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# Probabilty of Someone Winning a PB & MM Jackpot in the same Week!

Topic closed. 63 replies. Last post 10 years ago by Stack47.

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CA
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 Posted: March 3, 2007, 2:49 pm - IP Logged

johnph77

"The odds of winning if one buys two tickets are cut in half."

NOPE, they are not. Here's the flaw in this. Let's say there's a game with the odds of one million to one.

Lottery odds are based on the possible number of combinations.

So you guys with the odds are cut in half with each ticklet are saying this:

first ticket  one in one million

second ticket (same game) one in 500,000

third ticket: 1 in 250,000

fourth ticket:  1 in 125,000

fifth ticket: 1 in 62,500

sixth ticket: 1 in 31, 250

seventh ticket: 1 in 15, 625

eith ticket: 1 in 7,812.5

ninth ticket: 1 in 3,906.25

tenth ticket: 1 in  1,953.125

etc......

So, theorhetically, according to you and this reverse geometric progression drill on the odds, you could keep buying one more ticket until you'd "guarantee" a jackpot hit. Yeah, right.

All the next ticket does in any lottery is reduce the hundreds of millions (or millions - state lotteries) to one odds by one combination.

Jarasan

Yeah, I know, and thanks.

It just cracked me up to see something carried to 30 decimal places and then the comment about all figures being rounded off

Coin Toss -

The only way one can guarantee a win in any lottery is to buy enough tickets to cover every possible draw. Leave one combination out and all bets are off.

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.

=^.^=

Zeta Reticuli Star System
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 Posted: March 3, 2007, 3:42 pm - IP Logged

And even if you had that one missing combination, too, that couldn't guarantee a win becayse you can never assume a solo winner.

Cheers.

Those who run the lotteries love it when players look for consistency in something that's designed not to have any.

There is one and only one 'proven' system, and that is to book the action. No matter the game, let the players pick their own losers.

Wandering Aimlessly
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 Posted: March 3, 2007, 11:01 pm - IP Logged

"The odds of winning if one buys two tickets are cut in half. Whether those two tickets are purchased for the same or different drawings makes no difference, the odds are cut in half. Just as the odds are the odds, the mathematics are the mathematics. As was pointed out, 2::10,000,000 is the same as 1::5,000,000. That's math."  Johnph77

Well, I don't want to keep getting into long discussions about this, but I'll try one more time. Thanks, Coin Toss. I'm glad someone here is paying attention.  I am not a stubborn person (quite the opposite) but I am a logical one. In fact, I'm not that great at math, but I am using common sense. Sure, if you buy one ticket and your neighbor buys 2 tickets or even 100 tickets, he definitely has more opportunities to win a jackpot. Do you know how many people spend \$100 and don't get anything, not even 3 numbers?

Let's use a game where the odds are 1 in 140 million.  If there are 140 million combinations, how can buying just one more ticket wipe out the other 139,999,999 combinations?  To say that buying a second ticket cuts the odds from 1:140M to 1:70M doesn't make any sense to me at all. Mathematics is logical.  It is a science.  Buying one more ticket DOES NOT cut the odds in half, and this isn't just a matter of semantics - and if you think "that's math" it's fine with me.  I just hope you don't teach statistics or your students are all in big trouble.  Okay, I said I wasn't going to get into a lot of detail, but let me try one more time (as I have on so many other threads) to explain this step by step.

You have 140 million different combinations - buy 1 ticket

You have 139,999,999 different combinations left - buy another ticket

You have 139,999,998 different combinations left - buy a third ticket

You have 139,999,997 different combinations left - buy a fouth ticket

You have 139,999,996 different combinations left - buy a fifth ticket

You have 139,999,995 different combination lefts - buy a sixth ticket.

If you are crazy enough to spend \$999K on one drawing, you will still have over 139 million combinations that might come up that you didn't pick.

So now maybe you can tell me how buying a second ticket cuts the odds in half and makes it 1 in 70M.  What happened to the rest of the combinations?  When the odds are 1 in 70M it means there are 70 million combinations.  But I just showed you above that you have over 139,999,998 combinations left after buying 2 tickets. What happened to the other 69,999,998 ?

"John has it exactly right, assuming one ticket for each drawing.  The fact that a bunch of posters here don't understand the math doesn't change the facts. Having two tickets makes you twice as likely to win,"

KY Floyd, you can tell me that I don't understand a math equation and I might agree with you. But using convoluted logic to prove an invalid point won't change the fact that you are WRONG when you keep insisting that having 2 chances to win cuts the odds in half.  Using the example of 2/4 is silly.  Why not say the odds are 1 out of 1?  Let's keep it simple and make it 100.  A very easy game anyone can win, not MM or PB or even pick-3.  Buying 1 ticket gives you a 1 in 100 shot.  2 gives you a 2 in 100 shot, not 1 in 50.  Why? Because there are still 98 combinations you haven't bet on!   However, if you spend \$50, you are cutting the odds in half.  So the way to cut the odds in half in a game where the odds are 1 in 140 million is to spend \$70 million.

Zeta Reticuli Star System
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 Posted: March 4, 2007, 1:03 am - IP Logged

""John has it exactly right, assuming one ticket for each drawing.  The fact that a bunch of posters here don't understand the math doesn't change the facts. Having two tickets makes you twice as likely to win,""

Oh, the math is understood. But by some it's understood like the old Abbot and Costello routine where a 40 year old uncle has a 10 year old niece, so he's 4 times as old as she is.

Five years go by, now he's 45 and she's 15, so he's only three times as old as she is!

Another five years pass, finding him to be 50 and she's 20. Now he's only twa and a half times as old as she is!

Alas! Five more years, he's 60! She's 30! Only twice as old now

______________-

(my continuation of the story):

So they went to a lotto lounge, bout TWO tickets for the \$340,000,000 Mega Millions drawing, got out a calculator, and started figuring out how much longer they'd have to wait until they were the same age!

"Would you like a Mega Millions ticket, sir?"

"Two please, I'd like to cut the odds in half"

"And I'm sure in a few years you and your niece will be the same age. Good luck, sir."

Those who run the lotteries love it when players look for consistency in something that's designed not to have any.

There is one and only one 'proven' system, and that is to book the action. No matter the game, let the players pick their own losers.

Wandering Aimlessly
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 Posted: March 4, 2007, 1:26 am - IP Logged

""John has it exactly right, assuming one ticket for each drawing.  The fact that a bunch of posters here don't understand the math doesn't change the facts. Having two tickets makes you twice as likely to win,""

Oh, the math is understood. But by some it's understood like the old Abbot and Costello routine where a 40 year old uncle has a 10 year old niece, so he's 4 times as old as she is.

Five years go by, now he's 45 and she's 15, so he's only three times as old as she is!

Another five years pass, finding him to be 50 and she's 20. Now he's only twa and a half times as old as she is!

Alas! Five more years, he's 60! She's 30! Only twice as old now

______________-

(my continuation of the story):

So they went to a lotto lounge, bout TWO tickets for the \$340,000,000 Mega Millions drawing, got out a calculator, and started figuring out how much longer they'd have to wait until they were the same age!

"Would you like a Mega Millions ticket, sir?"

"Two please, I'd like to cut the odds in half"

"And I'm sure in a few years you and your niece will be the same age. Good luck, sir."

NY
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 Posted: March 4, 2007, 2:47 am - IP Logged

johnph77

"The odds of winning if one buys two tickets are cut in half."

NOPE, they are not. Here's the flaw in this. Let's say there's a game with the odds of one million to one.

Lottery odds are based on the possible number of combinations.

So you guys with the odds are cut in half with each ticklet are saying this:

first ticket  one in one million

second ticket (same game) one in 500,000

third ticket: 1 in 250,000

fourth ticket:  1 in 125,000

fifth ticket: 1 in 62,500

sixth ticket: 1 in 31, 250

seventh ticket: 1 in 15, 625

eith ticket: 1 in 7,812.5

ninth ticket: 1 in 3,906.25

tenth ticket: 1 in  1,953.125

etc......

So, theorhetically, according to you and this reverse geometric progression drill on the odds, you could keep buying one more ticket until you'd "guarantee" a jackpot hit. Yeah, right.

All the next ticket does in any lottery is reduce the hundreds of millions (or millions - state lotteries) to one odds by one combination.

Jarasan

Yeah, I know, and thanks.

It just cracked me up to see something carried to 30 decimal places and then the comment about all figures being rounded off

When I say that it's 2nd grade math I'm not being facetious. It really is simple, and I'm not sure what your block is that you can't understand something that really is simple.

There may be some morons who think it works the way you describe, but as far as I know nobody here has ever suggested that the odds are cut in half for each ticket played or that there's a geometric progression like the one you describe. The claim that is made by those of us who do understand how it works is that buying twice as many tickets doubles your chances of winning. I'm guessing that even you know that 3 isn't twice as much as 2, so I really have no clue where you came up with the scheme you've described.

Buying a second ticket cuts your odds in half because it doubles the number of tickets, and therefore the number of chances you have to win. You started with 1 ticket and you now have 2. 2 is twice as much as 1. Because it doubles the number of tickets you had the first additional ticket cuts your odds in half.

Buying 3 tickets means you have three chances to win instead of 1. 3 is three times as much as 1, not 4 times as much. That means that buying 3 tickets makes you 3 times as likely to win as you are with only 1 ticket. It's a simple linear progression. For your game with odds of 1 in 1 million 3 times as likely means 3 in 1 million or 1 in 333,333,333. It's a simple ratio, so dividing both sides by 3  doesn't change it's value.

If you buy 4 tickets you have four times as many chances to win so you are 4 times as likely to win.  That's 4 in 1 million or 1 in 250,000. 4 is twice as much as 2, so buying 4 tickets makes you twice as likely to win as if you bought 2 tickets.

Buying another ticket gives you 1 more ticket, but only doubles the number of tickets when the number of tickets goes from 1 to 2. Once you have more than 1 ticket buying another one doesn't double the number of tickets you have. That's why nobody is saying that buying "another" ticket doubles your chances unless they are talking about a second ticket.

If you have 1 ticket you need "another ticket" to have twice as many.

If you have 2 tickets you need 2  more to double your number of tickets to four.

Once you have 4 tickets you'll need 4 more to double your total to 8.

The way a geometric progression of halving your odds would work is this:

1 ticket   =  1 chance    =   1 in 1,000,000

2 tickets =  2 chances  =   2 in 1,000,000 which simplifies to 1 in 500,000

4 tickets =  4 chances  =   4 in 1,000,000 which simplifies to 1 in 250,000

8 tickets =  8 chances  =  8 in 1,000,000 which simplifies to 1 in 125,000

16 tickets = 16 chances = 16 in 1,000,000 which simplifies to 1 in 62,500

NY
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 Posted: March 4, 2007, 3:01 am - IP Logged

"The odds of winning if one buys two tickets are cut in half. Whether those two tickets are purchased for the same or different drawings makes no difference, the odds are cut in half. Just as the odds are the odds, the mathematics are the mathematics. As was pointed out, 2::10,000,000 is the same as 1::5,000,000. That's math."  Johnph77

Well, I don't want to keep getting into long discussions about this, but I'll try one more time. Thanks, Coin Toss. I'm glad someone here is paying attention.  I am not a stubborn person (quite the opposite) but I am a logical one. In fact, I'm not that great at math, but I am using common sense. Sure, if you buy one ticket and your neighbor buys 2 tickets or even 100 tickets, he definitely has more opportunities to win a jackpot. Do you know how many people spend \$100 and don't get anything, not even 3 numbers?

Let's use a game where the odds are 1 in 140 million.  If there are 140 million combinations, how can buying just one more ticket wipe out the other 139,999,999 combinations?  To say that buying a second ticket cuts the odds from 1:140M to 1:70M doesn't make any sense to me at all. Mathematics is logical.  It is a science.  Buying one more ticket DOES NOT cut the odds in half, and this isn't just a matter of semantics - and if you think "that's math" it's fine with me.  I just hope you don't teach statistics or your students are all in big trouble.  Okay, I said I wasn't going to get into a lot of detail, but let me try one more time (as I have on so many other threads) to explain this step by step.

You have 140 million different combinations - buy 1 ticket

You have 139,999,999 different combinations left - buy another ticket

You have 139,999,998 different combinations left - buy a third ticket

You have 139,999,997 different combinations left - buy a fouth ticket

You have 139,999,996 different combinations left - buy a fifth ticket

You have 139,999,995 different combination lefts - buy a sixth ticket.

If you are crazy enough to spend \$999K on one drawing, you will still have over 139 million combinations that might come up that you didn't pick.

So now maybe you can tell me how buying a second ticket cuts the odds in half and makes it 1 in 70M.  What happened to the rest of the combinations?  When the odds are 1 in 70M it means there are 70 million combinations.  But I just showed you above that you have over 139,999,998 combinations left after buying 2 tickets. What happened to the other 69,999,998 ?

"John has it exactly right, assuming one ticket for each drawing.  The fact that a bunch of posters here don't understand the math doesn't change the facts. Having two tickets makes you twice as likely to win,"

KY Floyd, you can tell me that I don't understand a math equation and I might agree with you. But using convoluted logic to prove an invalid point won't change the fact that you are WRONG when you keep insisting that having 2 chances to win cuts the odds in half.  Using the example of 2/4 is silly.  Why not say the odds are 1 out of 1?  Let's keep it simple and make it 100.  A very easy game anyone can win, not MM or PB or even pick-3.  Buying 1 ticket gives you a 1 in 100 shot.  2 gives you a 2 in 100 shot, not 1 in 50.  Why? Because there are still 98 combinations you haven't bet on!   However, if you spend \$50, you are cutting the odds in half.  So the way to cut the odds in half in a game where the odds are 1 in 140 million is to spend \$70 million.

The math and the logic are both very easy, but you're starting off with at least one major mistake, and logic only works if your premises are valid. Doubling your chances of winning and halving your chances of losing are not the same thing. If I had the vaguest idea of why it isn't fundamentally obvious to some people that doubling the number of tickets doubles the chances it might be easier to explain, but I really don't know what make it difficult for some people.

"When the odds are 1 in 70M it means there are 70 million combinations"

That is not what 1 in 70 million means. Odds of 1 in 70 million means there's a chance of winning once for *every* 70 million outcomes. That might sound like a subtle diference but it's not. Odds are a ratio. That means the two numbers have values that are relative  to one another. As long as the relationship remain the same the ratio is still the same. That's why the analogy of the fractions is a perfect example. Ratios and fractions are the same thing and I'm fairly confident that almost everyone here understands simple fractions. If you've got two buckets  and the ratio of the size of the big bucket to the little bucket is 4:1 then the little bucket will hold 1/4 of what the big bucket holds. If the big bucket holds 4 quarts then the little bucket holds 1 quart. That also means that the big bucket holds 8 pints and the little bucket holds 2 pints.  A ratio of 8:2 is exactly the same as 4:1, because the number on the left is 4 times the number on the right in both cases. By convention ratios are are expressed with one side reduced to the smallest number that still has a whole number on the other side. For many things that means that  the odds will be expressed as 1:n, but  2:2n and 3:3n are exactly the same ratio as 1:n.

Because they're ratios, odds work the same way.  Where possible they're usually expressed as 1:n, but 2:2n is still the same odds. That means that 1:70 million is exactly the same odds as 2:140 million.

"So now maybe you can tell me how buying a second ticket cuts the odds in half and makes it 1 in 70M."

I guess I can, and for me it doesn't seem hard to understand.

Wandering Aimlessly
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 Posted: March 4, 2007, 12:42 pm - IP Logged

"Buying a second ticket cuts your odds in half because it doubles the number of tickets, and therefore the number of chances you have to win. You started with 1 ticket and you now have 2. 2 is twice as much as 1. Because it doubles the number of tickets you had the first additional ticket cuts your odds in half."

Well, as I said before - no point in arguing.  I once worked with 2 men from Georgia who were laughing at me (making blonde jokes) because I said Atlanta was the capital city.  One of them said "I lived there for 15 years. Don't you think I'd know it's Savannah?"  I told him that unless he was over 100 years old, he didn't know the capital of his home state. (think it once was Savannah)  So the manager of the store got involved.  he was from Tennessee.  I said "That one is easy. It's Nashville."  Boy did they start to chuckle. "No, it's Chattanooga" this 72 year old manager said. "I should know. I brought up my family in Tennessee."  So my point is that there's no way to reason with people who think they "know" something even when it's dead wrong.

Zeta Reticuli Star System
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 Posted: March 4, 2007, 1:35 pm - IP Logged

Ky Floyd

"I guess I can, and for me it doesn't seem hard to understand."

Of course, it doesnt, for you, because after all you've seen "mathematical proof" that the uncle and his niece will one day be the same age.

No offense Floyd, but when anybody tells me lottery odds are not based on possivble combinations, that's all they've got to tell me about the lottery.

And I'll try to get away from that person before they start telling their stories about the wonderful time they had at the 1994 World Series.

Buono Fortuna

Those who run the lotteries love it when players look for consistency in something that's designed not to have any.

There is one and only one 'proven' system, and that is to book the action. No matter the game, let the players pick their own losers.

Greenwich, CT
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 Posted: March 4, 2007, 1:36 pm - IP Logged

This thread looks like it's going nowhere.

But I'm going to have to weigh in with KY Floyd and johnph77.

We are phrasing the question in terms of odds.  The other side phrases it in terms of probabilities.  Two different concepts that require different calculations.

Greenwich, CT
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 Posted: March 4, 2007, 1:42 pm - IP Logged

Ky Floyd

"I guess I can, and for me it doesn't seem hard to understand."

Of course, it doesnt, for you, because after all you've seen "mathematical proof" that the uncle and his niece will one day be the same age.

No offense Floyd, but when anybody tells me lottery odds are not based on possivble combinations, that's all they've got to tell me about the lottery.

And I'll try to get away from that person before they start telling their stories about the wonderful time they had at the 1994 World Series.

Buono Fortuna

The uncle and niece will never be the same age, but the proportion will continually get smaller, closer to one.

40/10 is greater than 60/30 is greater than 100/70.  Even when the uncle is 1,000 years old and the niece is 970, the uncle will still be older, but the proportion of 1000 to 970 is much smaller than 40 to 10.

Greenwich, CT
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 Posted: March 4, 2007, 1:49 pm - IP Logged

justxploring,

Let's revisit your 100 combo game, since it seems to be an easy example.

The key is that odds are ratios are fractions.  They are all the same.

You can reduce a fraction.  30/90 is the same as 9/27 is the same as 1/3.  Therefore you can also reduce odds.

If you buy 5 tickets for your game, you can express the odds of winning as 5 in 100.  Or 5/100...it's a fraction.  You can reduce this to 1/20.  It's perfectly valid to say that the odds are 1 in 20 of winning on five distinct tickets.  Or 5 in 100.  Same thing.

Wandering Aimlessly
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 Posted: March 4, 2007, 2:03 pm - IP Logged

Not really, but I do agree we'll never agree this way.  Saying 2 tickets changes the odds of a 1 in 140 million odd game to 1 in 70 million is assuming that each ticket is only gambling on 1/2 of the combinations.  But when you have 2 tickets that are betting on all 140 million combinations, the odds are NOT cut in half.  Sure, I agree if you had 2 separate games where there were 70 million possibilities (and this also applies to games with lesser odds) then buying a ticket for each game would prove KY Floyd and John correct.  But that's not what is happening.  A person is buying 2 tickets for the same 140 million choices.  So buying 2 tickets just leaves 139,999,998 combinations left.  If the odds were cut in half as you say, then each ticket purchased would represent an equal number of combinations.  A simple way to look at your theory or example is to use a lottery pool.  If 2 tickets cut the odds in half, why wouldn't a pool where members buy 100 tickets (and there are many that buy 1,000) win all the time? The reason is that you are still betting on the original 140 million combinations. The way you state it, you are betting 1 ticket on 1/2 the combinations and the second on the other 1/2 and so on.  Hope I am being clear, but maybe we are on such different planes here that we'll never see it the same way.

Zeta Reticuli Star System
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 Posted: March 4, 2007, 2:05 pm - IP Logged

JAG331

I had twenty years of casino work and most of the old time bosses said they'd rather have someone who just got a degree in probabilities buy in on one of their games than any other player.

Those who run the lotteries love it when players look for consistency in something that's designed not to have any.

There is one and only one 'proven' system, and that is to book the action. No matter the game, let the players pick their own losers.

New Jersey
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 Posted: March 4, 2007, 3:02 pm - IP Logged

The odds of winning if one buys two tickets are cut in half. Whether those two tickets are purchased for the same or different drawings makes no difference, the odds are cut in half. Just as the odds are the odds, the mathematics are the mathematics. As was pointed out, 2::10,000,000 is the same as 1::5,000,000. That's math.

The possible confusion here is with possibilities (and odds) vs. probabilities - they're two different ball games. Let's take a 1/10 lottery matrix, for example - draw 1 number from 10, 0 through 9. If I buy one ticket, my odds of winning are 1::10. If I buy two tickets (with different numbers for the same drawing, or with the same or different numbers for a subsequent drawing) my odds of winning are 2::10, or 1::5.

Does that mean that in every 10 drawings each number is guaranteed to appear once? And that, out of 10 people who buy tickets, all with differing numbers, all will win once? Nope. In the off chance, for instance, the numbers 1 through 9 appear in any order in 9 consecutive drawings, is 0 (zero) guaranteed to appear next? Nope. The odds of that zero appearing are exactly the same as any other number.

Take a look at the last 1,000 consecutive drawings of any Pick 3 game. Do all 1,000 possibilities appear? Not likely. I worked out a spreadsheet on the probability of such an event and the odds of that occurance are incredibly long. But here's the thing - if 999 different combinations should happen to appear in 999 consecutive draws, the odds of that 1,000th combination appearing in the very next draw would still be 1::1,000. No one is guaranteed an eventual win. That's why it's called gambling.

gl

j

John and KYFloyd are right.  But there is just too much confusion.

John sums it up pretty well.

"The possible confusion here is with possibilities (and odds) vs. probabilities - they're two different ball games."

Lets see if I can take a stab at explaining this.  Lets say for example I buy the combination 1, 2, 3, 4, 5 with every possible Mega Ball (46 in total).  The odds of matching 5 of 5 are 3,904,700.80 (not the same as how many different combinations).  The number of different combinations of 5 numbers out of a pool of 56 numbers is 3,819,816.  So if I buy the above combination with every Mega Ball then my odds of winning are brought down to 1 in 3,819,816.  To put it differently, my combination has to be selected from 3,819,816 different combinations.  So buying only 46 different lines, my odds of winning the jackpot are brought down to 1 in 3,819,816 (175,711,536 divided by 46).  But that just goes to show you how hard it is to win MegaMillions, even when my odds are reduced to 1 in 3,819,816.

Does it make sense now justxploring and Coin Toss?

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