Math Problem! Any Takers?

[Xenon]

I think the odds are as such:

(300-1) to the 12th power = 0.960725246
300

1-0.960725246 = .03927 or 3.93% numberically - it's 11.79 out of 300, nearly 12 in 300

It you were to remove each ticket drawn, the num and den of the fraction would decrease by 1 each time. If you had two tickets, it would be 300-2. Basically, this formula calculates the risk of losing and subtracts it from 100% for the win percentage

 

[Emperor]

1 in 25? (12 in 300 to the lowest)

I am retracting my support of this because of the multiple items available at the onset versus the the items decreasing as drawings continue, which has to decrease overall odds.

Remember; if you purchase a ticket when all 12 items are available, you have 12 items you have a chance of winning. If you purchase a ticket when say only 6 items were left, obviously you have less of a chance to win, because there are less items for you to have a chance at. You lose out on the benefit of 12 drawings as opposed to 6.

 

[Btl_Rkt_Sci]

It's the same as giving everyone 12 tickets and drawing 1 times from 3600. 12/3600=1/300.

 

[Btl_Rkt_Sci]

I am buying into this^. What if you bought mutiple tickets? Let's say 2.

As long as tickets are still re-entered, probability would double.

 

[Emperor]

As long as tickets are still re-entered, probability would double.

Now you are messing with me!

 

[Btl_Rkt_Sci]

I am retracting my support of this because of the multiple items available at the onset versus the the items decreasing as drawings continue, which has to decrease overall odds.

Remember; if you purchase a ticket when all 12 items are available, you have 12 items you have a chance of winning. If you purchase a ticket when say only 6 items were left, obviously you have less of a chance to win, because there are less items for you to have a chance at. You lose out on the benefit of 12 drawings as opposed to 6.

If one winner is drawn for each drawing and they get a prize, the number of prizes sitting of to the side is irrelevant. If I have 12 boxes of cheerios and my mom tells me to draw a straw but when i win I can only take one box of cheerios, my probability of winning is only determined by the number of straws she has in her hand. So as the amount of boxes of cheerios decreases to 11, if she has the same amount os sticks, same odds, 10, same odds, 9, same odds and so forth.

Think of it as having 12 of the exact same prize and just walking one out for each drawing. The basket with the tickets doesn't know that you have 11 more you'll bring out one at a time. Therefore they cannot impact the odds of winning ANY drawing.

 

[Emperor]

Allright, listen!

I am going home to eat and start drinking. I will come back on later and I expect one of you to have asked a math professor by then; because this is causing my head to hurt.

 

[Emperor]

If one winner is drawn for each drawing and they get a prize, the number of prizes sitting of to the side is irrelevant. If I have 12 boxes of cheerios and my mom tells me to draw a straw but when i win I can only take one box of cheerios, my probability of winning is only determined by the number of straws she has in her hand. So as the amount of boxes of cheerios decreases to 11, if she has the same amount os sticks, same odds, 10, same odds, 9, same odds and so forth.

Think of it as having 12 of the exact same prize and just walking one out for each drawing. The basket with the tickets doesn't know that you have 11 more you'll bring out one at a time. Therefore they cannot impact the odds of winning ANY drawing.

That doesn't add up. It sounds like it's that simple, but it can't be. If 299 of us buy a ticket when there are all 12 items, obvioulsy we have better odds odds of winning one of the items than the 300th person who buys a ticket when 6 of the items are gone. The 299 of us had 6 extra shots at items. #300 will pay the same as us, and only have 6 chances to win. We had 12. There is a difference!

 

[alex]

It's 1 in 300. That's the only way everyone's probability would add up to 1...which is a rule.

This!

 

[MRBULLRED]

It's 1 in 300. That's the only way everyone's probability would add up to 1...which is a rule.

I second 1 in 300...

 

[Btl_Rkt_Sci]

That doesn't add up. It sounds like it's that simple, but it can't be. If 299 of us buy a ticket when there are all 12 items, obvioulsy we have better odds odds of winning one of the items than the 300th person who buys a ticket when 6 of the items are gone. The 299 of us had 6 extra shots at items. #300 will pay the same as us, and only have 6 chances to win. We had 12. There is a difference!

By you said 300 people participate in all 12 drawings! This isn't explained by how many items are gone but by how many drawings remain. Would you agree that if 300 people participated in each drawing with one winner, the odds of winning each drawing is 1/300?

 

[MRBULLRED]

I don't think at any point he mentioned that you could win 12 items. He said each ticket had 12 chances to win. Now I'm really confused..

 

[Btl_Rkt_Sci]

You have to stop thinking about the prizes, they are only a way to keep track of how many drawings are left.

Imagine you have 12 fingers and your hands are wide open. Each time a drawing happens, no awards are given, the person just gets to hear they're name called. Each time a drawing is made, you close one finger and when you have two knuckles you tell everyone to go home. Did the position of your hands have any impact on the number of tickets in the basket or the number of tickets drawn for any of the twelve drawings?

 

[Richard in LA]

This is easy. If I bought the ticket, there's a 0% chance of winning.

 

[Btl_Rkt_Sci]

OK, upon further inspection, Xenon is right. You have to think of it as the equivalent of 12 simultaneous drawings each of which you have a 1/300 chance of winning (299/300 chance of not winning). Because no one drawing has any impact on the other you can think of it as 12 drawings done in parallel. When you deal with parallel systems, you analyze probabilities based on failure rather than success. When the probability of each component is equal, the probability of success (defined as not failing each at least once) is 1-(probablity of falling at each occurrence)^n where n is the number of parallel components.

The reason the 1/300 and all people odds adding up to 1 isn't valid is because the odds of every outcome must add up to one for each individual. Not for the experiment as a whole. If one person bought 200 million power ball tickets they have a 100% chance of winning but that doesn't mean other people can't buy tickets and when those probabilities are totaled they are greater than one, disproving that theory.

But Emperor, it still doesn't matter which drawing they are on. Only if you buy your ticket late.

So 11.79/300 is the answer.

 

[Armnhammer]

12:300 or 1:25

 

[Armnhammer]

Everytime a number is drawn you have a 1 in 300 chance so if their is 12 numbers drawn and each is returned after the drawing your ratio is 12:300. The ratings dont go down every time a number is drawn. If the drawn numbers were left out the mix then it would increase slightly after every draw. ex. 1:300(1st) 1:299(2nd) 1:298(3rd) ect...

 

[LA_Huntsman]

42

 

[returningliberty]

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)

 

[LA_Huntsman]

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)

I was just about to say that...