Help: Lotto odds

AndyD · 29th June 2009, 08:42 AM

If we have a lotto draw with 45 numbers to choose from and seven numbers to choose, the odds of winning are around 1:45Million.

I'm trying to calculate the reduction in odds if I "cheated" by effectively choosing 16 numbers instead of just seven?

If you need clarification of the problem, check out my blog article: Psychic Oz Lotto

The problem I'm interested in is noted near the bottom of the article but feel free to check my figures throughout.

Ta.

Myriad · 29th June 2009, 09:18 AM

The sixteen-number ticket will pay off if any seven numbers of the sixteen turn out to be the winning combination. So the question boils down to, how many different seven-number combinations are covered by the sixteen numbers.

The answer is, the number of different ways you could pick seven different numbers out of a set of sixteen different numbers, divided by the number of different orders you could put any set of seven different numbers in.

The number of different ways you could pick seven different numbers out of a set of sixteen different numbers is 16*15*14*13*12*11*10, which can also be written at 16!/9!. The number of different orders you could put any set of seven different numbers in is 7!. So the answer is 16! / (7! * 9!), which comes to 11,440.

So the 16-number ticket has 11,440 times the chance of winning (meaning, winning the full jackpot for matching all seven numbers, ignoring things like near-miss prizes for matching 6 numbers and so forth) that an ordinary 7-number ticket has. That's

11,440 / 45,379,620 = 1/3,966.75

so it has about a 1 in 4000 chance.

Another way to calculate it is: seven numbers are drawn. The 16-number ticket will lose if any number is drawn that is not one of the sixteen. Seven numbers are drawn. For the first number there are 45 possible numbers, so the ticket survives 16/45 of the time. If that happens, then for the second number there are only 44 possible numbers left and 15 safe numbers, so the ticket survives 15/44 of the time. And so forth for seven numbers drawn.

16/45 * 15/44 * 14/43 * 13/42 * 12/41 * 11/40 * 10/39 = 1/3966.75

If this were a weekly game, and 80 psychics were all making 16-number predictions, the chance expectation would be that one would hit every year.

Respectfully,
Myriad

Myriad · 29th June 2009, 09:43 AM

But of course, the psychic doesn't claim that all of the winning numbers come from that set of 16, only that those numbers are favored. Which means that the psychic can claim success if any of those numbers come up... and the probability of at least one of them coming up is 1 - (29/45 * 28/44 * 27/43 * 26/42 * 25/41 * 24/40 * 23/39) = 96.6%

If I were a phony psychic, here's what I'd do: instead of recommending specific favored numbers, I'd promote a method for choosing 16 favored numbers, based on a different "key number" that I'd predict each week. The procedure would involve the person's name and birthday and so forth, so it would generate different numbers (as randomly distributed as feasible, without making the procedure too hard) for each person. Then, I'd tell people to write the numbers down (very important!) and meditate over them to decide which seven feel the best.

If only 4,000 readers followed my advice, and only 1 in 10 of them bothered to notice the results, I'd still get about two readers a year reporting that the sixteen numbers predicted all 7 of the winners (though unfortunately, the reader played the wrong seven of them that time). With more readers it would happen more often -- and each time it did, I'd claim to have psychically predicted the numbers against odds of 1 in 45 million, yet again.

Respectfully,
Myriad

Molinaro · 29th June 2009, 09:49 AM

Originally Posted by Myriad

So the 16-number ticket has 11,440 times the chance of winning (meaning, winning the full jackpot for matching all seven numbers, ignoring things like near-miss prizes for matching 6 numbers and so forth) that an ordinary 7-number ticket has. That's

11,440 / 45,379,620 = 1/3,966.75

so it has about a 1 in 4000 chance.

Just to be clear, there is no such thing as a 16 number ticket is there? What this is saying is that if you bought 11,440 tickets, covering all combinations of 7 of the 16 numbers, you would have about a 1 in 4000 chance of winning.

Myriad · 29th June 2009, 09:58 AM

Originally Posted by Molinaro

Just to be clear, there is no such thing as a 16 number ticket is there? What this is saying is that if you bought 11,440 tickets, covering all combinations of 7 of the 16 numbers, you would have about a 1 in 4000 chance of winning.

Correct. The sixteen-number ticket represents a psychic "prediction" concerning the lottery, in which numbers with certain characteristics are said to be favored, and 16 of the 45 numbers have those characteristics.

If a real option to purchase a for a 16-number ticket were offered, it would have to be priced at about 11,440 times the price of a regular ticket. (A 45-number ticket would, of course, have to cost more than the value of the jackpot.)

Respectfully,
Myriad

Ashles · 29th June 2009, 10:09 AM

Originally Posted by AndyD

If we have a lotto draw with 45 numbers to choose from and seven numbers to choose, the odds of winning are around 1:45Million.

I'm trying to calculate the reduction in odds if I "cheated" by effectively choosing 16 numbers instead of just seven?

I'm not quite clear on the question.
As Molinaro says, you don't ever actually enter 16 numbers do you?

The odds for a single ticket will always remain the same.
If you buy two tickets with different numbers, you half the odds. It's not "cheating", just increasing your likelihood. And so on.
Buy enough tickets and you can reduce the probability to a certainty. But the money you'd need to spend (aside from the physical limitations rendering it impossible) would render it prohibitive.

Unless this is all theoretical, but then you'd need to recalculate the original odds to take into account 16 entry tickets.

(I'm assuming this all works like the UK lottery where you pick 6 numbers and win the jackpot if your 6 numbers match the random 6 numbers - except an equivalent with 7 numbers)

On a slight tangent, I find this interesting because I used to work at a Holistic healing centre and the owner claimed to be a member of Mensa. Yet she bought into those "Beat the lottery" scams in a big way whereby you "increase your chances" by covering off all the number combinations in a variety of elaborate ways. Watching her and her business partner lay out loads of lottery tickets across her desk on a Friday amused me no end.

Oddly she still never semed to win more than chance would suggest.

Molinaro · 29th June 2009, 10:15 AM

I am familiar with this startegy, thanks to my crazy neighbour, back when I was a teen.

He would spend $500 to $1500 every Wednesday and every Saturday on lotto tickets. He would bet on 10 numbers, buying many tickets with different combinations of those 10 numbers. That way, when say 3 of his 10 numbers came up, he would have many tickets with those 3. Likewise when 4 of his 10 came up he would have multiple winners again.

During the 15 years he was my neighbour he showed me checks for 29k, 12k, 9k, and the big one was 250k.

He claimed that on average he would finish each year up about 2k to 10k.

AndyD · 29th June 2009, 10:41 AM

Thanks for the detailed breakdown.

I've edited the article to reflect the advice.

To Molinaro and Ashles, no I'm not talking about a 16 number ticket. An Aussie "psychic, astrologer, numerologist" said multiples of three and the number 8 were significant in a big draw here tomorrow. With 45 numbers in total, that gives him 16 numbers in his theoretical selection which significantly increases his chances of getting hits as Myriad shows.

This was only one part of a very wide net he threw - to the point where the majority of people who enter should win - if he's right.

Ladewig · 29th June 2009, 11:04 AM

Originally Posted by AndyD

said multiples of three and the number 8 were significant in a big draw here tomorrow

A rookie mistake. Mentioning 8 gives you only four more numbers: 8, 18, 28, and 38.

Saying 1 will be significant gives the predictor fourteen outs: 1, 11, 21, 31, 41, 10, 12, 13, 14, 15, 16, 17, 18, and 19.

Ladewig · 29th June 2009, 11:22 AM

Originally Posted by Myriad

The number of different ways you could pick seven different numbers out of a set of sixteen different numbers is 16*15*14*13*12*11*10, which can also be written at 16!/9!. The number of different orders you could put any set of seven different numbers in is 7!. So the answer is 16! / (7! * 9!), which comes to 11,440.

The values for choosing 8 through 15 different numbers is available on the official website.

29th June 2009, 08:42 AM	#1
AndyD Critical Thinker Join Date: Apr 2008 Posts: 498	Help: Lotto odds If we have a lotto draw with 45 numbers to choose from and seven numbers to choose, the odds of winning are around 1:45Million. I'm trying to calculate the reduction in odds if I "cheated" by effectively choosing 16 numbers instead of just seven? If you need clarification of the problem, check out my blog article: Psychic Oz Lotto The problem I'm interested in is noted near the bottom of the article but feel free to check my figures throughout. Ta.
AndyD Critical Thinker Join Date: Apr 2008 Posts: 498	__________________ Psychics, astrology, creationism, skepticism and more Stop Sylvia Browne

29th June 2009, 09:18 AM	#2
Myriad Hyperthetical Moderator Join Date: Nov 2006 Posts: 3,773	The sixteen-number ticket will pay off if any seven numbers of the sixteen turn out to be the winning combination. So the question boils down to, how many different seven-number combinations are covered by the sixteen numbers. The answer is, the number of different ways you could pick seven different numbers out of a set of sixteen different numbers, divided by the number of different orders you could put any set of seven different numbers in. The number of different ways you could pick seven different numbers out of a set of sixteen different numbers is 16151413121110, which can also be written at 16!/9!. The number of different orders you could put any set of seven different numbers in is 7!. So the answer is 16! / (7! * 9!), which comes to 11,440. So the 16-number ticket has 11,440 times the chance of winning (meaning, winning the full jackpot for matching all seven numbers, ignoring things like near-miss prizes for matching 6 numbers and so forth) that an ordinary 7-number ticket has. That's 11,440 / 45,379,620 = 1/3,966.75 so it has about a 1 in 4000 chance. Another way to calculate it is: seven numbers are drawn. The 16-number ticket will lose if any number is drawn that is not one of the sixteen. Seven numbers are drawn. For the first number there are 45 possible numbers, so the ticket survives 16/45 of the time. If that happens, then for the second number there are only 44 possible numbers left and 15 safe numbers, so the ticket survives 15/44 of the time. And so forth for seven numbers drawn. 16/45 * 15/44 * 14/43 * 13/42 * 12/41 * 11/40 * 10/39 = 1/3966.75 If this were a weekly game, and 80 psychics were all making 16-number predictions, the chance expectation would be that one would hit every year. Respectfully, Myriad
	__________________ Never use a tool that's more intelligent than you are.

29th June 2009, 09:43 AM	#3
Myriad Hyperthetical Moderator Join Date: Nov 2006 Posts: 3,773	But of course, the psychic doesn't claim that all of the winning numbers come from that set of 16, only that those numbers are favored. Which means that the psychic can claim success if any of those numbers come up... and the probability of at least one of them coming up is 1 - (29/45 * 28/44 * 27/43 * 26/42 * 25/41 * 24/40 * 23/39) = 96.6% If I were a phony psychic, here's what I'd do: instead of recommending specific favored numbers, I'd promote a method for choosing 16 favored numbers, based on a different "key number" that I'd predict each week. The procedure would involve the person's name and birthday and so forth, so it would generate different numbers (as randomly distributed as feasible, without making the procedure too hard) for each person. Then, I'd tell people to write the numbers down (very important!) and meditate over them to decide which seven feel the best. If only 4,000 readers followed my advice, and only 1 in 10 of them bothered to notice the results, I'd still get about two readers a year reporting that the sixteen numbers predicted all 7 of the winners (though unfortunately, the reader played the wrong seven of them that time). With more readers it would happen more often -- and each time it did, I'd claim to have psychically predicted the numbers against odds of 1 in 45 million, yet again. Respectfully, Myriad
	__________________ Never use a tool that's more intelligent than you are. Last edited by Myriad; 29th June 2009 at 09:48 AM. Reason: to correct the probability calculation

29th June 2009, 09:49 AM	#4
Molinaro Graduate Poster Join Date: Dec 2005 Location: Toronto Posts: 1,534	Originally Posted by Myriad So the 16-number ticket has 11,440 times the chance of winning (meaning, winning the full jackpot for matching all seven numbers, ignoring things like near-miss prizes for matching 6 numbers and so forth) that an ordinary 7-number ticket has. That's 11,440 / 45,379,620 = 1/3,966.75 so it has about a 1 in 4000 chance. Just to be clear, there is no such thing as a 16 number ticket is there? What this is saying is that if you bought 11,440 tickets, covering all combinations of 7 of the 16 numbers, you would have about a 1 in 4000 chance of winning.
	__________________ 100% Cannuck!

29th June 2009, 09:58 AM	#5
Myriad Hyperthetical Moderator Join Date: Nov 2006 Posts: 3,773	Originally Posted by Molinaro Just to be clear, there is no such thing as a 16 number ticket is there? What this is saying is that if you bought 11,440 tickets, covering all combinations of 7 of the 16 numbers, you would have about a 1 in 4000 chance of winning. Correct. The sixteen-number ticket represents a psychic "prediction" concerning the lottery, in which numbers with certain characteristics are said to be favored, and 16 of the 45 numbers have those characteristics. If a real option to purchase a for a 16-number ticket were offered, it would have to be priced at about 11,440 times the price of a regular ticket. (A 45-number ticket would, of course, have to cost more than the value of the jackpot.) Respectfully, Myriad
	__________________ Never use a tool that's more intelligent than you are.

29th June 2009, 10:09 AM	#6
Ashles Pith Artist Join Date: Apr 2003 Location: The '80s Posts: 7,838	Originally Posted by AndyD If we have a lotto draw with 45 numbers to choose from and seven numbers to choose, the odds of winning are around 1:45Million. I'm trying to calculate the reduction in odds if I "cheated" by effectively choosing 16 numbers instead of just seven? I'm not quite clear on the question. As Molinaro says, you don't ever actually enter 16 numbers do you? The odds for a single ticket will always remain the same. If you buy two tickets with different numbers, you half the odds. It's not "cheating", just increasing your likelihood. And so on. Buy enough tickets and you can reduce the probability to a certainty. But the money you'd need to spend (aside from the physical limitations rendering it impossible) would render it prohibitive. Unless this is all theoretical, but then you'd need to recalculate the original odds to take into account 16 entry tickets. (I'm assuming this all works like the UK lottery where you pick 6 numbers and win the jackpot if your 6 numbers match the random 6 numbers - except an equivalent with 7 numbers) On a slight tangent, I find this interesting because I used to work at a Holistic healing centre and the owner claimed to be a member of Mensa. Yet she bought into those "Beat the lottery" scams in a big way whereby you "increase your chances" by covering off all the number combinations in a variety of elaborate ways. Watching her and her business partner lay out loads of lottery tickets across her desk on a Friday amused me no end. Oddly she still never semed to win more than chance would suggest.
	__________________ With extraordinary few exceptions no educated person in the history of Western Civilization from the third century B.C. onward believed that the earth was flat. - Jeffrey Burton Russell No one "proved" that a bumblebee can't fly. What was shown was that a certain simple mathematical model wasn't adequate or appropriate - Ivars Peterson

29th June 2009, 10:15 AM	#7
Molinaro Graduate Poster Join Date: Dec 2005 Location: Toronto Posts: 1,534	I am familiar with this startegy, thanks to my crazy neighbour, back when I was a teen. He would spend $500 to $1500 every Wednesday and every Saturday on lotto tickets. He would bet on 10 numbers, buying many tickets with different combinations of those 10 numbers. That way, when say 3 of his 10 numbers came up, he would have many tickets with those 3. Likewise when 4 of his 10 came up he would have multiple winners again. During the 15 years he was my neighbour he showed me checks for 29k, 12k, 9k, and the big one was 250k. He claimed that on average he would finish each year up about 2k to 10k.
	__________________ 100% Cannuck!

29th June 2009, 10:41 AM	#8
AndyD Critical Thinker Join Date: Apr 2008 Posts: 498	Thanks for the detailed breakdown. I've edited the article to reflect the advice. To Molinaro and Ashles, no I'm not talking about a 16 number ticket. An Aussie "psychic, astrologer, numerologist" said multiples of three and the number 8 were significant in a big draw here tomorrow. With 45 numbers in total, that gives him 16 numbers in his theoretical selection which significantly increases his chances of getting hits as Myriad shows. This was only one part of a very wide net he threw - to the point where the majority of people who enter should win - if he's right.
AndyD Critical Thinker Join Date: Apr 2008 Posts: 498	__________________ Psychics, astrology, creationism, skepticism and more Stop Sylvia Browne

29th June 2009, 11:04 AM	#9
Ladewig Philosopher Join Date: Dec 2001 Location: Gulf Coast Posts: 8,961	Originally Posted by AndyD said multiples of three and the number 8 were significant in a big draw here tomorrow A rookie mistake. Mentioning 8 gives you only four more numbers: 8, 18, 28, and 38. Saying 1 will be significant gives the predictor fourteen outs: 1, 11, 21, 31, 41, 10, 12, 13, 14, 15, 16, 17, 18, and 19.
	__________________ When I see all the kooky things posted on the JREF forums, I can't help but think of Max Bialystock's lament: "They come here, they all come here, how do they find us?"

29th June 2009, 11:22 AM	#10
Ladewig Philosopher Join Date: Dec 2001 Location: Gulf Coast Posts: 8,961	Originally Posted by Myriad The number of different ways you could pick seven different numbers out of a set of sixteen different numbers is 16151413121110, which can also be written at 16!/9!. The number of different orders you could put any set of seven different numbers in is 7!. So the answer is 16! / (7! * 9!), which comes to 11,440. The values for choosing 8 through 15 different numbers is available on the official website.
	__________________ When I see all the kooky things posted on the JREF forums, I can't help but think of Max Bialystock's lament: "They come here, they all come here, how do they find us?"