Powerball Odds and Statistics

[Arcana]

And, Arcana? You got to know I just shared that posting on game play with in our internal gameplay channel don't you? 

Joshex will be happy to know he's been immortalized in a game dev chat somewhere. 

[avelworldcreator]

Yes, the odds of *any* matching change with more trials, because the *any* changes with more trials.  But that's congruent to saying that the more entries there are, the greater the chance that there is a winner, that among all of them, there is at least one winner.  But that is not what you said.  You said the odds increase for a single player, that the odds of a specific ticket winning increase with more entries.  That is not the same thing you are saying here.

Because your wording is inconsistent and ambiguous, you should specify what you mean with precision.  I am a lottery player.  I buy a single lottery ticket with a single set of (potential) winning numbers.  One ticket, one entry.  Question: do the odds of me specifically winning change with the number of people who enter the lottery.  True or false.

It changes with the number of times you play as a certainty. But the situation here is if the chance of there being a winner increases with the size of the group does that affect the individual chance? For a group of a given size N in which there is a winner the chance of me being that winner is w=1/N. And if the group chance is P=p(N) then my chance is v=Pw or v=p(N)/N. So, yes, the number of players does affect my chance of winning because my situation is no longer independent of the other trials; it's part of an aggregate.

Edit: It affects my chance of being one of the winners, not my actual chance of winning, though the distinction is slight. If my chance of being one of the winners increases so does my chances of winning directly.

[blacksly]

Those were PEOPLE? You just making up names now aren't you?

I admit I heard of Posi, Castle, and I think I heard of Arcana while the game was up, but never heard of the others. And I played since just before IOs till the end.

Not sure if kidding. If not, well, the posters (Pilcrow, EvilGeko, Stupid_Fanboy) referenced all had well over 10k posts, some specializing in frequenting some forums more than others. It's quite easy to not know about them if you did not frequent the forums, but on the other hand if you did regularly visit the forums and did not notice them (or Arcanaville, and a few others), them most likely you were only frequenting a few specific forums such as the role-playing forum or a server specific forum. Even then there was a good chance to notice some highly active posters.

[Vee]

It changes with the number of times you play as a certainty. But the situation here is if the chance of there being a winner increases with the size of the group does that affect the individual chance? For a group of a given size N in which there is a winner the chance of me being that winner is w=1/N. And if the group chance is P=p(N) then my chance is v=Pw or v=p(N)/N. So, yes, the number of players does affect my chance of winning because my situation is no longer independent of the other trials; it's part of an aggregate.

Edit: It affects my chance of being one of the winners, not my actual chance of winning, though the distinction is slight. If my chance of being one of the winners increases so does my chances of winning directly.

My chance of winning is still one in 600whatever million, as is my chance of being one of the winners. More people buying tickets doesn't give me more number combinations or make them select more winning combinations or make them select fewer numbers, any of which would actually increase my chance of winning. In a lottery the number of winning and losing combinations is static and independent of the number of tickets sold and the number of tickets which have each combination.

Again, there's a language problem here. 'Increased odds of there being a winner' is just not equivalent to 'everyone's odds of winning go up.'

[Vee]

Those were PEOPLE? You just making up names now aren't you?

Whether internet people count as people is of course very debatable   

[Arcana]

Edit: It affects my chance of being one of the winners, not my actual chance of winning, though the distinction is slight. If my chance of being one of the winners increases so does my chances of winning directly.

I think your math is correct, right up to the part where you try to express it in English.  It seems you agree with my the odds of the ticket winning do not change, no matter how many other people enter.  That's correct, and there's no disagreement there.  But then you leave math, and extrapolate colloquially, and make a statement that is blatantly false.  Since my chance of winning never changes, my chance of being one of the winners also does not change.  Those two are mathematically distinct things, but you've calculated one and then simply asserted the two are the same.  Try calculating directly.  What are the odds not of winning, but of being one of the winners, if I am the only ticket entered.  Now calculate directly, not indirectly, what are the odds of being one of the winners, if a million other people enter.

The odds of a winner existing at all increases with more entries, but the odds I am one of them do not.  I'm beginning to wonder if you're making the same error I made earlier in the thread.  When I first calculated the odd of winning, I made the incorrect assumption that if there was no winner, all previous entries were still alive: that you could enter once, and if there was no winner you still had a chance to win until there was a winner.  That's not true for powerball.  If there is no winner, all previous entries lose and in effect there is a brand new drawing held *only* with the tickets purchased since the last drawing, with the prize rolling over into the new lottery.  That's the only explanation I have for why you would think the number of entries affects the odds of winning for a particular person.  If I'm the *only* entry and tickets actually rolled over with the prize, then my odds of winning are 100% - because there has to be a winner eventually and I'm the only entry.  It'll take about 40,000 years for me to win, but I will eventually win.  For more entries to *increase* the chances for me to be one of the winners requires a different set of lottery rules that don't match Powerball.

[Arcana]

Whether internet people count as people is of course very debatable   

Whether internet people count as people is debatable because you can count on internet people debating internet people about everything internet people count on.

[MM3squints]

I had no problem playing "anonymously" even on my home server.

Honestly I didn't know who you are either because I never went to the official forums after the death of i12 pvp. I only remembered you because I remember someone in the HUB channel with your name always trying to solo a Pylon quicker than the last time, put 2 and 2 together with your forum title then thought, "oh I remember you"

[avelworldcreator]

My chance of winning is still one in 600whatever million, as is my chance of being one of the winners. More people buying tickets doesn't give me more number combinations or make them select more winning combinations or make them select fewer numbers, any of which would actually increase my chance of winning. In a lottery the number of winning and losing combinations is static and independent of the number of tickets sold and the number of tickets which have each combination.

Again, there's a language problem here. 'Increased odds of there being a winner' is just not equivalent to 'everyone's odds of winning go up.'

Actually the number was barely over two million. 

Ok. If the number of players increases the chance of any of the players selecting the winning number goes up. I wrote that as P=p(N). This is a degenerate form of P=p(q, N) where q is the number of people of the population that have picked the random number. For my examples I simply treated q=1.  p(q,N,r) is just a variation of the binomial distribution nCk*r^k*(1-r)^(n-k).

If P is the chance of a win and someone wins then the chance of being that person is 1/N. You become a member of a group being selected from at random; you are a member of a trial - a random "value". So your odds are P/N of you being selected as the winner. Of course this isn't a linear function. I'm going to use a 4 sided dice here for simplicity... Ok. I got bit by my own math.
I just ran it through a spreadsheet.
(I hope this copy/paste doesn't look too crappy).
1   25.00%   25.00%
2   43.75%   21.88%
3   57.81%   19.27%
4   68.36%   17.09%
5   76.27%   15.25%
6   82.20%   13.70%
7   86.65%   12.38%
8   89.99%   11.25%
9   92.49%   10.28%
10   94.37%   9.44%
11   95.78%   8.71%
12   96.83%   8.07%
13   97.62%   7.51%
14   98.22%   7.02%
15   98.66%   6.58%
16   99.00%   6.19%
17   99.25%   5.84%
18   99.44%   5.52%
19   99.58%   5.24%
20   99.68%   4.98%
21   99.76%   4.75%
22   99.82%   4.54%
23   99.87%   4.34%
24   99.90%   4.16%
25   99.92%   4.00%
26   99.94%   3.84%
27   99.96%   3.70%
28   99.97%   3.57%
29   99.98%   3.45%
30   99.98%   3.33%
31   99.99%   3.23%
32   99.99%   3.12%
33   99.99%   3.03%
34   99.99%   2.94%
35   100.00%   2.86%
It looks like the odds of the individual go down as the win chances go up by increased number of players. Of course this is on a single trial. Repeated trials it would go up of course. Still, I stand corrected. For some reason I didn't think dividing by N would make for such a massive drop in odds.

[Fanta]

The last few days of reading this thread have been the most fun I've had in a few years. Keep it up!

[avelworldcreator]

The last few days of reading this thread have been the most fun I've had in a few years. Keep it up!

Heck I'm laughing at my own gaffe here. 
A number between 0 and 1 being divided by increasingly large whole numbers? Of course it gets smaller and smaller.

[worldweary]

I am eating dinner with my team Arcana shirt on

[avelworldcreator]

I am eating dinner with my team Arcana shirt on

She and I have both made different errors on the subject. I came to mine at about 3am in the morning and was overtired. I was just seeing the value in the numerator going up exponentially and the denominator "only" going up in a linear manner. It's a pattern I've seen often and I've come to expect certain results.  I neglected to take into account the range of values for both. I was just taking umbrage at people accusing me of an entirely different error. I didn't err on the win-chance being distributed or that the group's win chance increasing though. Or even that the instantaneous probabilities were being misused.

[Vee]

Your chances in powerball won't get smaller either. Odds of your number being the winner would still be one in however many possible combinations there are. The odds of there being a winner are variable. Your 'pot odds,' as they say in poker, are variable. Not the odds of your number being the winning number.

[avelworldcreator]

Your chances in powerball won't get smaller either. Odds of your number being the winner would still be one in however many possible combinations there are. The odds of there being a winner are variable. Your 'pot odds,' as they say in poker, are variable. Not the odds of your number being the winning number.

You make the assertion, but can you show how that is true? I've done the math and demonstrated the results. You are still using the odds of a specific value being generated in a single trial. It is part of the overall analysis but it's incomplete.

[Vee]

You make the assertion, but can you show how that is true? I've done the math and demonstrated the results. You are still using the odds of a specific value being generated in a single trial. It is part of the overall analysis but it's incomplete.

I'm not a math (that doesn't sound nearly as good as "i'm not a science") so I can't demonstrate anything mathematically other than that my checkbook is balanced (and demonstrations are always easier with really small numbers ). Arcana's shown it numerous times in her posts though.

But to attempt to answer your question anyway - powerball is a specific value being generated in a single trial - i.e. the ping pong ball drawing. So those are the odds we're using.

[avelworldcreator]

I'm not a math (that doesn't sound nearly as good as "i'm not a science") so I can't demonstrate anything mathematically other than that my checkbook is balanced (and demonstrations are always easier with really small numbers ). Arcana's shown it numerous times in her posts though.

But to attempt to answer your question anyway - powerball is a specific value being generated in a single trial - i.e. the ping pong ball drawing. So those are the odds we're using.

Actually it's one of several values being generated in multiple trials. The lottery organization's randomizer and the random selections of each of the players over multiple games. That's what I was talking about with that "red dice/blue dice" example earlier. The odds of a person being born and surviving to adulthood is even worse odds but we have a population in the billions (I think the number is somewhere in the order of 1 in 10^66 power against).

[Vee]

Actually it's one of several values being generated in multiple trials. The lottery organization's randomizer and the random selections of each of the players over multiple games. That's what I was talking about with that "red dice/blue dice" example earlier. The odds of a person being born and surviving to adulthood is even worse odds but we have a population in the billions (I think the number is somewhere in the order of 1 in 10^66 power against).

My last attempt, I long since should have given up and left it to smarter people. It's not multiple trials, it's a unique single trial multiple times. Only the money carries over from drawing to drawing. The tickets are valid for only one drawing.

[Arcana]

You make the assertion, but can you show how that is true? I've done the math and demonstrated the results.

The fact that you are doing math implies a problem.  The odds of a single ticket matching the results of the powerball draw, given that the ticket contains one possible number sequence out of a set of (presumably) equally likely values is equal to 1/N, where N is the number of possible winning combinations.  That's essentially an axiom of probability.  No other independent choice can change those odds: that's also axiomatically the definition of "independent" in statistics.  Given the first axiom and the second, there should never be a calculation that changes the odds of a single ticket winning.

[Arcana]

If P is the chance of a win and someone wins then the chance of being that person is 1/N. You become a member of a group being selected from at random; you are a member of a trial - a random "value". So your odds are P/N of you being selected as the winner.

Nope.  You are multiplying two statistical factors that are not independent factors.  Consider this: suppose the lottery asks the entrant to pick a number from 1 to 1000.  Suppose exactly one person enters, and guesses 12 as his entry.  What are the odds of someone winning this lottery?  One in a thousand.  What are the odds of you being that winner?  The same: one in a thousand.  It is not P/N.

If it was possible to know, out of all the people who entered the lottery, how many different possible winning numbers were chosen by at least one person and that number was n, and the total possible winning combinations was N, then the odds of *someone* winning would be n/N.  The odds that you would be one of them, if you put in only a single ticket, would be 1/n.  The actual odds of you being the winner would then be (1/n) * (n/N).  Notice that ends up being 1/N, which is also the odds of winning the lottery computed directly: you put in one entry, and the odds of that one entry matching the winning combination is 1/N.

You math is still correctly calculated but misapplied.  Somehow, you are invoking the set n, the total nunber of *different* combinations entered out of all the people who enter P.  That quantity has nothing to do with the discussion at hand, but its an improperly defined value in your calculation assumptions.