New York United States Member #103600 January 4, 2011 3988 Posts Offline

Posted: January 4, 2011, 12:14 pm - IP Logged

I am fascinated by the reduction of odds discussion. Each purchased single ticket has the same odds- 1 in 175+ million. If 13 tickets are purchased, the odds of winning are 13 in 175+ million. The reduction of this relationship on a fraction basis is 13 divided by 175 million, not 175 million divided by 13. Expressed as a percentage, 13 tickets have a 0.000007% chance of winning versus the odds of 0.000006% (decimal point moved one more space to the left) on a single ticket.

New York United States Member #103600 January 4, 2011 3988 Posts Offline

Posted: January 4, 2011, 12:26 pm - IP Logged

Having only registered today as a result of this discussion I apologize for taking a counter position to Todd, who truly has developed a meaningful website for lottery game players. The correct phrasing of the 1 purchased ticket out of 175+ million argument presented to counter CoinToss versus 13 purchased tickets is not from 1 in 175 million to 1 in ____________, rather it is from 1 in 175+ million to "13" in 175+ million. Makes sense in my mind- not sure it has translated well in written form.

New Jersey United States Member #1 May 31, 2000 23261 Posts Online

Posted: January 4, 2011, 12:31 pm - IP Logged

Please bear with me; I am going to post a lengthy description of odds calculations for multiple ticket purchases. Please follow the logic, and don't get frustrated.

Some folks here are having a difficult time with the mathematics of odds calculations. My guess is that dealing with huge numbers is causing a problem for them.

Maybe if we talk about small numbers, things will become clear.

After all, the subject of mathematics works the same way no matter how big or small the numbers are.

For example, 1 + 1 = 2, and 10 + 10 = 20, and 1,000,000 + 1,000,000 = 2,000,000. The same concept works, no matter how big the numbers are. It works the same for addition, subtraction, multiplication, and division.

A Small Example

Now then, let's take our odds calculation into the realm of "tens" instead of "millions" of calculations, and then we'll ramp up the discussion back to millions once we have the concepts down.

Let's say our lottery has a total of 10 numbers being selected from a drum, and you buy one ticket. Your odds of winning on that ticket are 1 in 10. I think everyone can agree with that statement.

It is the same as saying you have covered one-tenth of the possible combinations. In other words, an odds statement is a fraction — "one over 10", or 1/10. You can do the division (1 divided by 10) to say I have a "0.1" chance of winning. And as we also know, decimals can be expressed as percentages by simply moving the decimal place two positions to the right. So our "0.1" chance of winning becomes 10%.

That is all easy to understand when dealing with small numbers, especially tens. We have expressed our odds of winning in four different ways:

1 in 10 (a traditional odds statement)

1/10 (a fraction)

0.1 (a decimal)

10% (a percentage — we have covered 10% of the possible combinations)

All of those four ways should be easily and quickly agreed upon by all. The concept of 1 in 10 chances is very easy to understand and express.

Buying More Tickets

OK, so we buy 2 tickets for the drawing. Let's express those two tickets in the same four ways as we did for our 1-ticket purchase:

2 in 10

2/10

0.2

20% — we have covered 20% of the possible combinations

Just to be clear: We do NOT think of our lottery drawing as two separate "1 in 10" odds, because we would have to conduct two separate drawings for that to be true. We are still only conducting one drawing, but now two combinations are covered out of the possible 10 combinations.

Another way to look at it is visually. You can see how our two purchased numbers represents 20% of the possible combinations, and the ones we have NOT purchased is 80% of the combinations. Our math checks out perfectly.

Purchased

20%

Not purchased

80%

1

2

3

4

5

6

7

8

9

10

Expressing as "1 in ____"

One of the points I made previously was that all odds calculations can be expressed as "1 in _____", meaning you have "one chance in _______ (some number)" to win. We all know it's possible to express odds that way, because that is really the only way the lottery EVER speaks about odds, no matter how complicated the game is to play.

A great example of this is the Canada Lotto Max game. For $5, you get a ticket with 3 separate lines (combinations) on it. You get to pick your own numbers for the first line, but the second two lines are always quick picks. Regardless of the way the numbers are selected, each $5 purchase gets you the equivalent of buying three tickets. So how does the lottery express the odds for that 3-ticket purchase for $5? As "1 in _____" odds. They don't say "you have three separate chances of "1 in ____", they combine the three combinations and accurately tell you what your "1 in ____" odds of winning are. You can see it for yourself in the Prizes section at http://www.olg.ca/lotteries/games/howtoplay.do?game=lottomax.

Getting back to my little lottery drawing example here, we can see that when I bought a second ticket above, I expressed the odds as "2 in 10". So what is the "1 in _____" become?

That's pretty easy to calculate. Since one of the four ways to express the odds is a fraction, we can just simplify the fraction — something taught in elementary mathematics.

Simplifying our 2/10 fraction above, we divide the top and bottom of the fraction by the lowest common denominator, which is 2, and our new fraction becomes 1/5.

Now our four methods of expressing the 2-ticket purchase become:

1 in 5

1/5

0.2

20%

Notice something here! Although we have changed the first two expressions, the last two have stayed the same.

How is that possible? Because reducing a fraction does not change its value, it only changes the expression of it. 1/5 is the same number as 2/10. They both equate to 20%.

So hopefully the skeptics here will see that there is no "magic fraction reduction", or anything strange going on. We have not made the game somehow easier to win — all we have done is accurately reflect the mathematical way of stating of chances of winning.

When expressed visually, you can see how saying "1 in 5" instead of "2 in 10" does not magically make the game easier to win. It is still a 20% chance of winning.

Purchased

20%

Not purchased

80%

1

2

3

4

5

6

7

8

9

10

Purchased

20%

Chances of not purchased

80%

1

2

3

4

5

More Tickets

Now, let's buy 5 tickets for the drawing. The numbers become:

1 in 2 (or, 5 in 10 — same thing)

1/2 (or, 5/10 — same thing)

0.5

50%

Visually, it's easy to see this is correct:

Purchased

50%

Not purchased

50%

1

2

3

4

5

6

7

8

9

10

Scaling the Concept to Mega Millions

There is nothing magical about calculating the odds of multiple ticket purchases for Mega Millions. The same exact mathematical calculations are used, no matter how many combinations there are.

To review a 1-ticket purchase of Mega Millions, our four ways of expressing the odds are:

1 in 175,711,536

1/175,711,536

0.00000000569114597006311

0.000000569114597006311%

Wow, that's a pretty small chance of winning. Look at that percentage!

So let's use the example discussed in this thread of purchasing 13 tickets. Here's the new odds:

1 in 13,516,272 (same thing as 13 in 175,711,536)

1/13,516,272 (same thing as 13/175,711,536)

0.0000000739848976108205

0.00000739848976108205%

If this is where I lose you, then let's step back to the discussion above where I mentioned the Canada Lotto Max game. In that game you get 3 chances (tickets) per purchase, and the lottery expresses the odds as a "1 in ____" number. So let's see what the Mega Millions odds would be with a 3-ticket purchase.

Because Mega Millions has a larger number matrix than Lotto Max, we know that the odds must be steeper to win.

In other words, the Ontario Lottery has published the Lotto Max odds of winning the jackpot as 1 in 28,633,528, so our 3-ticket Mega Millions purchase, if calculated correctly, will have WORSE odds than that.

And here is the Mega Millions calculation for 3 tickets:

1 in 58,570,512 (same as 3 in 175,711,536)

1/58,570,512 (same as 3/175,711,536)

0.0000000170734379101893

0.00000170734379101893%

Well, we can see that our 1 in 58,570,512 chances are definitely worse than Lotto Max's 1 in 28,633,528 chances of winning, when the playing field is leveled by purchasing 3 Mega Millions tickets.

In fact, we would have to purchase 7 Mega Millions tickets in order to achieve approximately the same odds of winning Lotto Max (with the Lotto Max $5 purchase):

1 in 25,101,648 (same as 7 in 175,711,536)

1/25,101,648 (same as 7/175,711,536)

0.0000000398380217904418

0.00000398380217904418%

Summary

I really hope this sheds some light on the discussion of odds calculations for multiple ticket purchases.

I do not intend this information to reinforce some notion that buying multiple tickets somehow make the game much easier to win. In fact, if you think buying 13 tickets and making the odds 1 in 13,516,272 is "much easier to win", then you are deluding yourself. The odds against you are still astronomical either way.

However, one thing I am doing here is being FACTUAL. It IS possible to make your odds BETTER by buying more than one ticket. The mathematics I have described here accurately reflect exactly how much better your odds become. Frankly, to those who do not like the way it is being expressed, then it would be best for you to not think in terms of odds at all, because it will only become upsetting for you. Mathematics is a "black and white" subject. There are not two correct answers here, only the one answer I am giving you.

On the other hand, if you have always been confused by odds for multiple tickets, and this helps clarify the topic, then it is my pleasure to have helped!

Zeta Reticuli Star System United States Member #30470 January 17, 2006 10348 Posts Offline

Posted: January 4, 2011, 12:53 pm - IP Logged

Forget the top matrix for right now, let's look at the Mega ball number.

You choose one of forty -six. That leaves 45 that can beat you. So you say, "OK, I'll play another Mega number and reduce that by half (2/46), now there are only 23 numbers that can beat me".

Wrong. There are 44.

Now go back to the top matrix and apply the same thing.

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.

Indiana United States Member #48725 January 7, 2007 1953 Posts Offline

Posted: January 4, 2011, 12:59 pm - IP Logged

Quote: Originally posted by Coin Toss on January 4, 2011

Forget the top matrix for right now, let's look at the Mega ball number.

You choose one of forty -six. That leaves 45 that can beat you. So you say, "OK, I'll play another Mega number and reduce that by half (2/46), now there are only 23 numbers that can beat me".

Wrong. There are 44.

Now go back to the top matrix and apply the same thing.

No, because if you want to simplify a fraction, you have to change both the numerator and denominator. 2/46 = 1/23. To simplify the fraction, I had to divide both the numerator and denominator by 2.

Sunny part of town United States Member #102877 December 27, 2010 224 Posts Offline

Posted: January 4, 2011, 1:01 pm - IP Logged

Quote: Originally posted by Starr920 on January 4, 2011

******Breaking News*******

Channel 7 News in New York is reporting the jackpot amount is now $355 Million!

WOOOOOOOOOOOOOOOOOOOW there is still 9 more hours to purchase tickets!!! Before its all said and done it might hit $400 millions!!!!! Either i win or i dont!!!!!! I have never seen the mega millions being bumped up this many times, the winners might be some large group of co workers or lucky young me. Im just happy to be here and have this chance to win!!!!!!! I would not mind winning $250,000 2nd place!!!!!!!!!!!!!!!!!!!

New Jersey United States Member #1 May 31, 2000 23261 Posts Online

Posted: January 4, 2011, 1:01 pm - IP Logged

Quote: Originally posted by Coin Toss on January 4, 2011

Forget the top matrix for right now, let's look at the Mega ball number.

You choose one of forty -six. That leaves 45 that can beat you. So you say, "OK, I'll play another Mega number and reduce that by half (2/46), now there are only 23 numbers that can beat me".

Wrong. There are 44.

Now go back to the top matrix and apply the same thing.

Your logic is poor, and once again, despite the fact that I took great pains to explain things clearly, you have completely ignored -- or misstated -- what I said.

Where do you see that the Mega Ball would have only 23 number to beat?

Buying the second one would give you 2 chances in 46, or 1 chance in 23. It means out of every 23 draws, one of your two numbers should hit. It does not mean you've eliminated 23 possible numbers.

The Ville, FL United States Member #95879 August 19, 2010 1708 Posts Offline

Posted: January 4, 2011, 1:12 pm - IP Logged

Quote: Originally posted by Todd on January 4, 2011

Your logic is poor, and once again, despite the fact that I took great pains to explain things clearly, you have completely ignored -- or misstated -- what I said.

Where do you see that the Mega Ball would have only 23 number to beat?

Buying the second one would give you 2 chances in 46, or 1 chance in 23. It means out of every 23 draws, one of your two numbers should hit. It does not mean you've eliminated 23 possible numbers.

United States Member #31636 February 1, 2006 256 Posts Offline

Posted: January 4, 2011, 1:23 pm - IP Logged

Quote: Originally posted by Todd on January 4, 2011

Please bear with me; I am going to post a lengthy description of odds calculations for multiple ticket purchases. Please follow the logic, and don't get frustrated.

Some folks here are having a difficult time with the mathematics of odds calculations. My guess is that dealing with huge numbers is causing a problem for them.

Maybe if we talk about small numbers, things will become clear.

After all, the subject of mathematics works the same way no matter how big or small the numbers are.

For example, 1 + 1 = 2, and 10 + 10 = 20, and 1,000,000 + 1,000,000 = 2,000,000. The same concept works, no matter how big the numbers are. It works the same for addition, subtraction, multiplication, and division.

A Small Example

Now then, let's take our odds calculation into the realm of "tens" instead of "millions" of calculations, and then we'll ramp up the discussion back to millions once we have the concepts down.

Let's say our lottery has a total of 10 numbers being selected from a drum, and you buy one ticket. Your odds of winning on that ticket are 1 in 10. I think everyone can agree with that statement.

It is the same as saying you have covered one-tenth of the possible combinations. In other words, an odds statement is a fraction — "one over 10", or 1/10. You can do the division (1 divided by 10) to say I have a "0.1" chance of winning. And as we also know, decimals can be expressed as percentages by simply moving the decimal place two positions to the right. So our "0.1" chance of winning becomes 10%.

That is all easy to understand when dealing with small numbers, especially tens. We have expressed our odds of winning in four different ways:

1 in 10 (a traditional odds statement)

1/10 (a fraction)

0.1 (a decimal)

10% (a percentage — we have covered 10% of the possible combinations)

All of those four ways should be easily and quickly agreed upon by all. The concept of 1 in 10 chances is very easy to understand and express.

Buying More Tickets

OK, so we buy 2 tickets for the drawing. Let's express those two tickets in the same four ways as we did for our 1-ticket purchase:

2 in 10

2/10

0.2

20% — we have covered 20% of the possible combinations

Just to be clear: We do NOT think of our lottery drawing as two separate "1 in 10" odds, because we would have to conduct two separate drawings for that to be true. We are still only conducting one drawing, but now two combinations are covered out of the possible 10 combinations.

Another way to look at it is visually. You can see how our two purchased numbers represents 20% of the possible combinations, and the ones we have NOT purchased is 80% of the combinations. Our math checks out perfectly.

Purchased

20%

Not purchased

80%

1

2

3

4

5

6

7

8

9

10

Expressing as "1 in ____"

One of the points I made previously was that all odds calculations can be expressed as "1 in _____", meaning you have "one chance in _______ (some number)" to win. We all know it's possible to express odds that way, because that is really the only way the lottery EVER speaks about odds, no matter how complicated the game is to play.

A great example of this is the Canada Lotto Max game. For $5, you get a ticket with 3 separate lines (combinations) on it. You get to pick your own numbers for the first line, but the second two lines are always quick picks. Regardless of the way the numbers are selected, each $5 purchase gets you the equivalent of buying three tickets. So how does the lottery express the odds for that 3-ticket purchase for $5? As "1 in _____" odds. They don't say "you have three separate chances of "1 in ____", they combine the three combinations and accurately tell you what your "1 in ____" odds of winning are. You can see it for yourself in the Prizes section at http://www.olg.ca/lotteries/games/howtoplay.do?game=lottomax.

Getting back to my little lottery drawing example here, we can see that when I bought a second ticket above, I expressed the odds as "2 in 10". So what is the "1 in _____" become?

That's pretty easy to calculate. Since one of the four ways to express the odds is a fraction, we can just simplify the fraction — something taught in elementary mathematics.

Simplifying our 2/10 fraction above, we divide the top and bottom of the fraction by the lowest common denominator, which is 2, and our new fraction becomes 1/5.

Now our four methods of expressing the 2-ticket purchase become:

1 in 5

1/5

0.2

20%

Notice something here! Although we have changed the first two expressions, the last two have stayed the same.

How is that possible? Because reducing a fraction does not change its value, it only changes the expression of it. 1/5 is the same number as 2/10. They both equate to 20%.

So hopefully the skeptics here will see that there is no "magic fraction reduction", or anything strange going on. We have not made the game somehow easier to win — all we have done is accurately reflect the mathematical way of stating of chances of winning.

When expressed visually, you can see how saying "1 in 5" instead of "2 in 10" does not magically make the game easier to win. It is still a 20% chance of winning.

Purchased

20%

Not purchased

80%

1

2

3

4

5

6

7

8

9

10

Purchased

20%

Chances of not purchased

80%

1

2

3

4

5

More Tickets

Now, let's buy 5 tickets for the drawing. The numbers become:

1 in 2 (or, 5 in 10 — same thing)

1/2 (or, 5/10 — same thing)

0.5

50%

Visually, it's easy to see this is correct:

Purchased

50%

Not purchased

50%

1

2

3

4

5

6

7

8

9

10

Scaling the Concept to Mega Millions

There is nothing magical about calculating the odds of multiple ticket purchases for Mega Millions. The same exact mathematical calculations are used, no matter how many combinations there are.

To review a 1-ticket purchase of Mega Millions, our four ways of expressing the odds are:

1 in 175,711,536

1/175,711,536

0.00000000569114597006311

0.000000569114597006311%

Wow, that's a pretty small chance of winning. Look at that percentage!

So let's use the example discussed in this thread of purchasing 13 tickets. Here's the new odds:

1 in 13,516,272 (same thing as 13 in 175,711,536)

1/13,516,272 (same thing as 13/175,711,536)

0.0000000739848976108205

0.00000739848976108205%

If this is where I lose you, then let's step back to the discussion above where I mentioned the Canada Lotto Max game. In that game you get 3 chances (tickets) per purchase, and the lottery expresses the odds as a "1 in ____" number. So let's see what the Mega Millions odds would be with a 3-ticket purchase.

Because Mega Millions has a larger number matrix than Lotto Max, we know that the odds must be steeper to win.

In other words, the Ontario Lottery has published the Lotto Max odds of winning the jackpot as 1 in 28,633,528, so our 3-ticket Mega Millions purchase, if calculated correctly, will have WORSE odds than that.

And here is the Mega Millions calculation for 3 tickets:

1 in 58,570,512 (same as 3 in 175,711,536)

1/58,570,512 (same as 3/175,711,536)

0.0000000170734379101893

0.00000170734379101893%

Well, we can see that our 1 in 58,570,512 chances are definitely worse than Lotto Max's 1 in 28,633,528 chances of winning, when the playing field is leveled by purchasing 3 Mega Millions tickets.

In fact, we would have to purchase 7 Mega Millions tickets in order to achieve approximately the same odds of winning Lotto Max (with the Lotto Max $5 purchase):

1 in 25,101,648 (same as 7 in 175,711,536)

1/25,101,648 (same as 7/175,711,536)

0.0000000398380217904418

0.00000398380217904418%

Summary

I really hope this sheds some light on the discussion of odds calculations for multiple ticket purchases.

I do not intend this information to reinforce some notion that buying multiple tickets somehow make the game much easier to win. In fact, if you think buying 13 tickets and making the odds 1 in 13,516,272 is "much easier to win", then you are deluding yourself. The odds against you are still astronomical either way.

However, one thing I am doing here is being FACTUAL. It IS possible to make your odds BETTER by buying more than one ticket. The mathematics I have described here accurately reflect exactly how much better your odds become. Frankly, to those who do not like the way it is being expressed, then it would be best for you to not think in terms of odds at all, because it will only become upsetting for you. Mathematics is a "black and white" subject. There are not two correct answers here, only the one answer I am giving you.

On the other hand, if you have always been confused by odds for multiple tickets, and this helps clarify the topic, then it is my pleasure to have helped!

Good luck in tonight's big drawing!

Todd,

Very good post. Odds are just hype. What kind of percent you have to win is all that matters.

NEW YORK United States Member #90535 April 29, 2010 11978 Posts Offline

Posted: January 4, 2011, 2:05 pm - IP Logged

Quote: Originally posted by Todd on January 4, 2011

Please bear with me; I am going to post a lengthy description of odds calculations for multiple ticket purchases. Please follow the logic, and don't get frustrated.

Some folks here are having a difficult time with the mathematics of odds calculations. My guess is that dealing with huge numbers is causing a problem for them.

Maybe if we talk about small numbers, things will become clear.

After all, the subject of mathematics works the same way no matter how big or small the numbers are.

For example, 1 + 1 = 2, and 10 + 10 = 20, and 1,000,000 + 1,000,000 = 2,000,000. The same concept works, no matter how big the numbers are. It works the same for addition, subtraction, multiplication, and division.

A Small Example

Now then, let's take our odds calculation into the realm of "tens" instead of "millions" of calculations, and then we'll ramp up the discussion back to millions once we have the concepts down.

Let's say our lottery has a total of 10 numbers being selected from a drum, and you buy one ticket. Your odds of winning on that ticket are 1 in 10. I think everyone can agree with that statement.

It is the same as saying you have covered one-tenth of the possible combinations. In other words, an odds statement is a fraction — "one over 10", or 1/10. You can do the division (1 divided by 10) to say I have a "0.1" chance of winning. And as we also know, decimals can be expressed as percentages by simply moving the decimal place two positions to the right. So our "0.1" chance of winning becomes 10%.

That is all easy to understand when dealing with small numbers, especially tens. We have expressed our odds of winning in four different ways:

1 in 10 (a traditional odds statement)

1/10 (a fraction)

0.1 (a decimal)

10% (a percentage — we have covered 10% of the possible combinations)

All of those four ways should be easily and quickly agreed upon by all. The concept of 1 in 10 chances is very easy to understand and express.

Buying More Tickets

OK, so we buy 2 tickets for the drawing. Let's express those two tickets in the same four ways as we did for our 1-ticket purchase:

2 in 10

2/10

0.2

20% — we have covered 20% of the possible combinations

Just to be clear: We do NOT think of our lottery drawing as two separate "1 in 10" odds, because we would have to conduct two separate drawings for that to be true. We are still only conducting one drawing, but now two combinations are covered out of the possible 10 combinations.

Another way to look at it is visually. You can see how our two purchased numbers represents 20% of the possible combinations, and the ones we have NOT purchased is 80% of the combinations. Our math checks out perfectly.

Purchased

20%

Not purchased

80%

1

2

3

4

5

6

7

8

9

10

Expressing as "1 in ____"

One of the points I made previously was that all odds calculations can be expressed as "1 in _____", meaning you have "one chance in _______ (some number)" to win. We all know it's possible to express odds that way, because that is really the only way the lottery EVER speaks about odds, no matter how complicated the game is to play.

A great example of this is the Canada Lotto Max game. For $5, you get a ticket with 3 separate lines (combinations) on it. You get to pick your own numbers for the first line, but the second two lines are always quick picks. Regardless of the way the numbers are selected, each $5 purchase gets you the equivalent of buying three tickets. So how does the lottery express the odds for that 3-ticket purchase for $5? As "1 in _____" odds. They don't say "you have three separate chances of "1 in ____", they combine the three combinations and accurately tell you what your "1 in ____" odds of winning are. You can see it for yourself in the Prizes section at http://www.olg.ca/lotteries/games/howtoplay.do?game=lottomax.

Getting back to my little lottery drawing example here, we can see that when I bought a second ticket above, I expressed the odds as "2 in 10". So what is the "1 in _____" become?

That's pretty easy to calculate. Since one of the four ways to express the odds is a fraction, we can just simplify the fraction — something taught in elementary mathematics.

Simplifying our 2/10 fraction above, we divide the top and bottom of the fraction by the lowest common denominator, which is 2, and our new fraction becomes 1/5.

Now our four methods of expressing the 2-ticket purchase become:

1 in 5

1/5

0.2

20%

Notice something here! Although we have changed the first two expressions, the last two have stayed the same.

How is that possible? Because reducing a fraction does not change its value, it only changes the expression of it. 1/5 is the same number as 2/10. They both equate to 20%.

So hopefully the skeptics here will see that there is no "magic fraction reduction", or anything strange going on. We have not made the game somehow easier to win — all we have done is accurately reflect the mathematical way of stating of chances of winning.

When expressed visually, you can see how saying "1 in 5" instead of "2 in 10" does not magically make the game easier to win. It is still a 20% chance of winning.

Purchased

20%

Not purchased

80%

1

2

3

4

5

6

7

8

9

10

Purchased

20%

Chances of not purchased

80%

1

2

3

4

5

More Tickets

Now, let's buy 5 tickets for the drawing. The numbers become:

1 in 2 (or, 5 in 10 — same thing)

1/2 (or, 5/10 — same thing)

0.5

50%

Visually, it's easy to see this is correct:

Purchased

50%

Not purchased

50%

1

2

3

4

5

6

7

8

9

10

Scaling the Concept to Mega Millions

There is nothing magical about calculating the odds of multiple ticket purchases for Mega Millions. The same exact mathematical calculations are used, no matter how many combinations there are.

To review a 1-ticket purchase of Mega Millions, our four ways of expressing the odds are:

1 in 175,711,536

1/175,711,536

0.00000000569114597006311

0.000000569114597006311%

Wow, that's a pretty small chance of winning. Look at that percentage!

So let's use the example discussed in this thread of purchasing 13 tickets. Here's the new odds:

1 in 13,516,272 (same thing as 13 in 175,711,536)

1/13,516,272 (same thing as 13/175,711,536)

0.0000000739848976108205

0.00000739848976108205%

If this is where I lose you, then let's step back to the discussion above where I mentioned the Canada Lotto Max game. In that game you get 3 chances (tickets) per purchase, and the lottery expresses the odds as a "1 in ____" number. So let's see what the Mega Millions odds would be with a 3-ticket purchase.

Because Mega Millions has a larger number matrix than Lotto Max, we know that the odds must be steeper to win.

In other words, the Ontario Lottery has published the Lotto Max odds of winning the jackpot as 1 in 28,633,528, so our 3-ticket Mega Millions purchase, if calculated correctly, will have WORSE odds than that.

And here is the Mega Millions calculation for 3 tickets:

1 in 58,570,512 (same as 3 in 175,711,536)

1/58,570,512 (same as 3/175,711,536)

0.0000000170734379101893

0.00000170734379101893%

Well, we can see that our 1 in 58,570,512 chances are definitely worse than Lotto Max's 1 in 28,633,528 chances of winning, when the playing field is leveled by purchasing 3 Mega Millions tickets.

In fact, we would have to purchase 7 Mega Millions tickets in order to achieve approximately the same odds of winning Lotto Max (with the Lotto Max $5 purchase):

1 in 25,101,648 (same as 7 in 175,711,536)

1/25,101,648 (same as 7/175,711,536)

0.0000000398380217904418

0.00000398380217904418%

Summary

I really hope this sheds some light on the discussion of odds calculations for multiple ticket purchases.

I do not intend this information to reinforce some notion that buying multiple tickets somehow make the game much easier to win. In fact, if you think buying 13 tickets and making the odds 1 in 13,516,272 is "much easier to win", then you are deluding yourself. The odds against you are still astronomical either way.

However, one thing I am doing here is being FACTUAL. It IS possible to make your odds BETTER by buying more than one ticket. The mathematics I have described here accurately reflect exactly how much better your odds become. Frankly, to those who do not like the way it is being expressed, then it would be best for you to not think in terms of odds at all, because it will only become upsetting for you. Mathematics is a "black and white" subject. There are not two correct answers here, only the one answer I am giving you.

On the other hand, if you have always been confused by odds for multiple tickets, and this helps clarify the topic, then it is my pleasure to have helped!