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# A dollar and a dream true for PB and MM?

Topic closed. 109 replies. Last post 9 years ago by Perfect Timing.

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Chief Bottle Washer
New Jersey
United States
Member #1
May 31, 2000
23352 Posts
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 Posted: September 1, 2007, 8:24 pm - IP Logged

I'm just wondering if any winners ever won the big one on PB and MM with just a \$1 dollar ticket.  In the news, all I hear are ones with \$3 or \$5 tickets.

Reason I'm asking is because I might reduce the amount I spend on lottery.  Everyone's telling me "One is all you need".

It has absolutely happened many times before (winning on one pick), and if you search the News forum for past winners you will uncover some.  However, more people win with multiple picks.  That is odds at work, plain and simple.  It is the same reason why more winners tend to come from states with large populations.

Jack Whittaker won with \$98 in Powerball tickets.  (To use an infamous example.)

Check the State Lottery Report Card

Sign the Petition for True Lottery Drawings
Help eliminate computerized drawings!

Indiana
United States
Member #48725
January 7, 2007
1958 Posts
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 Posted: September 1, 2007, 8:24 pm - IP Logged

If you bought all the combinations (over 3.5 million) of the 55 numbers, it would give you a 1 in 42 chance of hitting the jackpot.

Actually there are 3,478,761 white ball combinations. The reason why the PowerBall website lists your odds of getting 5/5 + 0 as 1 in 3,563,608.83 has to do with this statement from the PowerBall website:

YOUR ODDS / PROBABILITIES ARE WRONG.

A common question is why the odds for winning the prize for matching 1 red ball out of 42 is not 1 in 42. The answer is that you must match the red ball ALONE to win the prize. If you match one or more white balls, you win some other prize, but not this prize.

Some persons who enjoy statistics will (they do really exist) come up with odds of 1 in 17 billion. Remember that you don't need to match in exact order - we use combinations to determine the probabilities for the first five white balls and not permutations.

So basically when they calculate your odds of winning a certain prize, they're not only taking into consideration the odds of receiving AT LEAST that prize, but also the odds of NOT winning some OTHER prize. This is the math to get the number of white ball combinations:

(55 x 54 x 53 x 52 x 51) / 5!

5! means 5 factorial. To find the factorial of a number, you take all the numbers between 1 and that number and multiply them. So 5! = (1 x 2 x 3 x 4 x 5) = 120. So 5! = 120.

(55 x 54 x 53 x 52 x 51) / 120

which is

417,451,320 / 120 = 3,478,761. Then to find out what the odds are for winning the jackpot are, you just multiply that by the number of PowerBalls, which is 42. That comes out to 146,107,962.

To find the odds of winning a Pick 6/48 game:

(48 x 47 x 46 x 45 x 44 x 43) / 6!

(48 x 47 x 46 x 45 x 44 x 43) / (1 x 2 x 3 x 4 x 5 x 6)

(48 x 47 x 46 x 45 x 44 x 43) / 720

8,835,488,640 / 720

1 in 12,271,512.

Gonna win.

New Jersey
United States
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June 28, 2005
51138 Posts
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 Posted: September 1, 2007, 8:34 pm - IP Logged

You're odds are not cut in half for every ticket you buy. It's fractions.

5 in 146,107,962 is the same as 1 in 29,221,592.4

100 in 146,107,962 is the same as 1 in 1,461,079.62

1,000 in 146,107,962 is the same as 1 in 146,107.962

10,000 in 146,107,962 is the same as 1 in 14,610.7962

100,000 in 146,107,962 is the same as 1 in 1,461.07962

You can call that painting a more optimistic picture if you want, but that's the way it is. Even if you spend \$1,000,000, your odds are still 1 in 146.107962, which is still pretty small.

If I buy 1 ticket my chance of Winnng is:
1 in 146,107,962 (146,107,961 remaining)

If I buy a 2nd ticket my chance of Winnng is:
1 in 146,107,961 (146,107,960 remaining)

If I buy a 3rd ticket my chance of Winnng is:
1 in 146,107,960 (146,107,959 remaining)

If I buy a 4th ticket my chance of Winnng is:
1 in 146,107,959 (146,107,958 remaining)

If I buy a 5th ticket my chance of Winnng is:
1 in 146,107,958 (146,107,957 remaining)

So buying more Tickets doesn't improve your chance of Winning by much. (It's infinitesimal.)

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

Indiana
United States
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January 7, 2007
1958 Posts
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 Posted: September 1, 2007, 9:07 pm - IP Logged

If I buy 1 ticket my chance of Winnng is:
1 in 146,107,962 (146,107,961 remaining)

If I buy a 2nd ticket my chance of Winnng is:
1 in 146,107,961 (146,107,960 remaining)

If I buy a 3rd ticket my chance of Winnng is:
1 in 146,107,960 (146,107,959 remaining)

If I buy a 4th ticket my chance of Winnng is:
1 in 146,107,959 (146,107,958 remaining)

If I buy a 5th ticket my chance of Winnng is:
1 in 146,107,958 (146,107,957 remaining)

So buying more Tickets doesn't improve your chance of Winning by much. (It's infinitesimal.)

If I buy a 2nd ticket my chance of Winnng is:
1 in 146,107,961 (146,107,960 remaining)

If I buy a 3rd ticket my chance of Winnng is:
1 in 146,107,960 (146,107,959 remaining)

If I buy a 4th ticket my chance of Winnng is:
1 in 146,107,959 (146,107,958 remaining)

If I buy a 5th ticket my chance of Winnng is:
1 in 146,107,958 (146,107,957 remaining)

Those are incorrect because you also have to take into consideration the previous ones that you have purchased. If you have 5 tickets, your odds of winning the jackpot are 1 in 29,221,592.4.

Gonna win.

Harbinger
D.C./MD.
United States
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July 30, 2006
5587 Posts
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 Posted: September 1, 2007, 9:29 pm - IP Logged

If I buy a 2nd ticket my chance of Winnng is:
1 in 146,107,961 (146,107,960 remaining)

If I buy a 3rd ticket my chance of Winnng is:
1 in 146,107,960 (146,107,959 remaining)

If I buy a 4th ticket my chance of Winnng is:
1 in 146,107,959 (146,107,958 remaining)

If I buy a 5th ticket my chance of Winnng is:
1 in 146,107,958 (146,107,957 remaining)

Those are incorrect because you also have to take into consideration the previous ones that you have purchased. If you have 5 tickets, your odds of winning the jackpot are 1 in 29,221,592.4.

The tix you purchased are all in play at the same time, pretend you don't know anything and read this:

Lets try another way of looking at this discussion of odds, it hasn't popped up in a while so I will attempt another way to look at this topic.

This is what JX, Raven and the Pac are saying and I agree.

It is more of a set theory perspective than say: odds or fractions: it is much simpler.

I am sure we all agree that each draw is ONE event, the event (Mega rounding off) has 150,000,000 possible outcomes, only of which 1 of those 150M will fall on any draw.  These 150M possible outcomes of that event is set A.

The  lines you play  let's say is 2 lines that is set B.

Remember set A has 150M possible outcomes,  set B has 2 of those 150M possibilities. Because the draw is ONE event all 150M are in play at ONCE, 149,999,998 of the possibilities do not belong to set B, therefore the size of set B must approach the size of set A in order to even start thinking of a jackpot hit.

It is a mistake to say you double your chances by buying 2 tickets, because what you are doing is falling into a math trap.

You see when you fractionalize by half, you are now saying there are now TWO distinct sets of possible outcomes per draw.  In other words, instead of set A being 150M possible outcomes, it is now two sets of possible outcomes, set A is broken into two separate sets of 75,000,000 possibilities each and the 2 tickets you are playing can't be applied this way because it is a SINGLE EVENT (SINGLE SET) of 150M possible outcomes.  So which of the 2 separate 75M sets do you apply your two tickets to?

You see the problem?  Which set would you apply the tickets? You can't because you don't have the luxury of choosing where the outcome will come from, it is a single set of 150M possible outcomes. It is 150M  to 1 ONCE,  not  75M to 1, twice.

Remember it is one draw, one set of possible outcomes, not 1 draw with two possible sets of outcomes, buying more tickets doesn't change the size of SET A (150M) significantly relative to SET B (your tickets).

Kentucky
United States
Member #32652
February 14, 2006
7344 Posts
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 Posted: September 1, 2007, 9:34 pm - IP Logged

1 thru 55 is only 3.5 million?

Funny, 1 thru 53 in Floriduh is about 23 million combos. What Am I missing here?

Afterall you did say ALL combos.

That's because there are only 5 numbers in each combo; (55 x 54 x 53 x 52 x 51) / (5 x 4 x 3 x 2 x 1) = 3,478,761 combos. Multiply that by the 42 pb numbers to get the total number of combos (146,107,962). If somebody played all the combos with 1 pb number they would have a 1 in 42 chance of hitting the jackpot. Or they could play all the pb numbers and have a 1 in 3,478,761 chance of hitting the jackpot.

The Florida Lotto has 6 numbers in each combo so it's (53 x 52 x 51 x 50 x 49 x 48) / (6 x 5 x 4 x 3 x 2 x 1) = 22,957,480 combos. For comparison it would be similar to keying one number in the Florida Lotto (like playing all the combos with the number 1).

I doubt even if somebody had a spare \$3.5 they would want to make that bet and playing all the pb numbers for \$42 isn't all that great either because you're only guaranteed to get back 3 or 4 bucks. A lottery pool might consider playing \$420 (each pb number 10 times) to get a 1 in 347,876.9 chance of hitting the jackpot.

Harbinger
D.C./MD.
United States
Member #44103
July 30, 2006
5587 Posts
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 Posted: September 1, 2007, 9:51 pm - IP Logged

That's because there are only 5 numbers in each combo; (55 x 54 x 53 x 52 x 51) / (5 x 4 x 3 x 2 x 1) = 3,478,761 combos. Multiply that by the 42 pb numbers to get the total number of combos (146,107,962). If somebody played all the combos with 1 pb number they would have a 1 in 42 chance of hitting the jackpot. Or they could play all the pb numbers and have a 1 in 3,478,761 chance of hitting the jackpot.

The Florida Lotto has 6 numbers in each combo so it's (53 x 52 x 51 x 50 x 49 x 48) / (6 x 5 x 4 x 3 x 2 x 1) = 22,957,480 combos. For comparison it would be similar to keying one number in the Florida Lotto (like playing all the combos with the number 1).

I doubt even if somebody had a spare \$3.5 they would want to make that bet and playing all the pb numbers for \$42 isn't all that great either because you're only guaranteed to get back 3 or 4 bucks. A lottery pool might consider playing \$420 (each pb number 10 times) to get a 1 in 347,876.9 chance of hitting the jackpot.

Set theory applied here works, If you are going for the PBALL, your played set has all 42 PBALL possible outcomes --> SET B (your plays for the PBALL) has all the possible outcomes for SET A, the 42 balls in play for that draw.

SET B approached SET A  ;  SET B = SET A  a guaranteed match.

Indiana
United States
Member #48725
January 7, 2007
1958 Posts
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 Posted: September 1, 2007, 9:51 pm - IP Logged

The tix you purchased are all in play at the same time, pretend you don't know anything and read this:

Lets try another way of looking at this discussion of odds, it hasn't popped up in a while so I will attempt another way to look at this topic.

This is what JX, Raven and the Pac are saying and I agree.

It is more of a set theory perspective than say: odds or fractions: it is much simpler.

I am sure we all agree that each draw is ONE event, the event (Mega rounding off) has 150,000,000 possible outcomes, only of which 1 of those 150M will fall on any draw.  These 150M possible outcomes of that event is set A.

The  lines you play  let's say is 2 lines that is set B.

Remember set A has 150M possible outcomes,  set B has 2 of those 150M possibilities. Because the draw is ONE event all 150M are in play at ONCE, 149,999,998 of the possibilities do not belong to set B, therefore the size of set B must approach the size of set A in order to even start thinking of a jackpot hit.

It is a mistake to say you double your chances by buying 2 tickets, because what you are doing is falling into a math trap.

You see when you fractionalize by half, you are now saying there are now TWO distinct sets of possible outcomes per draw.  In other words, instead of set A being 150M possible outcomes, it is now two sets of possible outcomes, set A is broken into two separate sets of 75,000,000 possibilities each and the 2 tickets you are playing can't be applied this way because it is a SINGLE EVENT (SINGLE SET) of 150M possible outcomes.  So which of the 2 separate 75M sets do you apply your two tickets to?

You see the problem?  Which set would you apply the tickets? You can't because you don't have the luxury of choosing where the outcome will come from, it is a single set of 150M possible outcomes. It is 150M  to 1 ONCE,  not  75M to 1, twice.

Remember it is one draw, one set of possible outcomes, not 1 draw with two possible sets of outcomes, buying more tickets doesn't change the size of SET A (150M) significantly relative to SET B (your tickets).

I know I'm correct.

5/146,107,962 = 1/29,221,592.4

the same as

5/20 = 1/4.

Gonna win.

Harbinger
D.C./MD.
United States
Member #44103
July 30, 2006
5587 Posts
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 Posted: September 1, 2007, 10:03 pm - IP Logged

I know I'm correct.

5/146,107,962 = 1/29,221,592.4

the same as

5/20 = 1/4.

Here it is, you have to try and conceptualize differently and focus:

Actually when you play the lottery it is this:  It helps if you turn your calculator off.

5/146,107,962 = 1/146,107,962 + 1/146,107,962 + 1/146,107,962 + 1/146,107,962 + 1/146,107,962 .

Not this:

5/146,107,962 = 1/29,221,592.4

In which 29,221,592.4 tickets do you put each of your 5 tickets?

Kentucky
United States
Member #32652
February 14, 2006
7344 Posts
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 Posted: September 1, 2007, 10:06 pm - IP Logged

The tix you purchased are all in play at the same time, pretend you don't know anything and read this:

Lets try another way of looking at this discussion of odds, it hasn't popped up in a while so I will attempt another way to look at this topic.

This is what JX, Raven and the Pac are saying and I agree.

It is more of a set theory perspective than say: odds or fractions: it is much simpler.

I am sure we all agree that each draw is ONE event, the event (Mega rounding off) has 150,000,000 possible outcomes, only of which 1 of those 150M will fall on any draw.  These 150M possible outcomes of that event is set A.

The  lines you play  let's say is 2 lines that is set B.

Remember set A has 150M possible outcomes,  set B has 2 of those 150M possibilities. Because the draw is ONE event all 150M are in play at ONCE, 149,999,998 of the possibilities do not belong to set B, therefore the size of set B must approach the size of set A in order to even start thinking of a jackpot hit.

It is a mistake to say you double your chances by buying 2 tickets, because what you are doing is falling into a math trap.

You see when you fractionalize by half, you are now saying there are now TWO distinct sets of possible outcomes per draw.  In other words, instead of set A being 150M possible outcomes, it is now two sets of possible outcomes, set A is broken into two separate sets of 75,000,000 possibilities each and the 2 tickets you are playing can't be applied this way because it is a SINGLE EVENT (SINGLE SET) of 150M possible outcomes.  So which of the 2 separate 75M sets do you apply your two tickets to?

You see the problem?  Which set would you apply the tickets? You can't because you don't have the luxury of choosing where the outcome will come from, it is a single set of 150M possible outcomes. It is 150M  to 1 ONCE,  not  75M to 1, twice.

Remember it is one draw, one set of possible outcomes, not 1 draw with two possible sets of outcomes, buying more tickets doesn't change the size of SET A (150M) significantly relative to SET B (your tickets).

It's not to argue but if I buy one ticket with each of the 42 pb numbers, each ticket has a 1 in 3,478,761 chance of winning the jackpot.

But  that if I buy 100 tickets I have 100 chances in 146,107,962 possible outcomes and not 1 chance out of every 1,461,079.62 combinations.

Indiana
United States
Member #48725
January 7, 2007
1958 Posts
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 Posted: September 1, 2007, 10:10 pm - IP Logged

Here it is, you have to try and conceptualize differently and focus:

Actually when you play the lottery it is this:  It helps if you turn your calculator off.

5/146,107,962 = 1/146,107,962 + 1/146,107,962 + 1/146,107,962 + 1/146,107,962 + 1/146,107,962 .

Not this:

5/146,107,962 = 1/29,221,592.4

In which 29,221,592.4 tickets do you put each of your 5 tickets?

Are you saying I'm wrong? It's fractions man! You don't get it! If you want to change the numerator of a fraction, then you have to change the denominator as well. To change the 5 to a 1, you divide by 5, then you have to divide 146,107,962 by the same amount which is 29,221,592.4.

Gonna win.

Harbinger
D.C./MD.
United States
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July 30, 2006
5587 Posts
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 Posted: September 1, 2007, 10:11 pm - IP Logged

It's not to argue but if I buy one ticket with each of the 42 pb numbers, each ticket has a 1 in 3,478,761 chance of winning the jackpot.

But  that if I buy 100 tickets I have 100 chances in 146,107,962 possible outcomes and not 1 chance out of every 1,461,079.62 combinations.

No argument, it is an excellent approach.

Harbinger
D.C./MD.
United States
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July 30, 2006
5587 Posts
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 Posted: September 1, 2007, 10:26 pm - IP Logged

Are you saying I'm wrong? It's fractions man! You don't get it! If you want to change the numerator of a fraction, then you have to change the denominator as well. To change the 5 to a 1, you divide by 5, then you have to divide 146,107,962 by the same amount which is 29,221,592.4.

Mathematically you are correct, you know fractions. But we are talking lottery,  the denominator doesn't change until the numerator begins approaching the denominator.

You are not applying those 5 numbers to five unique sets of possibilities, you are not covering 5 sets of 29,22,592.4 possibilities uniquely,  again I will ask differently:

What set of  29,22,592.4 possibilities are you applying each of your 5 tickets to the draw that has a total of 146,107,962 possibilities?

It is hard to see at first but think: LOTTERY.

Indiana
United States
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January 7, 2007
1958 Posts
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 Posted: September 1, 2007, 10:41 pm - IP Logged

Mathematically you are correct, you know fractions. But we are talking lottery,  the denominator doesn't change until the numerator begins approaching the denominator.

You are not applying those 5 numbers to five unique sets of possibilities, you are not covering 5 sets of 29,22,592.4 possibilities uniquely,  again I will ask differently:

What set of  29,22,592.4 possibilities are you applying each of your 5 tickets to the draw that has a total of 146,107,962 possibilities?

It is hard to see at first but think: LOTTERY.

I did not say 5 sets of 29,221,592.4 possibilities. You have 5/146,107,962 OR 1/29,221,592.4 chance of winning the jackpot. You still have 5 sets of 146,107,962 possibilities. I'm just converting it so that the numerator is 1.

Gonna win.

Kentucky
United States
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February 14, 2006
7344 Posts
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 Posted: September 1, 2007, 10:56 pm - IP Logged

Are you saying I'm wrong? It's fractions man! You don't get it! If you want to change the numerator of a fraction, then you have to change the denominator as well. To change the 5 to a 1, you divide by 5, then you have to divide 146,107,962 by the same amount which is 29,221,592.4.

You're saying you have 1 chance to win out of every 29,221.592.4 possible outcomes, but I'm assuming each chance is a different combination and only 1 of those 5 chances can win so why not just compare that 1 chance to win to the 146,107,957 combinations you don't have.

If you want fractions, divide your 5 chances by 146,107,962 and you'll get the exact percentage.

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