Mog Bonanza Imminent! (05/08/2009)

[Sev]

If they were paying attention and understood that the Bonanza numbers are 0~99,999 then they'd notice that numbers like 53004, 800, 00000, 99999 are all equally valid and equally likely.

If the conspiracy outlook is correct, then the numbers were chosen to give a certain number of winners, so smart money would have been on things that people "obviously" wouldn't pick, like 00's.

Actually if you have noticed you cant win rank one with a number < 10000, rank two < 1000, rank three with < 100 or rank 4 with <10. The combos have never started with a 0.

[Weeks]

With 10 data points?

[Suiram]

The combos have never started with 5, 6, 7, or 9 either. What's your point?

[Max™]

That would be a good argument for a re-do in the name of fairness if they seriously didn't consider having 0xxxx as a viable Rank 1 winner... except they can just say "fuck you, it's our game, our rules, deal with it or go away."

[Not not Mattaru]

Here ya go: http://i152.photobucket.com/albums/s.../Nightmare.jpg

無事購入できました
ぶじこうにゅう
長らくお騒がせして失礼します
ながらくおさわがせして失礼します
これ以上は買える余力ありませんが

I made the purchase without problem
Pardon the trouble I have caused for a long time (this is probably not referring to him being a douche, rather - intruding on people looking to make the purchase)
But now I am completely out of money and can not buy anything beyond this

Put the hiragana in because IME tends to allow for use of Kanji not always super common

[loial]

tldr; math stuff /yawn

In regards to the comments of Max and Day:

The numbers do not lie. According to playonline 6344623 marbles were sold. Based purely on the odds of any one randomly picked number to win a particular rank prize, we should have seen this:

Rank1: ~63 winners
Rank2: ~634 winners
Rank3: ~6344 winners
Rank4: ~63446 winners
Rank5: ~634462 winners

Of course, you'd expect some variance off those results. What we got instead was:

Rank1: 74
Rank2: 468
Rank3: 4813
Rank4: 68268
Rank5: 642517

Rank2 and Rank3 have 25% less winners than under the scenario of players just picking random numbers alone (and not zipcodes, birthdays, etc).

So, I would agree that those people that picked their own numbers probably didn't pick stuff that ended in 2004 or 800. And it shows.

Edit:
Sev makes a good point. I bet that they simply do not allow there to be a marble that could win 2 ranks. Nothing is certain, but consider that rank5 this year ended in a 5. So, they pulled a winning number with 5. Is it possible that the rank4 number they originally pulled was a 25 instead of 27? So they throw it back in the hat and tried again?

Consider that if you pick a 5 the first time for rank5, they have a 9/10 chance of picking a two-digit number that does not end in 5 the next time. But, for them to not pull a 3-digit, 4-digit or 5-digit number that ended in 5 the other times... 65% chance. Odds are pretty good.

Same deal happened last year. Rank5 was a "1" and they didn't pull another two/three/four/five digit number ending in a 1 either. Chances of that happening (compounded with what happened this year)? I think is ~38%.

But like others said, 10 data points is a little scarce to work with. In my opinion, next year, I'd play numbers all greater than 10000.

[Xanthe]

Sev makes a good point. I bet that they simply do not allow there to be a marble that could win 2 ranks. Nothing is certain, but consider that rank5 this year ended in a 5. So, they pulled a winning number with 5. Is it possible that the rank4 number they originally pulled was a 25 instead of 27? So they throw it back in the hat and tried again?

Consider that if you pick a 5 the first time for rank5, they have a 9/10 chance of picking a two-digit number that does not end in 5 the next time. But, for them to not pull a 3-digit, 4-digit or 5-digit number that ended in 5 the other times... 65% chance. Odds are pretty good.

Same deal happened last year. Rank5 was a "1" and they didn't pull another two/three/four/five digit number ending in a 1 either. Chances of that happening (compounded with what happened this year)? I think is ~38%.

But like others said, 10 data points is a little scarce to work with. In my opinion, next year, I'd play numbers all greater than 10000.

Doubtful, considering the following in the rules:

* The winning number for each prize rank will be drawn separately. In the event that winning numbers overlap for multiple prize ranks, the higher-ranked prize will be given precedence. For example, should the winning numbers for ranks 4 and 5 be “11” and “1” respectively, the holder of a bonanza marble etched with the number “11111” will be awarded with a rank 4 prize.

[Max™]

I'd be really fucking pissed if like 2 and 3 were 4300 and 300, and I had 54300 so I could get Vbelt but get jipped out of my rank 3 homam pantz win as well.

[Valisk]

I'd be really fucking pissed if like 2 and 3 were 4300 and 300, and I had 54300 so I could get Vbelt but get jipped out of my rank 3 homam pantz win as well.

Greedytaru!

[Max™]

Well, I imagine that is part of the selection for the numbers.

If you seriously think they just totally randomly pick the numbers, you're smoking crack.

IF there is any randomness, it is in generating sets of numbers to determine viability, but even then... meh. Computers don't do random.

[Olo]

Fast cast pants ftmfw!

[Deimos]

IF there is any randomness, it is in generating sets of numbers to determine viability, but even then... meh. Computers don't do random.

And that's why they said they don't use computers to generate the numbers.

[Pyree]

If my math is correct, the probability that all five winning numbers are completely disjoint, meaning there are no numbers that win more than one rank, is 63.53%. This number is far too high to conclusively say that SE didn't allow for marbles to win multiple ranks, even if SE hadn't already stated what happens if your marble does win multiple prizes.

And for reference, my math to get this number was:
(9/10)^4 * (99/100)^3 * (999/1,000)^2 * (9999/10,000) = .635277

Also, the number of winners per rank is not off by an unreasonable amount. Mathematical probability and expirimental probability aren't going to be exactly equal.

[Max™]

It would be simple enough for them to calculate the number of winners per rank by marbles sold that would be expected, and then pick sets of numbers that A: don't overlap, and B: don't skew away from those numbers significantly.

The two 00's showing up skews it because as was said, most people wouldn't pick them (I actually had a few marbles with 00's in them, just not in winning combos), but the other numbers were higher than the strict average sales/winners would predict.

[Valisk]

The two 00's showing up skews it because as was said, most people wouldn't pick them.

I actually didn't even let the mog pick them. If it gave me a stupid looking number I would have it random again. I'm sure I passed up a couple 00s. Also rerolled if it was EEEEE or OOOOO. You live, you learn.

[Max™]

I did xxx46 > xxx75 trying to cover a swath of rank 4 options like I did last year, then rolled 3 d10 for the first three digits for all 30.

[Narol]

Mathematical probability and expirimental probability aren't going to be exactly equal.

This. A million times this. People need to realize that experimental probability is only going to be truly equal to mathematical probability over an infinite number of trials. Over a statistically significant number of trials, it'll be pretty damn close (you can even figure out how many are needed to expect a certain percent of accuracy in experimental probability), but two years of Bonanza isn't enough for that.

[Max™]

Strictly speaking, the finiteness of nature means you do not need an infinite number of trials to match up, particularly when you include the limitations from the uncertainty principle and sample sizes large enough for those to become important, but that's the physics geek in me talking.

[Narol]

If you really want to split hairs, flip a fair coin 10000000001 times. You're not getting a perfect 50/50 split, even though it'll be damned close (and I wouldn't be surprised at a few more one way or the other on an even number of trials). But that's just the hair-splitter in me talking.

[Max™]

*refers to the ultimate fair coin problem, the baffling imbalance between matter-antimatter amounts in the Universe*

Doesn't make sense for C(harge)P(arity) Symmetry to be violated like that.

For every 1,000,000,000 particles of anti-matter there were 1,000,000,001 particles of matter.

Course, I have a possible explanation involving dark matter being composed of pairs of tangled quark-antiquark mesons which are composed in such a way to leave them all but completely non-interactive.