OT: PA Millionaire Raffle. Questionable math?

[artsandletters]

Rules pertinent to my question:

4. Ticket Characteristics: Each Pennsylvania Millionaire Raffle XX lottery game ticket will contain one chance consisting of one unique computer-generated eight-digit number between 00000001 and 00500000, the drawing date, amount bet, and validation data.

 5. Prizes: The prizes that can be won in this game are $100, $1,000, $100,000 and $1,000,000. A player may only win one time on each ticket or chance.

 6. Maximum Number of Tickets Printed and Sold for the Game: There will be no more than 500,000 tickets printed and sold for the Pennsylvania Millionaire Raffle XX lottery game. The chances will be sequentially issued on a statewide basis from the range of individual unique numbers representing the chances available for the game.

 7. Conduct of Drawing: The results of the Pennsylvania Millionaire Raffle XX will be televised on July 11, 2015, at or about 7:00 p.m. A computer-generated randomizer will be used to conduct the drawing. Six-thousand (6,000) unique eight-digit numbers will be drawn from the range of numbers representing the chances sold. The first four unique eight-digit numbers drawn will be the first-prize-tier winning numbers. The fifth through eighth unique eight-digit numbers drawn will be the second-prize-tier winning numbers. The ninth through 108th unique eight-digit numbers drawn will be the third-prize-tier winning numbers. The 109th through 6,000th unique eight-digit numbers drawn will be the fourth-prize-tier winning numbers.

 8. Determination of Prize Winners:

 (a) Holders of tickets upon which the unique eight-digit number matches exactly one of the first-prize-tier numbers selected by the Lottery shall be entitled to a prize of $1,000,000.

 (b) Holders of tickets upon which the unique eight-digit number matches exactly one of the second-prize-tier numbers selected by the Lottery shall be entitled to a prize of $100,000.

Odds are listed as follows for the million bucks ( 4 winners):

Top Prize 4 prizes of $1,000,000 = odds of 1:125,000 (on PALottery website).

My Question -- Shouldn't the odds for the million bucks be 1 : 500,000 for the first million, 1: 499,999 for the second million and so on? If I'm right, where did they come up with the advertised odds of 1: 125,000 (I understand the 500,000/4 = 125,000) ?

 

[ClarkstonMark]

odds and probability are two different things?

 

[artsandletters]

odds and probability are two different things?

PALottery website says 'ODDS'. Is the probability (chances) 1:500,000 (I'm 42 years removed from school)?

 

[JR4PSU]

odds and probability are two different things?

They are, however, directly linked. If the odds of winning are 1 in 500,000 or 1 to 499,999 ... then the probability of winning is 1/500,000.

More generally:

Probability
P(A) = {Number of Event A} / {Total Number of Events}

Odds
{Number of Event A} to {Number of Events Not A}
where {Number of Event A} + {Number of Events Not A} = {Total Number of Events}

 

[artsandletters]

They are, however, directly linked. If the odds of winning are 1 in 500,000 or 1 to 499,999 ... then the probability of winning is 1/500,000.

More generally:

Probability
P(A) = {Number of Event A} / {Total Number of Events}

Odds
{Number of Event A} to {Number of Events Not A}
where {Number of Event A} + {Number of Events Not A} = {Total Number of Events}

Regarding my original question -- isn't the 1: 125,000 listed by the PALottery incorrect? Shouldn't it be 1: 500,000?

 

[jjsocrates]

Top Prize 4 prizes of $1,000,000 = odds of 1:125,000 (on PALottery website).

My Question -- Shouldn't the odds for the million bucks be 1 : 500,000 for the first million, 1: 499,999 for the second million and so on? If I'm right, where did they come up with the advertised odds of 1: 125,000 (I understand the 500,000/4 = 125,000) ?

The odds quoted (1:125,000) are the odds of having ONE ticket picked in the first 4. There are 4 possible tickets that will be drawn from the 500,000 sold. Therefore, 4 in 500,000 chance.

You're thinking in terms of winning ALL 4 drawings, not just a single drawing.
1:500,000 + 1:499,999 + 1:498,998 + 1,498,997 is your odds of winning all 4 tickets, as opposed to your odds of winning 1 out of the 4 tickets. Which is 1:125,000

This is similar to playing a game of cards and figuring out what your probability of picking an aces from a deck of 52 cards is. 4 aces out of 52 cards is 4/52. (This is NOT the same as picking all 4 aces out of a deck of 52 cards when you draw 4 cards).

Or, if you will, to simplify things: take a deck of 50 cards, write your name on 4 of them. Shuffle and pick 4 cards. Your odds of drawing your name ONCE is 4/50 or 1/12.5 (for simplicity sake, we'll call this 1:12.5 although this isn't the actual calculation).

HOWEVER, if you want to win see what your chances are of getting ALL 4 of the cards you wrote your name on, thats a much higher probability. In that case its (1/50) * (1/49) * (1/48) * (1/47) which is roughly 1/(5.5M)

 

[jjsocrates]

Regarding my original question -- isn't the 1: 125,000 listed by the PALottery incorrect? Shouldn't it be 1: 500,000?

You're confusing these two scenarios...

Probability of having AT LEAST ONE TICKET drawn first in the first four out of 500,000 tickets = (1)/(500,000) + (1)/(499,999) + (1)/(499,998) + (1)/(499,997). This roughly equals 1/125,000.

Probability of having ALL FOUR TICKET drawn first in the first four out of 500,000 tickets = (1)/(500,000) * (1)/(499,999) * (1)/(499,998) * (1)/(499,997). This roughly equals Avogadro's Number.

 

[ClarkstonMark]

OP buys 1 ticket
Probability of winning one of 4 first Prizes is 1/500000+1/499999+1/499998+1/499997
Whatever that adds up to
If I am wrong let me know

 

[LionDeNittany]

OP buys 1 ticket
Probability of winning one of 4 first Prizes is 1/500000+1/499999+1/499998+1/499997
Whatever that adds up to
If I am wrong let me know

You're buying 1 ticket. Not 4. So the 1/499,999 doesn't matter.
4 $1mm prizes
500k total tickets

1/125k

LdN

 

[fairgambit]

Let's cut to the chase. I have bought 2 tickets in every Millionaire Raffle since it started. I can tell you the chances of winning are 0.00 and I have the losing tickets to prove it.

 

[ClarkstonMark]

You're buying 1 ticket. Not 4. So the 1/499,999 doesn't matter.
4 $1mm prizes
500k total tickets

1/125k

LdN

Dude, that is exactly what I said. 1 ticket 4 winners 1/500000+1/499999+1/499998+1/499997=1/125k

 

[LionJim]

Dude, that is exactly what I said. 1 ticket 4 winners 1/500000+1/499999+1/499998+1/499997=1/125k

Not quite. The numbers round off because the 500K is big. Someone else made a comparison to picking an ace out of a deck of cards. The probability of picking an ace out of a deck of cards is 4/52=1/13=.0769. But 1/52 + 1/51 +1/50 +1/49 = .0800. It seems to work for you because the 500K is big, but because 52 is relatively small you can see that the two calculation methods differ.

 

[ralpieE]

Let's cut to the chase. I have bought 2 tickets in every Millionaire Raffle since it started. I can tell you the chances of winning are 0.00 and I have the losing tickets to prove it.

The closest I came was the first drawing. I had all digits in order and was off by 15 on the last 2. Still have the ticky.

 

[Op2]

Re. the question at the end of the OP, here's how it works.

Probability the 1st million dollar ticket of the 500,000 they draw is yours is 1 / 500,000.

Probability the 2nd million dollar ticket of the 500,000 they draw is yours is

(499,999 / 500,000) x (1 / 499,999) = 1 / 500,000

In other words, the probability that the 2nd ticket they draw is yours is the probability that the 1st ticket they drew WASN'T yours (499,999 / 500,000) times the probability that the 2nd ticket they draw is yours (1 / 499,999).

You can't just go straight to the 1 / 499,999 chance of the 2nd ticket drawn being yours because you're ignoring the (small) possibility that the 2nd ticket drawn can't be yours if the 1st ticket drawn was yours.

 

[fairgambit]

The closest I came was the first drawing. I had all digits in order and was off by 15 on the last 2. Still have the ticky.

I don't buy lottery tickets as a rule, but I somehow I got sucked into this one. I've never come remotely close to a millionaire winning number, and have never even won $100.00. Like most lottery players, I keep thinking surely after all the money I've spent, this time I will win something but, of course, it never happens.

 

[LionDeNittany]

Re. the question at the end of the OP, here's how it works.

Probability the 1st million dollar ticket of the 500,000 they draw is yours is 1 / 500,000.

Probability the 2nd million dollar ticket of the 500,000 they draw is yours is

(499,999 / 500,000) x (1 / 499,999) = 1 / 500,000

In other words, the probability that the 2nd ticket they draw is yours is the probability that the 1st ticket they drew WASN'T yours (499,999 / 500,000) times the probability that the 2nd ticket they draw is yours (1 / 499,999).

You can't just go straight to the 1 / 499,999 chance of the 2nd ticket drawn being yours because you're ignoring the (small) possibility that the 2nd ticket drawn can't be yours if the 1st ticket drawn was yours.

But the odds aren't around "1st 1mm ticket" the odds are of A 1mm ticket being drawn.

There are 4 of those. 500k total tickets. 1:125000

 

[Op2]

But the odds aren't around "1st 1mm ticket" the odds are of A 1mm ticket being drawn.

There are 4 of those. 500k total tickets. 1:125000

If you say, there are 500,000 tickets and 4 will be winners, then the chances are 1 in 125,000. But at the end of the OP the poster was going through it one at a time, saying the odds for the first ticket was 1 in 500,000 and the odds for the second ticket were 1 in 499,999 and so he didn't understand why doing it that way resulted the odds of 4 in 500,000. I was noting that the chance that the 1st, 2nd, 3rd or 4th being his were each 1 in 500,000, which comes out to exactly (not approximately) 1 in 125,000 for the four tickets.

The odds that the 1st, 2nd, 3rd and 4th million dollar ticket being your ticket are each 1 in 500,000. And the odds that any of the four tickets are yours is 1 in 125,000.

 

[jjsocrates]

You're buying 1 ticket. Not 4. So the 1/499,999 doesn't matter.
4 $1mm prizes
500k total tickets

1/125k

LdN

This is wrong. You may be buying 1 ticket, but there are MORE THAN ONE tickets being drawn. So after the first ticker is drawn, your chance of winning the second prize is 1/499,999

 

[Op2]

This is wrong. You may be buying 1 ticket, but there are MORE THAN ONE tickets being drawn. So after the first ticker is drawn, your chance of winning the second prize is 1/499,999

The 1 in 499,999 chance of you winning on the 2nd ticket drawn is conditional on you not winning on the 1st ticket drawn.

ETA: To try to be more clear, let me write it this way.

If they do not draw your ticket on the 1st draw then the chance they'll draw it on the 2nd draw is 1 in 499,999.

If they do draw your ticket on the 1st draw then the chance they'll draw it on the 2nd draw is 0.

If you ask before the lottery starts "What is the chance they'll draw my ticket on the 2nd draw?" the answer is 1 in 500,000.

If you ask that same question after the 1st draw then the answer is either 1 in 499,999 or 0, depending on whether they drew your ticket on the first draw.

 

[jjsocrates]

The 1 in 499,999 chance of you winning on the 2nd ticket drawn is conditional on you not winning on the 1st ticket drawn.

ETA: To try to be more clear, let me write it this way.

If they do not draw your ticket on the 1st draw then the chance they'll draw it on the 2nd draw is 1 in 499,999.

If they do draw your ticket on the 1st draw then the chance they'll draw it on the 2nd draw is 0.

If you ask before the lottery starts "What is the chance they'll draw my ticket on the 2nd draw?" the answer is 1 in 500,000.

If you ask that same question after the 1st draw then the answer is either 1 in 499,999 or 0, depending on whether they drew your ticket on the first draw.

The question asks what are your chances of winning one of the top 4 prizes with one ticket? That means, to find that answer, you must sum up all of your chances of that ticket being selected in each of the 4 draws.

(1)/(500,000) + (1)/(499,999) + (1)/(499,998) + (1)/(499,997).

This is NOT a conditional probability statement. You don't know any of the outcomes yet. A conditional probability statement would be: What are the chances of me winning one of the top 4 prizes GIVEN that the first ticket was drawn and it was/wasn't mine.

And it doesn't ask WHICH drawing of the four you won, just your chances of having it drawn in the top 4.

Nevermind, you're going back on ingore where you belong.

 

[Op2]

The question asks what are your chances of winning one of the top 4 prizes with one ticket? That means, to find that answer, you must sum up all of your chances of that ticket being selected in each of the 4 draws.

(1)/(500,000) + (1)/(499,999) + (1)/(499,998) + (1)/(499,997).

This is NOT a conditional probability statement. You don't know any of the outcomes yet. A conditional probability statement would be: What are the chances of me winning one of the top 4 prizes GIVEN that the first ticket was drawn and it was/wasn't mine.

And it doesn't ask WHICH drawing of the four you won, just your chances of having it drawn in the top 4.

Nevermind, you're going back on ingore where you belong.

The unconditional probability that they draw your ticket on the 2nd draw is 1 in 500,000.

The conditional probability that they draw your ticket on the 2nd draw given that they did not draw it on the 1st draw is 1 in 499,999.

The conditional probability that they draw your ticket on the 2nd draw given that they did draw it on the 1st draw is 0.

 

[91Joe95]

It depends how the rules define some.