Should I run a raffle for the VIR trackday?

[oreo_n2]

it has everything to do with the problem definition. Not the numbers themselves.

and degs... if you dont replace the drawn number, that reduces the set by one and odds go to 1/29... then 1/28 etc... dont keep halving it! (thats assuming no conditional probabilities)

Honcl... i have to think about that, i'll get back to you tonight when i have more time.

[Degsy]

No, the number was replaced each time.

When I was halving it, I was over simplifying it by stating that 2 numbers were then drawn out of a set of 30. It was a tongue-in-cheek analogy anyway. I wasn't serious.

[oreo_n2]

honcl.. you're an evil chicken effer... i couldn't concentrate on work after reading your problem...

If all door choices and problem setup are independant (doubtful in reality, especially since if you dont get it right on your first guess, the game organizer will not show you the door with the prize ... he will rig his display to show you an empty door)

anyways... in simplest forms, with lots of independance assumptions...

it works out your odds of winning by switching are 2/3
odds of winning by not switch 1/3

you should switch.

[oreo_n2]

Originally posted by Degsy
No, the number was replaced each time.

When I was halving it, I was over simplifying it by stating that 2 numbers were then drawn out of a set of 30. It was a tongue-in-cheek analogy anyway. I wasn't serious.

thats ok, my argument was mostly bs... quoting stats i wanted to use to improve my odds... you were right initially... nevermind

baaa.....

we need to stop this discussion... too nerdy for a public forum.

[Honclfibr]

Originally posted by oreo_n2
honcl.. you're an evil chicken effer... i couldn't concentrate on work after reading your problem...

If all door choices and problem setup are independant (doubtful in reality, especially since if you dont get it right on your first guess, the game organizer will not show you the door with the prize ... he will rig his display to show you and empty door)

anyways... in simplest forms, with lots of independance assumptions...

it works out your odds of winning by switching are 2/3
odds of winning by not switch 1/3

you should switch.

Hehe, NEEERD!

If it's any consolation, I also couldn't concentrate at work after trying to work through your problem.

If I am thinking about this right, well, the odds of winning if 16 out of 30 tickets were sold assuming a random set of 16 would be 5/16 = .3125.

Whereas if you assume a random set of 30, the odds would be, for n = 0 to infinity, Sigma((14/30)^n*(5/30)). Which from running through the first few of the series, approaches .3125 as n approaches infinity...

So you're wrong, it is statistically fair I think

But you got the Monty Hall problem right, which is pretty damn commendable.

[Honclfibr]

Originally posted by oreo_n2
thats ok, my argument was mostly bs... quoting stats i wanted to use to improve my odds... you were right initially... nevermind

Fucker, no, you're wrong! I proved you wrong (maybe, assuming I didn't fuck it up)! Don't admit it, that takes all the fun away!

Man, sometimes I miss the squid, he never admitted he was wrong

[oreo_n2]

Originally posted by Honclfibr
Hehe, NEEERD!

If it's any consolation, I also couldn't concentrate at work after trying to work through your problem.

If I am thinking about this right, well, the odds of winning if 16 out of 30 tickets were sold assuming a random set of 16 would be 5/16 = .3125.

Whereas if you assume a random set of 30, the odds would be, for n = 0 to infinity, Sigma((14/30)^n*(5/30)). Which from running through the first few of the series, approaches .3125 as n approaches infinity...

So you're wrong, it is statistically fair I think

Blast You!!!

Its simpler than that...
one drawing would be ideal.

my odds... initially 5/30 banking on 30 tickets being sold (~1/6).
17 tickets sold. for "fairness"... make the drawing for a lesser prize based on the lesser prize sales... so the sample set becomes 17!
then my 5 tickets odds are 5/17... (~1/3) almost doubled my odds!

this would be an example of the kind of math politicians use.


now... the method used for drawing... the random number generator based on a 30 sample set... as long as its really random (each drawing completely independant of all others and a really high seed value), and since numbers are being replaced... actually shows no bias to any of the 30 numbers, as you noted.


i am not the squid, i dont come here to argue... but i do bust balls on a regular basis!

[TTD]

--------------
anyways... in simplest forms, with lots of independance assumptions...

it works out your odds of winning by switching are 2/3
odds of winning by not switch 1/3

you should switch.
-------------

Explain this further, please.

Either here or in an email direct to me.



My brain is not grasping why your odds increase by switching.

[Honclfibr]

It seems to make more intuitive sense if there are 100 doors, with a prize behind only one door.

You pick one, and your odds of being right are 1/100.

Then, the host opens 98 out of the 100 doors to show you that they are empty. There are two doors left, and you are giving a chance to switch or stay. If you were right in the first place, the prize is behidn the door you've picked, otherwise it's behidn the other door. But there's only a 1/100 change that you picked right in the first place, so there's a 99/100 chance that the prize is behind the other door. Therefore, you switch, and that reverses your odds of getting a prize from 1/100 to 99/100.

It's the same thing, only with occurences of 100 replaced with 3, 99 with 2, and 98 with 1. But when it's 3 doors rather than 100, it's harder to see intuitively.