you quote 39x38x37

that is correct if you want to get the exact order of drawn numbers, ie 1st ball drawn is a 5, it is indeed 1 in 39, but if the order isnt important then there are 5/39 odds for 1st ball. and then 4/38 for second ball. 3 of 37 balls. etc.

math is something like, 39x38x37x36x35  / 5x4x3x2x1

" Still swinging, still missing "
2014 = -1016; 2015= -1409; 2016  = -1171; 2017 = ?  TOT =  -3596: JAN= -

keno historic = -2291 ; 2015= -603; 2016= -424; 2017 = ? TOT = - 3318: JAN= -39

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