Big Giveaway Maths

[marcownz747]

I just did some looking around, and found that there are 45 possible prizes for the 2015 Big Giveaway. There are also 288 pages of entries (When you display at 10 posts per page). Assuming that Austin has finished entering mail in entries, from mere we could infer the number of entries.

Assuming that there are NO repeats, deleted posts, or non-members saying "I wish I could do this", or members whose memberships expired between the posting and the drawing, there would be 2880 posts.

Divide 2880 entries by the 45 drawings, and you get 64.

This means that there is AT LEAST a 1 in 64 chance that you will take home a prize, if you entered and still hold a membership at the time of the drawing.


Them's some pretty nice odds.

Best of luck to everybody!

 

[motobrewer]

miller...64?

 

[treacheroustexan]

Good luck all.

 

[Psylocide]

Fingers crossed for the fast rack wine combo set.

 

[brewprint]

Those are good odds. When is the drawing?

 

[treacheroustexan]

Those are good odds. When is the drawing?

Not sure, Austins cleaning the thread up now. Sometime later today!

 

[Psylocide]

No idea... anyone know?

 

[brewprint]

LOL didn't even notice.

 

[IslandLizard]

1/64

Those odds are way better than any previous drawing I've seen. Not just here on HBT.

Still realize though, the probability of winning is 1 in 2880 in each of the 45 times a number is drawn...

 

[bacgabe]

Good point IslandLizard, does 45 shots at 1 in 2880 odds equate to 1 in 64 odds? Been too long since I took statistics. Either way, not bad for the swag. Thanks marcownz747 for the number crunching.

 

[IslandLizard]

Good point IslandLizard, does 45 shots at 1 in 2880 odds equate to 1 in 64 odds? Been too long since I took statistics. Either way, not bad for the swag. Thanks marcownz747 for the number crunching.

Most statisticians will simply reduce 45/2880 to 1/64 without a 2nd thought, but I tend to differ. For example, look at Poisson's theories.

 

[ZebulonBrewer]

Soooooo, in reality we would have 1/2880 for the first, 1/2879 for the second, and so on (assuming you can't win twice!). Give me a minute and I'll check the maths.

 

[orangehero]

You realize the posts are numbered, right?

There's a number in the upper right corner of each post. That's also how you should post links to specific posts as it will work for everyone regardless of posts per page settings.

There are 2873 entries.

 

[max384]

Good luck all.

Screw that! Good luck to me!!!!


:rockin:

 

[ZebulonBrewer]

Ok. Stay with me. This article has a pretty good explanation

http://math.stackexchange.com/questions/91998/probability-of-winning-a-prize-in-a-raffle

I'm using the easier equation, the one that says 'assume that the prizes are drawn with replacement'

In that case, the math is based on the probabiity of losing, as in, you have 2779/2880 chance of losing the first drawing. Over 45 drawings, it approximates to -> P-loss = (2779/2880)^45 = 0.9845

We all have a 98.5% chance of losing! yaaaaay. Probability of winning is (1-0.9845)*100 = 1.55%

How close is this to the initial 1/64?

1/64 = 0.0156 .......or, very very close.

 

[SnakeRidge]

look at Poisson's theories.

Look, but don't make eye contact.

 

[Psylocide]

Well, if one's name is drawn, does it go back into the pool? Or are they taken out?

That would come into play in this scenario.

 

[kenoglass]

the odds are better this year last there were over 3700 entries.... Good Luck!!!

 

[Thedutchtouch]

No, it doesnt. After the first prize is drawn, theres 2879 possible winners so the odds are always 1 in 2880ish not 1 in 64.

 

[max384]

No, it doesnt. After the first prize is drawn, theres 2879 possible winners so the odds are always 1 in 2880ish not 1 in 64.

Those are the individual odds for each drawing. Those are NOT the odds of winning any prize in the contest overall, which is what the OP was referring to.

 

[firerat]

Those are the individual odds for each drawing. Those are NOT the odds of winning any prize in the contest overall, which is what the OP was referring to.

Correct!!

You have 1/2880ish odds of winning a SPECIFIC prize.

You have 1/64 chance of winning ANY of the prizes.

Right? I hope I'm right. I think I'm right. But I', married so I'm never actually right.

 

[Hello]

Correct!!

You have 1/2880ish odds of winning a SPECIFIC prize.

You have 1/64 chance of winning ANY of the prizes.

Right? I hope I'm right. I think I'm right. But I', married so I'm never actually right.

Pretty sure that is correct.

 

[BGBC]

You are also all assuming that everyone claims their prizes, which never seems to happen (remarkably).

I'll take a shot in the dark and say that there will be 7 add'l names drawn before everything is claimed.

 

[IslandLizard]

I wouldn't be surprised if Austin plugs into the generator:

Pull 45 random numbers from a pool of 2880-ish, whatever the number of unique qualifying entrants.​

There won't be doubles and the deck remains the same for all 45 prizes.

 

[firerat]

Pretty sure that is correct.

Which part, my math or the fact that I'm never right?

 

[Frilock]

Which part, my math or the fact that I'm never right?

Yes.

 

[BGBC]

I wouldn't be surprised if Austin plugs into the generator:

Pull 45 random numbers from a pool of 2880-ish, whatever the number of unique qualifying entrants.​

There won't be doubles and the deck remains the same for all 45 prizes.

By saying there can't be duplicates, the deck cannot remain the same for all 45 pulls.

 

[IslandLizard]

By saying there can't be duplicates, the deck cannot remain the same for all 45 pulls.

Yes it can.

 

[BGBC]

Yes it can.

I'd like to hear your explanation.

Think of a literal deck of cards. Even if you pull multiple cards at one time (the top 10 cards in the deck, for example), it's basically the same as pulling them one after the other (and not replacing them after). So, after each pull, the total number of cards decreases.

 

[TheCADJockey]

You lost me at Maths.