Math Problem! Any Takers?

[Btl_Rkt_Sci]

Lol. You guys are over complicating this.
1 winner out of 300 tickets With Replacement 12 times.

12games * 1/300 odds = 12/300
=6/150
=3/75
=1/25 to win at the beginning of the drawings.

Therefore we can derived our own equation!
Let n= # of games already played

(12-n) / 300 = p(x)

Right, my dissertation were on the odds of winning one prize. And this formula works for of you buy a ticket after drawings started.

11.79/300 are the odds of winning ONE prize. 12/300 or 1/25 are the odds of winning PERIOD (that is any number of prizes). 12-11.79=0.21 are the odds of winning MULTIPLE prizes and 300-12=288 are the odds of winning NO prizes.

 

[Btl_Rkt_Sci]

And really, we're all wrong. Odds are written as A:B where A and B are outcomes, so really

12:288 are the odds of winning
11.79:288.21 are the odds of winning once

 

[jimdana1942]

300/12=25
300x25=7500

1 chance in 7500

 

[Btl_Rkt_Sci]

300/12=25
300x25=7500

1 chance in 7500

It is statistically impossible for your probability to be worse than 1/300

 

[XD-GEM]

42

This.

Or maybe 87, but don't hold me to that one.

 

[Armnhammer]

And really, we're all wrong. Odds are written as A:B where A and B are outcomes, so really

12:288 are the odds of winning

This is the ratio of wins:loses if the numbers werent thrown back into the mix. Every time a number is drawn its 1 in 300 so 12 numbers drawn is 12 out of 300. Which in turn makes your chance of winning one of the 12 prizes 1 in 25 or 4%.

 

[alex]

So an airplane is on a treadmill...

 

[Btl_Rkt_Sci]

This is the ratio of wins:loses if the numbers werent thrown back into the mix. Every time a number is drawn its 1 in 300 so 12 numbers drawn is 12 out of 300. Which in turn makes your chance of winning one of the 12 prizes 1 in 25 or 4%.

Yes, the 4% is the probability, a percentage of the whole. Odds are expressed as a ratio of outcomes so if you are 4% likely to win and 96% likely to lose, 4% probability is written as 4:96 or 1:24 in odds format. It's just a stupid detail I was using to make fun of us, even when we found the formula we didn't answer the question correctly.

 

[Emperor]

Lol. You guys are over complicating this.
1 winner out of 300 tickets With Replacement 12 times.

12games * 1/300 odds = 12/300
=6/150
=3/75
=1/25 to win at the beginning of the drawings.

Therefore we can derived our own equation!
Let n= # of games already played

(12-n) / 300 = p(x)

This makes the most sense so far. In this raffle, there is the opportunity to continue to buy tickets after the drawing has started and some of the prizes are already awarded.

So in your scenario, you are basing your answers on the decreasing number of prizes left, even though you joined the raffle later in the game. Am I understanding that correctly?

 

[Btl_Rkt_Sci]

This makes the most sense so far. In this raffle, there is the opportunity to continue to buy tickets after the drawing has started and some of the prizes are already awarded.

So in your scenario, you are basing your answers on the decreasing number of prizes left, even though you joined the raffle later in the game. Am I understanding that correctly?

See that's where I misunderstood you. You are correct that if you come in later, your odds of winning are lower. I didn't know that you wanted to consider the scenario that someone would come in late though...I thought you were trying to use that as an example that which drawing you were on would impact the outcome of just that drawing.

I've had enough.

 

[Emperor]

Okay, explaining it to me as I was a 5 year old;

At the beginning of the raffle (with all 12 prizes available and only 300 tickets to be sold), the odds of winning a prize are 1 in 25.
rliberty, do you agree with that?

If all 300 tickets were going to be eventually accounted for, and after 6 prizes have been awarded the chances of any of the 300 tickets has a 1 in 150 chance of winning.
Does that make sense?

 

[Emperor]

See that's where I misunderstood you. You are correct that if you come in later, your odds of winning are lower. I didn't know that you wanted to consider the scenario that someone would come in late though...I thought you were trying to use that as an example that which drawing you were on would impact the outcome of just that drawing.

I've had enough.

Puss!

 

[Emperor]

There really is a valid reason that I would like to know this; and I will share it with you all when I get this answer. And I WILL get this answer from a mathematician eventually, but in the interim and more specifically; here is what I would like to ascertain from any math whiz, professor, lottery guru, or physicist, on here;

There will be only 300 tickets to the raffle.
There will be 12 items up for grabs in the raffle (let's say 1 drawing a day for 12 days).
Every ticket has the potential to win all 12 items (meaning each ticket that wins, can continue on to win mutliple times).
What I need to know:

The odds of winning at the start of the raffle (Done) 1/25
The odds of winning at each stage of the raffle if you bought a ticket after a prize or prizes were already awarded. In other words, if there were 1 ticket left after 6 of the prizes were awarded; what is my odds of winning at that time if I bought that last ticket?

If I knew this answer, I would not even consider asking this board. My mind cannot compute this crap any longer. I can tell you how much wine I have in my glass right now, but this math is 20 years past my prime.

 

[Storm52]

Where you "f" things up are by adding tickets after the raffle begins. If you have 100 tickets sold at the beginning there are 100 chances of a win the first pull. The 100 ticket holders have a 1 in 100 chance of winning. The next pull you still have the 100 chances but you add another 50 tickets or 50 chances. Holders have a 1/150 chance of winning...etc, etc.

 

[Armnhammer]

How much are the tickets? Lol

 

[returningliberty]

There really is a valid reason that I would like to know this; and I will share it with you all when I get this answer. And I WILL get this answer from a mathematician eventually, but in the interim and more specifically; here is what I would like to ascertain from any math whiz, professor, lottery guru, or physicist, on here;

There will be only 300 tickets to the raffle.
There will be 12 items up for grabs in the raffle (let's say 1 drawing a day for 12 days).
Every ticket has the potential to win all 12 items (meaning each ticket that wins, can continue on to win mutliple times).
What I need to know:

The odds of winning at the start of the raffle (Done) 1/25
The odds of winning at each stage of the raffle if you bought a ticket after a prize or prizes were already awarded. In other words, if there were 1 ticket left after 6 of the prizes were awarded; what is my odds of winning at that time if I bought that last ticket?

If I knew this answer, I would not even consider asking this board. My mind cannot compute this crap any longer. I can tell you how much wine I have in my glass right now, but this math is 20 years past my prime.

Ok, I'm not fully understanding your question, BUT:
The probability of a successful ticket is only determinate when all the tickets are sold for a particular drawing, meaning:
if you sell 200 tickets for the first game, the odds would be 1/200. if you Then sold the Other 100 for the next day's drawing (and any drawing thereafter, because you cant Unsell a ticket), the odds would go Down to 1/300; unless of course you Set the odds at 1/300 regardless (by entering any unsold tickets into the drawing, which is kind of mean, imo).

Since it's a simple probability with replacement, you can have a very easy equation:
N= total # of games. n= games already played
x=# of tickets you bought
Y= # of tickets sold total @ time of particular game.
u = # of prizes to win (by you)
p(x) = { [ x * (N-n) ] / Y } ^ u

This gives you the probability of winning U prizes.

It is really late though, so my math may be off, but i'm thinking that's the correct equation. Yay physics majors!

 

[Request Dust Off]

What are the odds of someone asking a questions on here, getting the correct answer, not believing it and then changing the original question?

 

[Storm52]

40 to 1

 

[JadeRaven]

If there are X tickets, and Y drawings of those X tickets. . .

For each drawing each ticket has a one in X chance of winning.

If there are Y drawings left then you have Y opportunities at that one in X chance.

It's that simple.

 

[JadeRaven]

What are the odds of someone asking a questions on here, getting the correct answer, not believing it and then changing the original question?

Big.