[Devilboy]

What's the under / over for when the Sin City chips will first arrive? I'll take bets up to $100 for either side depending on the odds.

[Nexttime]

DId they get out sourced to BCC? Because would really make a difference on choosing a date. You might want to add 2010 option.

[aquaman]

I'm gonna go ahead and make my vote public since I don't want to appear as though I'm voting this way since it's a private vote. Not that it'll never happen, I just think post January '09 is not unrealistic.

[jdunford]

Originally Posted by aquaman

I'm gonna go ahead and make my vote public since I don't want to appear as though I'm voting this way since it's a private vote. Not that it'll never happen, I just think post January '09 is not unrealistic.

Well, I voted for that option primarily because it has the largest bin size. In fact, I'd argue that it is "infinitely more probable", given the other bins are finite...
(Histogram - Wikipedia, the free encyclopedia)

[Jonah 99]

funny, after reading through the group buy post today, I was thinking about the over/under as well.

hope things turn out well but geez, having big bucks tied up for a year or more is scary.

[Devilboy]

Originally Posted by jdunford

In fact, I'd argue that it is "infinitely more probable", given the other bins are finite...

Just because the other possibilities are finite in timespan and the last one is not does NOT make them 'infinitely more probable'

Why?

The total probability of all the options added together are 100% (by design). If the first 5 options have a combined probability > 0 then we know that the last option has a probability < 100%

Mathematically:

Per definition:
P(option 1-5) >= 0
P(option 6) >= 0
P(1-5) + P(6) = 100%

Assumption:
P(1-5) > 0 (we assume there is a non-zero chance that chips will arrive this year)

Proof that P(6) is not infinitely more likely than P(1-5):
Proof by counter-example

Let R be how much more likely P(6) is than P(1-5)
thus R = P(6) / P(1-5)

if we set: R = oo (infinity)
then: oo = P(6) / P(1-5)
which means that either
1) P(6) = oo
2) P(1-5) = 0

But we know 1) to be false because P(6) cannot be more than 100% (per definition) and we know 2) to be false (assumption 1)

QED

[AlbinoDragon]

I guess I'm more optimistic than most. I figured 1 year after the original estimated delivery date (for SCS).

That said, I have no direct interest as I didn't order any. I feel for those of you who have money tied up and not much to show for it.

[jdunford]

Originally Posted by Devilboy

...

QED

Nicely done. Fair enough. But I was thinking:

In the limit that P(1-5) -> 0, P(6)/[P(1)+P(2)+P(3)+P(4)+P(5)] -> oo.
As the clock ticks, each of probabilities P(1-5) will shrink towards 0 and this limit will be approached.

[nelohalo]

They'll have plenty to show when those chips arrive

[Devilboy]

Originally Posted by jdunford

Nicely done. Fair enough. But I was thinking:

In the limit that P(1-5) -> 0, P(6)/[P(1)+P(2)+P(3)+P(4)+P(5)] -> oo.
As the clock ticks, each of probabilities P(1-5) will shrink towards 0 and this limit will be approached.

Ah yea of course I forgot to analyse how P(x) varies as time passes.