Welcome Guest
You last visited December 10, 2016, 2:50 am
All times shown are
Eastern Time (GMT-5:00)

# Statistics question

Topic closed. 12 replies. Last post 9 years ago by RJOh.

 Page 1 of 1
New Member
Ohio
United States
Member #53971
July 29, 2007
2 Posts
Offline
 Posted: July 29, 2007, 7:52 pm - IP Logged

Hey all...here's a challenge out there for anyone in to the statistical nature of the lottery:

Let's say we are talking about a standard 6-ball lotto, 49 balls. I'm pretty sure I read somewhere that the chances of winning said lotto is 1 in 13,983,816. Well let's say that a particular lotto has reached a respectable \$115 million and everyone and their brother is purchasing tickets. Here's the big question: what are the odds that 14 (yes, that's FOURTEEN) winning tickets will be drawn for this lotto?

Yeah, I know, seems impossible, and who cares anyway? We all know that anything is possible statistically, and I care in BIG way. If someone can answer this for me CORRECTLY and show how they reached their answer, I will tell you why I needed to know. The story will fascinate you! Please email me at bykergrrl at hotmail dot com....and good luck!

New Member
Ohio
United States
Member #53971
July 29, 2007
2 Posts
Offline
 Posted: July 29, 2007, 8:24 pm - IP Logged

EDIT: Change the type of drawing in this scenerio to this:

The lotto is played by drawing 11 numbers, from 1 through 80. Players have to match only seven of the 11 numbers drawn to win a jackpot.

Now, this makes a big difference, right? Two people can win and have completely different sets of numbers. But it makes the challenge of figuring the statistically possibility of 14 winnings tickets being drawn even sweeter.

mid-Ohio
United States
Member #9
March 24, 2001
19831 Posts
Offline
 Posted: July 29, 2007, 9:24 pm - IP Logged

EDIT: Change the type of drawing in this scenerio to this:

The lotto is played by drawing 11 numbers, from 1 through 80. Players have to match only seven of the 11 numbers drawn to win a jackpot.

Now, this makes a big difference, right? Two people can win and have completely different sets of numbers. But it makes the challenge of figuring the statistically possibility of 14 winnings tickets being drawn even sweeter.

You would figure the odds like you would any keno type game. There would be 3,176,716,400 possible outcomes if you could pick seven numbers.

 MATCH COMBOS OF 7 FROM 11of80 NUMBERS ODDS 7 330 1 : 9626413 6 31878 1 : 99652 5 1083852 1 : 2931 4 17290020 1 : 184 3 142642665 1 : 22 2 618118215 1 : 5 1 1318652192 1 : 2 0 1078897248 1 : 3
mid-Ohio
United States
Member #9
March 24, 2001
19831 Posts
Offline
 Posted: July 29, 2007, 9:42 pm - IP Logged

Hey all...here's a challenge out there for anyone in to the statistical nature of the lottery:

Let's say we are talking about a standard 6-ball lotto, 49 balls. I'm pretty sure I read somewhere that the chances of winning said lotto is 1 in 13,983,816. Well let's say that a particular lotto has reached a respectable \$115 million and everyone and their brother is purchasing tickets. Here's the big question: what are the odds that 14 (yes, that's FOURTEEN) winning tickets will be drawn for this lotto?

Yeah, I know, seems impossible, and who cares anyway? We all know that anything is possible statistically, and I care in BIG way. If someone can answer this for me CORRECTLY and show how they reached their answer, I will tell you why I needed to know. The story will fascinate you! Please email me at bykergrrl at hotmail dot com....and good luck!

There could be 14 winners in any lottery that players are free to chose any combination, but I doubt if any 6/49 game would have a jackpot that big if it had a cash value of half that amount as most lotteries jackpot do now days.

Lotteries with odds of 1:10M or more seldom have a cash value greater than the odds of winning them since that investment syndicate won the Virgina lottery over ten years ago as far as I know.

Pick5 games with odds of 1:1M or less of winning the jackpot get jackpots close to \$1M several times each year.

* you don't need to buy more tickets, just buy a winning ticket *

NY
United States
Member #23835
October 16, 2005
3475 Posts
Offline
 Posted: July 29, 2007, 11:04 pm - IP Logged

If you want an answer with an explantion you should provide the necessary information. The matrix and jackpot amount aren't enough for a definite answer, bu here's a rough estimate.

Yes, the change makes a big difference, but I'm not sure you know what the difference is. In the first version there was an extremely slim chance it might happen, but it's almost a slikely as the chance  that there will even be 1 winner inth enew version.

RJ: You lost a zero somewhere.

mid-Ohio
United States
Member #9
March 24, 2001
19831 Posts
Offline
 Posted: July 29, 2007, 11:17 pm - IP Logged

If you want an answer with an explantion you should provide the necessary information. The matrix and jackpot amount aren't enough for a definite answer, bu here's a rough estimate.

Yes, the change makes a big difference, but I'm not sure you know what the difference is. In the first version there was an extremely slim chance it might happen, but it's almost a slikely as the chance  that there will even be 1 winner inth enew version.

RJ: You lost a zero somewhere.

If you find the missing zero let me  know, I calculated that by hand using some information I found on the Internet about calculating Keno Odds.  I plan to write the routine in GWBasic and add it to my ODDS calculation program since Ohio is coming out with a Keno game called "Ten OH!" next week.

* you don't need to buy more tickets, just buy a winning ticket *

mid-Ohio
United States
Member #9
March 24, 2001
19831 Posts
Offline
 Posted: July 31, 2007, 2:38 pm - IP Logged

If you find the missing zero let me  know, I calculated that by hand using some information I found on the Internet about calculating Keno Odds.  I plan to write the routine in GWBasic and add it to my ODDS calculation program since Ohio is coming out with a Keno game called "Ten OH!" next week.

I didn't find a missing zero in my calculations, so I added the routine to my odds calculator which allows me to make other calculations about regular lotteries that are useful also.  I'll never have to calculated it by hand again.

* you don't need to buy more tickets, just buy a winning ticket *

CA
United States
Member #2987
December 10, 2003
832 Posts
Offline
 Posted: August 2, 2007, 9:51 am - IP Logged

Hey all...here's a challenge out there for anyone in to the statistical nature of the lottery:

Let's say we are talking about a standard 6-ball lotto, 49 balls. I'm pretty sure I read somewhere that the chances of winning said lotto is 1 in 13,983,816. Well let's say that a particular lotto has reached a respectable \$115 million and everyone and their brother is purchasing tickets. Here's the big question: what are the odds that 14 (yes, that's FOURTEEN) winning tickets will be drawn for this lotto?

Yeah, I know, seems impossible, and who cares anyway? We all know that anything is possible statistically, and I care in BIG way. If someone can answer this for me CORRECTLY and show how they reached their answer, I will tell you why I needed to know. The story will fascinate you! Please email me at bykergrrl at hotmail dot com....and good luck!

Before you could calculate the odds of there being 14 winners in any lottery game the number of tickets sold would be necessary. "Everybody and their brother" doesn't quite cut it.

RJOh's figures of x of 7 if 11 of 80 are correct - no missing zeroes. I do my calcs with a spreadsheet.

 7 Spot - 11/80 Bet Hits Possibilities Odds - 1:: 3176716400 1 0 1078897248 2.94 1 1318652192 2.41 2 618118215 5.14 3 142642665 22.27 4 17290020 183.73 5 1083852 2930.95 6 31878 99652.31 7 330 9626413.33

gl

j

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

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

=^.^=

mid-Ohio
United States
Member #9
March 24, 2001
19831 Posts
Offline
 Posted: August 2, 2007, 12:52 pm - IP Logged

Thanks John, I always appreciate my calculations being confirmed by a serious mathematician.

* you don't need to buy more tickets, just buy a winning ticket *

United States
Member #16612
June 2, 2005
3493 Posts
Offline
 Posted: August 4, 2007, 2:06 am - IP Logged

What are the odds where a Mega Millions winning numbers [5/56+1/46] can become a California Super Lotto Plus pick [5/47+1/27]? How about Powerball winning numbers [5/55+1/42] to become a California Super Lotto Plus pick [5/47+1/27]

CA
United States
Member #2987
December 10, 2003
832 Posts
Offline
 Posted: August 4, 2007, 2:33 am - IP Logged

What are the odds where a Mega Millions winning numbers [5/56+1/46] can become a California Super Lotto Plus pick [5/47+1/27]? How about Powerball winning numbers [5/55+1/42] to become a California Super Lotto Plus pick [5/47+1/27]

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

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

=^.^=

NY
United States
Member #23835
October 16, 2005
3475 Posts
Offline
 Posted: August 7, 2007, 2:27 pm - IP Logged

If you find the missing zero let me  know, I calculated that by hand using some information I found on the Internet about calculating Keno Odds.  I plan to write the routine in GWBasic and add it to my ODDS calculation program since Ohio is coming out with a Keno game called "Ten OH!" next week.

Rather than you losing a zero I apparently found an extra zero, or at least an extra place. I forget what I came up with but I rounded a number that was the same (or at least similar) to your figure of 3,176,716,400.  Maybe I  punched a 3rd 0 before saving it in the calculator's memory, since I came up with odds around 1 in 90 million twice. I think I did the math right, but that doesn't matter if you start with the wrong numbers. The Windows calculator is a really useful feature, but having commas would help when you're using  numbers with 9 or more digits.

I suppose the OP isn't that interested, anyway, since we haven't heard back. Or maybe they just keep running to their Hotmail box and wondering why it's empty.

mid-Ohio
United States
Member #9
March 24, 2001
19831 Posts
Offline
 Posted: August 7, 2007, 11:11 pm - IP Logged

"Rather than you losing a zero I apparently found an extra zero"

I figured as much after I rechecked my calculations.  I've now written the routine into my odds calculations program and have checked it with several casino sites for accuracy.

* you don't need to buy more tickets, just buy a winning ticket *

 Page 1 of 1