Author Topic: Powerball Odds and Statistics  (Read 25676 times)

avelworldcreator

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Re: Powerball Odds and Statistics
« Reply #100 on: January 24, 2016, 06:34:04 PM »
Unless you can figure out how to program a "luck" stat which increases likelihood of things going in your favor and make a luck power which scales up based on the number of enemies/players in range. as a developer such things are open to our hands to put into a game world, but not the real world.

would be funny though, you max enhance the luck of the luck power and walk into the golden giza and as you merely pass slot machines they pour out coins and the people at the tables shiver in sight of you.

Huh! I'll pass that on. I think I might have seen something of that nature in the works but I'll ask. Thanks for that idea. May not be used and it may not be unique, but still worth pondering.
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Arcana

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Re: Powerball Odds and Statistics
« Reply #101 on: January 24, 2016, 06:56:05 PM »
I never said directly "V" itself was a set. I said " But there is a chance that two or more people may choose the same set of numbers as well. ".

No.  I quote:

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Your first assertion is that the odds of an individual ticket winning is 1/V no matter what. The error here is that the chance any member of member of V must first be a winning number.

There are no members of V in the context of the post you quoted.  V was a number.  You yourself stated as such when you acknowledged that the probability was being asserted as "1/V" so I know you read it right.  Then you asserted there were members of V.

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The lotter involves each player selecting a set of numbers i.e. {15, 27, 33,34,15}, by an arbitrary method, with each number distinctive (with the possible exception of a "powerball" number), and each selected from a finite range of integral values. How did you conclude that I was equating those independent sets with a variable used to enumerate. I was pointing out the possibility of duplicate selection which does affect the distribution. (I can see I was also getting very tired at close to 2am when I said "any member of member ").  How about about the set of selected sets "V" where the "v" is the quantity of chosen sets? Does this word salad satisfy you?. The lottery picks are a set of sets each game. The size of the set containing the other sets is finite and quantifiable.

That's just obfuscation.  The powerball lottery is mathematically equivalent to this situation, which is actually a textbook-class homework assignment:

An integer between one and 292,201,338 is selected at random.  A player is asked to guess the number, given the range of possibilities. 

Question one: What are the odds of the player guessing the correct number, given one guess.
Question two: Do these odds change if other people make guesses, if no one is allowed to know whether any of the guesses are correct until all guesses are made?

The answer to the first question is: "one in 292,201,338."  The answer to the second question is: "no."  Period the end.


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The lottery device produces 292,201,338 possible set combinations but that is not the set of POTENTIALLY winning combinations. THAT was previously generated by the players. This means the odds of there even being a winning ticket is not independent of the previous actions of the players.

That's correct.  The odds of there being a winning combination is n/N, where n is the number of different combinations generated by the players, and N is 292,201,338.  No one has said otherwise.

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If the device does not select one of the previously generated combinations the odds of a person being a winner isn't 1/v - it's ZERO. It's he win chance of zero divided by the number of unique ticket combinations - which is still ZERO.  Tell me again how the odds of winning are always either 1/v or 1/292201338? Tell me again how the odds of winning are independent of the previously selected set of combinations?

That is also true.  IF the device does not select a number chosen by one of the players, then the odds of any particular person winning are zero.

Quote
Let me use that word again to drive my point home: ZERO. Those are your base odds. ZERO or 100%. It's a coin toss. Tell me why it's not such. This should be good.  But wait! That's only if the odds of a winning are even. This is a WEIGHED coin toss. Did you get that part? "Weighed"? How is it weighed? By the number of unique sets chosen previously by the players and the weighing is calculated by the binomial distribution formula. Again, I want it explained why this is not the case. So the actual  base odds are x% lose and (1-x)% win. It also means the base odds are NOT solely dependent on the possible lottery combinations. 1/v? Those are the odds of a player having a picked a specific combination. It's NOT the odds of the player having picked a WINNING combination. Again I want a direct explanation why that is NOT the case.

Sure.  The odds of *at least one player* picking the winning combination, IF the device has actually picked one of the combinations actually entered, is 100%.  That's because the premise is: the lottery picked one of the entries actually chosen by at least one player.  However NO ONE IS TALKING ABOUT THAT PROBABILITY, INCLUDING YOU.  The statement being argued, and you've repeated yourself, is that the odds of A SPECIFIC PERSON winning change based on the number of other players entering.  THAT IS STILL FALSE.

The odds of a specific ticket winning are still 1/N.  Why?  The odds of a specific ticket winning IF THE LOTTERY MACHINE ACTUALLY PICKS A WINNING NUMBER are 1/n, where n is the number of different combinations actually entered into the lottery.  Since one of them was actually picked GIVEN THE PREMISE THAT THE LOTTERY MACHINE PICKED A WINNING NUMBER, the odds of that ticket winning are 1/n, and n depends on the number of people who entered.

HOWEVER, THAT ASSUMES THE MACHINE PICKED A NUMBER ACTUALLY ENTERED.  The odds of THAT happening itself are not 100%.  Therefore, you cannot say "the odds of a ticket winning depend on how many other people enter."  You can only say "the odds of a ticket BEING THE WINNING TICKET DURING A DRAWING IN WHICH THERE IS A DECLARED WINNER depends on the number of people entered.  That's obvious: if you are the only entrant at all, and the lottery declares that there is a winner, it has to be you.  But the odds of a ticket winning are, according to basic probability, the odds of there actually being a winner, multiplied by the odds of you being that winner if there is a winner.  The second part, 1/n, is the part you are claiming are the odds of a ticket being a winner.  THAT IS FALSE.  The true odds of a ticket being a winner are the odds of there actually being a winner - the odds of your premise being true - multiplied by the odds of that winner being that ticket.  And the odds of your premise being true - that the lottery machine actually picked a winner - is easy to determine.  Given that n was the number of different combinations entered, the odds of one of them being picked are n/N.  So the odds of the ticket being the winner are n/N times 1/n.  In other words, 1/N.  And that's invariant to the number of entries.

The direct explanation for your error is that you performed calculations presuming there was a winner, and calculated what the odds were of a single ticket being that winner.  But the odds of there being a winner at all are not 100%.  Sometimes there is no winner at all.  And obviously, when there is no winner, no ticket wins, and that affects the overall odds of a specific ticket actually winning.  You are not calculating the odds of a ticket winning over all possible lottery outcomes.  You are calculating the odds of a ticket winning over a subset of them.  And if you do that, then you have to state what calculation you are doing correctly. 

What you are trying to say, incorrectly, is that in all powerball lottery draws in which there is a declaration of a winning ticket, the odds that your ticket is the winning ticket are affected by how many people entered..  And everyone would agree with that, because that's obvious.  IF there is a winner at all, the odds that it is your ticket rise if there are less entrants, and fall if there are more of them.  In the degenerate case where you are the only entrant, the odds you are the winner are 100% - it has to be you.  In the opposite degenerate case where there are an extremely large number of entries, the odds that you will be one of them fall to a low of 1 in 292,201,338.  Obviously, the odds can't get any lower than that.  If the lottery says there was a winner, the odds of your ticket being the winner are higher if less people entered.  HOWEVER, before the drawing is held and before you know if there actually was a winner or not, the odds of a particular ticket being the winning ticket are still 1 in 292,201,338.  Your odds of winning don't change as more people enter.

worldweary

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Re: Powerball Odds and Statistics
« Reply #102 on: January 24, 2016, 07:22:33 PM »
For some reason after every post from Arcana the response I keep hearing in my head is Vizzini from The Princess Bride saying "INCONCEIVABLE".

Soul Resonance

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Re: Powerball Odds and Statistics
« Reply #103 on: January 24, 2016, 07:24:23 PM »
Okay, so..let me get this straight..avel comes in first speaking about how limited and confounded the old engine for CoH/The topic of 'just how dedicated are you to this "legacy" CoH and die hard fans(with the added mix of passive-aggressive subversion of the CoH fans here to support/play COT(mind you, I will support it, but not above CoH i23. I feel others will most likely act in the same manner.) Then, someone speaks about something math-based, Arcana is mentioned, someone brings up the lottery/chances of winning, and then avel spirals into this argument with Arcana. Someone then accuses avel of not playing/being active in CoH because they didn't know arcana, and they get offended.  :-\(I know I'm missing abit here and there but I think I got the jest of things).

P.S: Avel, I know you mean well and like was posted before, I'm sure your an absolutely swell person! :D, but, coming into a CoH revival thread, talking about how old and decreped CoH's engine was(very true on that one)/proceeding to question, if not directly, why stay with CoH legacy when COT will offer a much more modern engine, then(and, this is just how I took your post on CoH, I'm sure this is absolutely incorrect in founding) using subtle passive-aggressive undertones and diction to describe our dedication to a game that is a bit dated to divert our attention over to COT.(The deal as it stands, if I am correct, is that the legacy won't even be on that old engine(disk image), but instead will be on a new engine used by a third-company that is yet to be decided(at the time.) As well as APR making use of UE4 and the IP(along with y'all's very own character creator as well :D))Forgive me for this assertion, but, you must understand how this makes you look to me, and, presumably from the post that followed after your initial post, others in this forum as well.
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Arcana

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Re: Powerball Odds and Statistics
« Reply #104 on: January 24, 2016, 07:50:27 PM »
Arcanaville Statistics 201: how to calculate probabilities for people who don't like math.

The basic idea behind calculating probabilities comes down to two basic principles and very simple math.  The first one is what I call the counting rule, and it is this:

If there is a set of equally likely occurrences, and you are interested in a subset of them, the odds of what you're interested happening are just p/N, where p is the number of different things you are interested in, and N is the total number of possible occurrences.

That's simple enough: it is just counting.  What are the odds of rolling a two on a six sided die?  One in six.  Six ways for the die to roll, only one of them is a two, so the odds are 1:6, or about 16.7%.  What are the odds of rolling an even number?  Three in six, or 3/6, or 1/2, or 50%.  Easy.  Just count.

But what if it is difficult to count?  What if the different things that can happen are not all equally likely?  The second rule is what I call the divide and conquer rule:

If the odds of a set of things happening is p, and the odds of any one of those things in the set is q, then the odds of any one of those things happening is p * q

You attack something with an attack that has a 10% chance to crit.  What are the odds of getting a critical hit, if your tohit chance is 85%?  It is just 0.85 * 0.1 = 0.085, or 8.5%.  First you have to get a hit at all, then you have to actually trigger the critical.

Most of statistical analysis is just about applying these rules correctly, and sometimes it can be non-obvious.  Statistics is actually less about math in terms of calculation, and more about applying the rules correctly.  And it is an open secret in the world of statistics that the rules are often not applied correctly.  The fact that a large percentage of math professors get the Monte Haul paradox wrong is a testament to the fact that even though they know the rules, they often let their intuition override proper mathematical reasoning.  Even though that paradox has a trivial mathematical resolution every math professor should know.

How do we analyze something like the lottery.  We apply the rules, without letting intuition or word games get in the way.  What are the odds of a single ticket matching the powerball draw?  Well, first lets figure out how many possibilities there are.  The powerball lottery draws five numbers from a set of 69 balls labelled one through sixty-nine, and a special "powerball" from a separate set of balls labeled one through twenty-six.  How do we determine how many ways there are to do that?

Lets start with the first five.  Obviously there are 69 possible ways to draw the first ball, because there are 69 of them.  Once that first ball is draw, how many possibilities are there for the second draw?  68, because in every case there are only 68 balls left.  Then 67, then 66, then 65.  Using the two rule above, you might guess there are 69*68*67*66*65 ways to draw those balls, and you'd be correct.

However the Powerball lottery does not require you to pick all of the balls in the correct order.  It only requires you to get all the numbers right.  In other words, according to the rule of powerball, the draw "1,2,3,4,5" is considered identical to the draw "5,4,3,2,1."  So actually, the number we calculated above is larger than the actual number of possibilities, because we've double counted many of them (more than double, actually).  Now what?  Well, actually we can still figure this out, because we can use our rules in reverse.  The first rule, the counting rule, says that if you want to know what the odds of something happening are, you just count up the possibilities.  In this case, we've counted too high.  What we need to figure out is, for every "real" possibility, how many times have we repeatedly counted the same thing, and reduce our number by the overage.

Consider a simple case where we pick two numbers.  We could pick 1,2, or 2,1.  Both are the same "powerball" draw of a one and a two.  We've counted the same possibility twice.  So we'd divide by two to get the true number of possibilities.  We'd normally calculate the total possibilities as 69* 68 = 4692, but because we counted everything twice the real number is 69*68/2 = 2346.  There are two thousand three hundred forty six ways to draw two powerball balls.  Math people would say there are 4692 permutations of two balls, but only 2346 combinations.  Permutations are when order matters, combinations are when order doesn't matter.

So what happens when we pick five balls.  Well, we can play a neat trick here.  We want to know how many ways we overcounted those five ball sequences.  And we know how to do that already.   In effect, it is like we have five balls in a bag, and we want to know how many possible ways - permutations in this case - there are to pull those balls out.  And that's just 5 * 4 * 3 * 2 * 1.  So there are 120 ways you can order five balls.  So for every possible combination of powerball, we've overcounted by 120 times.  So we take 69*68*67*66*65 and divide by 120, and we get 11,238,513.  That's the number of possible ways to draw the first five balls.  The actual "powerball" is a separate ball picked from a separate set of 26 balls.  There are exactly 26 ways to do that.  So the total number of possibilities is 11,238,513 * 26 = 292,201,338.  Only *one* of those possibilities is selected at drawing time.  So there's only one winning sequence out of 292,201,338 possible sequences they could draw.  So if you enter a ticket and that ticket has a specific sequence of numbers on it, the odds that that specific sequence will match the drawn sequence is one in 292,201,338.  That's where those numbers come from.

This is all following the rules for probability.  Notice that number doesn't rely on anything besides counting possibilities.  In effect, we've mostly just used rule one.  But what if you want to look at more complicated things, things that can confuse people.  What if you wanted to answer the question: if the powerball authority says there's a winning ticket out there somewhere, what are the odds that it is my ticket?

That's more complicated, but the rules still work.  It is just that now, you'll likely need to do more math because there are too many possibilities to count easily.  You could, in theory, but the numbers are too big.  Basically, if you know there is a winner somewhere, then the total number of possibilities for what came out of the powerball machine aren't 292,201,338 anymore, at least not necessarily.  The new piece of information we have - that there was a winner - means the machine could have only picked a sequence that someone has entered.  If there are a billion entries into the powerball, the odds are pretty good that every single sequence was played by at least one person, but if only a million entries went in then it is impossible that all 292 million were played, because there aren't enough entries to do that.  If we knew how many tickets were entered, and if we made the assumption that all ticket sequences were selected randomly (they aren't in reality, but it is not an unreasonable approximation most of the time) we could then calculate how many sequences are *likely* to have been played, out of all of them.  Unfortunately, that's Arcanaville statistics 401.  Let's just assume we can.  If big N is 292,201,338, we call little n the number of sequences that were actually played by players, and we do what mathematicians do when they don't know something: use the letter as if we knew what it was.

Since the total possibilities for the draw are no longer big N but little n, the odds of one specific ticket being the winner is no longer 1/N, but 1/n.  Less possibilities, so the odds change.  But does this mean that the odds of a ticket winning depend on how many tickets are entered?  No, and that's because we started by assuming there was a winner.  What do the rules say about what the odds of winning are overall?

Well, they are 1/N, because the rules say so.  But what if we want to analyze this the hard way?  Well then why are you reading this post: go read a stats text book.  Okay, fine.  The rules say that to figure out what the odds of a specific ticket winning, we can try to count up all possibilities.  First, there is the case where there is actually a winner.  We know there are n possibilities there.  Of them, one out of n is a specific ticket sequence.  The odds of a specific ticket winning *if there is a winner* is one out of n.  But that's not the set of all possibilities.  What about the case where there is no winner?  Well, if n of the draws picks a winner, then the rest don't.  That's N-n.  In all of them, that specific ticket doesn't win, obviously.  So the odds of the ticket winning, using the counting rule, is that it is one out of n + (N-n) = N.  There are n possibilities when there is a winner, N-n where there is no winner, and the total is the sum of the two.  That's obviously N, but that's how that works.

Notice that no matter how we choose to do it, the easy way or the long way, the odds are the same.  That's because they have to be.  The point of view shouldn't change the odds.  When two different mathematical methods of calculating something generate different answers, the problem is generally not with the math, but with the mathematician.  In fact, it can be a valuable way to double check yourself.  If you know two ways to do something and you do them both and they agree, that would tend to suggest you haven't made an error.  If they don't, pretty sure you did.

Man, this got longer than I thought it would.  But it is still shorter than a stats textbook.  If there any City of Heroes players out there still awake, consider this final thing.  We used to calculate the odds of hitting a target all the time.  Well, I did anyway.  We needed to know the tohit chance, which is kind of like asking how many possible tohit rolls are hits.  We knew how many possibilities there were in total, because we roll 0-100 (there's a teeny tiny complication here that isn't important to this discussion, lets leave that alone for now).  We then calculate the odds of hitting as all the ways you can hit, divided by all the ways there are to roll tohit total.  "75% chance to hit" was just another way of saying 75 rolls hit out of 100 rolls.  Or, because the tohit system seemed to round off to the nearest .0001, you could say 7500 rolls hit out of 10000.  Did we ever ask whether anything else was rolling tohit rolls at the same time?  Of course not.  Because that doesn't matter.  All that matters in this case is how many possibilities hit, and how many total.  Divide, done.

brothermutant

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Re: Powerball Odds and Statistics
« Reply #105 on: January 24, 2016, 08:12:02 PM »
You majored in English twice?
Maybe it was English and American?

Void Huntress

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Re: Powerball Odds and Statistics
« Reply #106 on: January 24, 2016, 10:02:57 PM »
But what if we want to analyze this the hard way?  Well then why are you reading this post: go read a stats text book. 

... Have you actually encountered a stats text in the wild written remotely as comprehensibly as your writeups have been over the years? Because textbooks generally suck pretty harshly, and statistics hasn't in my experience been so much an exception as an exemplar.

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Re: Powerball Odds and Statistics
« Reply #107 on: January 24, 2016, 10:07:29 PM »
... Have you actually encountered a stats text in the wild written remotely as comprehensibly as your writeups have been over the years? Because textbooks generally suck pretty harshly, and statistics hasn't in my experience been so much an exception as an exemplar.

Strongly agreed. I have read exactly one good book on statistics (Lady Luck, by Warren Weaver), and everything else has been utterly atrocious.

avelworldcreator

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Re: Powerball Odds and Statistics
« Reply #108 on: January 24, 2016, 10:43:53 PM »
No.  I quote:
"Your first assertion is that the odds of an individual ticket winning is 1/V no matter what. The error here is that the chance any member of member of V must first be a winning number."

There are no members of V in the context of the post you quoted.  V was a number.  You yourself stated as such when you acknowledged that the probability was being asserted as "1/V" so I know you read it right.  Then you asserted there were members of V.
I also pointed out in the same paragraph that I was tired and apparently hadn't spotted that error in my proofreading. This is a strawman and you know it.

"The lotter involves each player selecting a set of numbers i.e. {15, 27, 33,34,15}, by an arbitrary method, with each number distinctive (with the possible exception of a "powerball" number), and each selected from a finite range of integral values. How did you conclude that I was equating those independent sets with a variable used to enumerate. I was pointing out the possibility of duplicate selection which does affect the distribution. (I can see I was also getting very tired at close to 2am when I said "any member of member ").  How about about the set of selected sets "V" where the "v" is the quantity of chosen sets? Does this word salad satisfy you?. The lottery picks are a set of sets each game. The size of the set containing the other sets is finite and quantifiable."

That's just obfuscation.
What am I supposed to be hiding by just describing mechanical structure? Because that is what "obfuscation" is. It's a deliberate act to hide something. You accused me of being deceitful. You made the accusation now support it or withdraw the claim.

The powerball lottery is mathematically equivalent to this situation, which is actually a textbook-class homework assignment:

An integer between one and 292,201,338 is selected at random.  A player is asked to guess the number, given the range of possibilities. 

First of all at NO point is it a single integer in the range from 1 to 292,201,338. It's a set of much smaller integers in combination. Worse.  The number you are quoting? I just checked it:
How many Powerball combinations are there?
Quote
There are 175,223,510 combinations of Powerball numbers. Five numbers are drawn from a pool of 59 white balls, and the final Powerball is drawn from a separate pool of 35 red balls
I suspect that value is 35*59*58*57*56*55 but I haven't checked it. (Update: checked it: 21,026,821,200 is the result of my calculations. 59C5 * 35 to be exact.)

I don't know which lottery you are describing but it's not one I have ever heard of!

That is NOT how the lottery is done. The number is NOT selected in advance and hidden from those choosing.  What happens is the players create a "bag" of of numbers sets each being one of 175,223,510 combinations. The "pick" is then generated from the range of combinations in a public, televised, drawing and if it is "in the bag" then, and only then, is there a winner. What does that mean? It means there are v+1 possible outcomes NOT 175,223,510 each drawing. Either one of v or none is the outcome.

Don't give me that "textbook model or assignment" garbage. Not only is that an appeal to authority, but you haven't even bothered to check your authorities for accuracy. Textbooks often simplify their examples or assignments for clarity or for the expected capacity of the student to complete the assignment in a reasonable amount of time - and they have been found to have gotten them wrong before. I've actually found a solution presented in a TEACHER'S edition of a math text to be in error and I've had my math teachers also discover such over the years.

Question one: What are the odds of the player guessing the correct number, given one guess.
Question two: Do these odds change if other people make guesses, if no one is allowed to know whether any of the guesses are correct until all guesses are made?

The answer to the first question is: "one in 292,201,338."  The answer to the second question is: "no."  Period the end.

This is correct only within the applicable frame of reference but you have not shown the frame of reference you are using is the one that applies. The evidence is to the contrary.

That's correct.  The odds of there being a winning combination is n/N, where n is the number of different combinations generated by the players, and N is 292,201,338.  No one has said otherwise.

That is also true.  IF the device does not select a number chosen by one of the players, then the odds of any particular person winning are zero.

Sure.  The odds of *at least one player* picking the winning combination, IF the device has actually picked one of the combinations actually entered, is 100%.  That's because the premise is: the lottery picked one of the entries actually chosen by at least one player.  However NO ONE IS TALKING ABOUT THAT PROBABILITY, INCLUDING YOU.  The statement being argued, and you've repeated yourself, is that the odds of A SPECIFIC PERSON winning change based on the number of other players entering.  THAT IS STILL FALSE.

Support that claim with evidence. Basic classroom instruction - show your work. Show that the odds of a given, arbitrary person winning of the possible candidates is restricted to a single factor. You haven't done that yet. You've only made blanket assertions. It's not false because you claim it to be false, but only because you have taken the steps to do that. That's not math; it's the practice of being accountable.

The odds of a specific ticket winning are still 1/N.  Why?  The odds of a specific ticket winning IF THE LOTTERY MACHINE ACTUALLY PICKS A WINNING NUMBER are 1/n, where n is the number of different combinations actually entered into the lottery.  Since one of them was actually picked GIVEN THE PREMISE THAT THE LOTTERY MACHINE PICKED A WINNING NUMBER, the odds of that ticket winning are 1/n, and n depends on the number of people who entered.

In a single paragraph you have both asserted the outcome is both independent of the lottery machine and also dependent. Which will it be? The odds of a specific ticket winning is not 1/N. That's only the odds of it being independently selected from a set of possibilities of size N. It only becomes a chance of being a winning ticket if, and only if, the set in question actually contains a winning entry. This is a dependency created by separate source. You are forgetting an important detail. Percentages aren't merely numbers. They represent fractions of a range. What is the range here? The set of 175,223,510 base combinations. That range is modified by arbitrary points marked by the players. The lottery machine picks a member from that set at random. The base odds of a winner is the chance of the marked combination being selected. It's the same mechanics as a roulette wheel.

HOWEVER, THAT ASSUMES THE MACHINE PICKED A NUMBER ACTUALLY ENTERED.  The odds of THAT happening itself are not 100%.

Didn't I already say that?

Therefore, you cannot say "the odds of a ticket winning depend on how many other people enter."
Why not?
You can only say "the odds of a ticket BEING THE WINNING TICKET DURING A DRAWING IN WHICH THERE IS A DECLARED WINNER depends on the number of people entered.
Uh, no. That's not a complete statement. The odds of there even being a declared winner is directly affected by the number of unique entries.  See "roulette wheel". These odds are distributed binomially.
  That's obvious: if you are the only entrant at all, and the lottery declares that there is a winner, it has to be you.
Yep, and the odds of that happening in a Powerball drawing are one in 175,223,510.
But the odds of a ticket winning are, according to basic probability, the odds of there actually being a winner, multiplied by the odds of you being that winner if there is a winner.
Which is exactly what I said.  ;D
The second part, 1/n, is the part you are claiming are the odds of a ticket being a winner.
You might want to recheck that against what I actually said. I notice this time you didn't directly quote me. Why not?
THAT IS FALSE.
And it would be false - if that's what I actually claimed. I gave as my derived formula of individual win chance p(N)/N. I even ran it through a spreadsheet and published the results and pointed out an error I had made. Funny thing about a public forum - claims about what was said or not said can be directly checked.
  The true odds of a ticket being a winner are the odds of there actually being a winner - the odds of your premise being true - multiplied by the odds of that winner being that ticket.  And the odds of your premise being true - that the lottery machine actually picked a winner - is easy to determine.  Given that n was the number of different combinations entered, the odds of one of them being picked are n/N.  So the odds of the ticket being the winner are n/N times 1/n.  In other words, 1/N.  And that's invariant to the number of entries.

Good! You FINALLY address the issue of whether my use of the binomial is flawed. When I first brought it in to play I was considering not a single lottery drawing but multiple. In a single drawing p(N) = n/N is correct and your conclusions are true (That assumes that n is limited to unique entries). In multiple drawings it is not and the binomial formula applies instead.
The direct explanation for your error is that you performed calculations presuming there was a winner, and calculated what the odds were of a single ticket being that winner.
Where did you get that idea? I did my calculations presuming there was a selection process between the existence of any winner and there being none as the first step.
But the odds of there being a winner at all are not 100%.  Sometimes there is no winner at all.  And obviously, when there is no winner, no ticket wins, and that affects the overall odds of a specific ticket actually winning.  You are not calculating the odds of a ticket winning over all possible lottery outcomes.  You are calculating the odds of a ticket winning over a subset of them.  And if you do that, then you have to state what calculation you are doing correctly.

That's odd. Show me exactly where I did that. I rather distinctly remember a heavy use of the word "zero". I even made a point of emphasizing it. I also spoke the base odds were the result of a weighed "coin-toss". In short I first considered there being NO winner, and then I considered the chance of there being a winner. Even when I was describing my binomial calculations I considering in terms of the chance of failure over a number of trials.
What you are trying to say, incorrectly, is that in all powerball lottery draws in which there is a declaration of a winning ticket, the odds that your ticket is the winning ticket are affected by how many people entered..
Again, that is NOT what I said. And no point did I make such a qualified statement. That would be in direct contradiction to the formula I actually used of p(N)/N. In fact my original formula was only p(N). I introduced 1/N as a coefficient later.
And everyone would agree with that, because that's obvious.  IF there is a winner at all, the odds that it is your ticket rise if there are less entrants, and fall if there are more of them.  In the degenerate case where you are the only entrant, the odds you are the winner are 100% - it has to be you.  In the opposite degenerate case where there are an extremely large number of entries, the odds that you will be one of them fall to a low of 1 in 292,201,338. Obviously, the odds can't get any lower than that.
Actually they can. The number of entries can exceed the number of possible lottery combinations. There is nothing to cap that number except player availability. That's a factor entirely independent of the lottery.
If the lottery says there was a winner, the odds of your ticket being the winner are higher if less people entered.  HOWEVER, before the drawing is held and before you know if there actually was a winner or not, the odds of a particular ticket being the winning ticket are still 1 in 292,201,338.  Your odds of winning don't change as more people enter.
The odds of there even being a winning ticket is not independent of the number of players. And the odds of a ticket being a winning ticket depends on the size of the pool of potentially winning tickets 1/n (or 1/v or whatever you want to use to reflect the size of that pool).  When one, and only one person, enters the drawing the odds are at their worst at 1 in whatever the base number of combinations are. The odds change with additional unique entries. If I am rolling a standard dice and there is only one possible matching value the odds are 1/6. If there are 2 unique values it's 1/3, and so on.
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Vee

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Re: Powerball Odds and Statistics
« Reply #109 on: January 24, 2016, 11:07:20 PM »
It would help if you looked at the right year's powerball format before going off on Arcana's numbers.

Back to  :-X Really good things seem to happen when I don't talk.

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Re: Powerball Odds and Statistics
« Reply #110 on: January 24, 2016, 11:24:32 PM »
Maybe still tired?

Twisted Toon

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Re: Powerball Odds and Statistics
« Reply #111 on: January 24, 2016, 11:28:44 PM »
Just to change the subject :). I am a bit confused with the compatibility with windows. Being in networking I am unsure if we covered it and I just forgot, but why does some old games seem to have an issue running on newer windows? Such as I was able to play Star Wars: Rebellion on windows 98 and no issue, but as soon as I put in on 8 or 8.1 it cannot run. This is with toying around with the compatibility settings.

I have run into this problem with several games when I upgraded from Win98 to XP. The main cause is, I believe, because Win98 was on a Windows platform, and XP (and everything after it) is on an NT platform (basically the internal structure, as was states in a previous post by someone else). And, DOSBox is the way to go to run most of those games pre-XP. Although, a few still have issues running. I still can't seem to get Battlespire to run, even with DOSBox.

Um, AvelWolrdCreator.

The chances of choosing the same numbers that the person pulling little white balls out of a cage gets won't change depending on whether that person pulling the little white balls out of the cage before or after everyone else picks a number. If you were to take all the possible combinations of the little numbered balls (lets just use 292,201,338 as an example) and ask 500,000 people to pick a single number between 1 and 292,201,338. The chances that any one individual picking the same number that the Guy in Charge reads from the random number generator on his computer will be, I'm sure you would have guessed that I would say, 1 in 292,201,338. That will not change for any of the 500,000 people that were asked to pick a number. It won't change for any individual if the number of people asked to pick was increased by a factor of 100, or decreased by a factor of 1000. The fact that the number of possible outcomes is based on the combinations of a number of little white balls is irrelevant. One out of the total number of combinations is any one individual's chance of wining the lottery.

Personally, I think you and Arcana are discussing two almost entirely different points of almost similar issues. You, I believe are talking about the chances of having a winning ticket (with the assumption that there is a winning ticket). Arcana is talking about the chance of a specific ticket having the same number sequences as the person pulling balls out of a cage.

You're comparing Golden Delicious apples to Granny Smith apples.

As always, correct me if I'm wrong.

Dangit people. Let me post!!!  >:(
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avelworldcreator

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Re: Powerball Odds and Statistics
« Reply #112 on: January 24, 2016, 11:29:56 PM »
It would help if you looked at the right year's powerball format before going off on Arcana's numbers.

Back to  :-X Really good things seem to happen when I don't talk.

It was a quick Google search. I was curious where she was getting her numbers. But her saying that only a single number was being picked was definitely not very accurate.  ;D
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avelworldcreator

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Re: Powerball Odds and Statistics
« Reply #113 on: January 25, 2016, 12:00:43 AM »
I have run into this problem with several games when I upgraded from Win98 to XP. The main cause is, I believe, because Win98 was on a Windows platform, and XP (and everything after it) is on an NT platform (basically the internal structure, as was states in a previous post by someone else). And, DOSBox is the way to go to run most of those games pre-XP. Although, a few still have issues running. I still can't seem to get Battlespire to run, even with DOSBox.

Um, AvelWolrdCreator.

The chances of choosing the same numbers that the person pulling little white balls out of a cage gets won't change depending on whether that person pulling the little white balls out of the cage before or after everyone else picks a number.

Well, that depends on if you know what's being drawn or not.  :P And actually the odds do change. The purpose of the drawing isn't to generate the winning combination but rather to pick one of the tickets already bought by the players as a winner (I'm only considering the case of all the numbers being generated being used for selection - the game includes lesser prize categories too).

As the number of possible tickets increase the chance of any of them being a winner increases directly.

If you ever go to a casino with a lottery game you can actually watch the whole process in action (I've actually done this).

The lottery is closer to Bingo or a roulette wheel than just a number guessing game.

Never said that. If you were to take all the possible combinations of the little numbered balls (lets just use 292,201,338 as an example) and ask 500,000 people to pick a single number between 1 and 292,201,338. The chances that any one individual picking the same number that the Guy in Charge reads from the random number generator on his computer will be, I'm sure you would have guessed that I would say, 1 in 292,201,338. That will not change for any of the 500,000 people that were asked to pick a number. It won't change for any individual if the number of people asked to pick was increased by a factor of 100, or decreased by a factor of 1000. The fact that the number of possible outcomes is based on the combinations of a number of little white balls is irrelevant. One out of the total number of combinations is any one individual's chance of wining the lottery.
This would be true if was just a guessing game. Roulette is based on guessing. Bingo is just being lucky enough to have the right card with the right numbers on it (which makes it even closer than roulette to being like the lottery). With lottery you get to "choose your card" or just let the computer pick one out for you. I think this is even allowed in some versions of Bingo.

Personally, I think you and Arcana are discussing two almost entirely different points of almost similar issues. You, I believe are talking about the chances of having a winning ticket (with the assumption that there is a winning ticket). Arcana is talking about the chance of a specific ticket having the same number sequences as the person pulling balls out of a cage.
You got it! Almost. I'm working first with the odds of their being a winning ticket. That's that "p(N)" part. I'm not actually assuming there is even a winning ticket at all, just the possibility. Then I'm applying that possibility to the chances of any person having that ticket if it exists - that's the "1/N" part. Two parts - one problem. And I'm not limiting things to a single lottery drawing but a number of such drawings over time. It's really just the Law of Large Numbers but I'm actually doing the math.

You're comparing Golden Delicious apples to Granny Smith apples.

As always, correct me if I'm wrong.

Dangit people. Let me post!!!  >:(
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Codewalker

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Re: Powerball Odds and Statistics
« Reply #114 on: January 25, 2016, 12:10:27 AM »
First of all at NO point is it a single integer in the range from 1 to 292,201,338. It's a set of much smaller integers in combination.

It's a pretty basic problem to show that they are mathematically equivalent, which is what proofs are based on. Arcana detailed the steps to showing them as equivalent earlier in the thread.

Quote
That range is modified by arbitrary points marked by the players. The lottery machine picks a member from that set at random. The base odds of a winner is the chance of the marked combination being selected. It's the same mechanics as a roulette wheel.

The whole "set of entries defined by the players" is a massive unnecessary overcomplication of the problem. The lottery machine picks a number (or a set of numbers, they're equivalent), whether anyone bought any tickets or not.

Arcana attempted to show that the complex overcomplication does cancel itself out (N ends up in both the numerator and the denominator), but you rejected the logical progression out of hand.

Quote
When I first brought it in to play I was considering not a single lottery drawing but multiple.

Here's the root of the problem. There is only one drawing. Why are you considering multiple drawings? A ticket is only valid for a single drawing. It almost seems like you have a fundamental misunderstanding of how lotteries work.

That's the reason that it can also be shown as equivalent if the winning numbers are determined before or after everyone buys tickets.

The odds of lotteries are an extremely well-researched topic for which there is much information available, not least of all from the lottery operators themselves. They need to know the exact odds in order to avoid losing money on it; just like casino operators do.

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Re: Powerball Odds and Statistics
« Reply #115 on: January 25, 2016, 12:13:28 AM »

Well, that depends on if you know what's being drawn or not.  :P And actually the odds do change. The purpose of the drawing isn't to generate the winning combination but rather to pick one of the tickets already bought by the players as a winner

No, that would be a raffle. It's been stated multiple times in the thread that there doesn't have to be a winner, even if everyone didn't already know that from the increasing jackpot that made everyone start talking about powerball in the first place.

Quote
It was a quick Google search. I was curious where she was getting her numbers. But her saying that only a single number was being picked was definitely not very accurate.  ;D

funny because when i type 'powerball odds' into google I get

"On October 4, 2015, the Powerball format changed again; the white-ball pool increased from 59 to 69 while the Powerball pool decreased from 35 to 26. While this improved the chance of winning any prize to 1 in 24, it also lengthened the jackpot odds to 1 in 292,201,338."

in a little box at the top of the page, prior to any of the links to sites. And it did that the first time I ever googled the odds yesterday so it's not just a tailored result caused by me never clearing my history.

and as long as I'm talking again, damn me, Arcana saying something is a textbook example, which is a pretty common idiomatic phrase, then proceeding to state it in the format of an actual textbook example, is not an appeal to authority.

Vee

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Re: Powerball Odds and Statistics
« Reply #116 on: January 25, 2016, 12:15:41 AM »
The odds of lotteries are an extremely well-researched topic for which there is much information available, not least of all from the lottery operators themselves. They need to know the exact odds in order to avoid losing money on it; just like casino operators do.

and also to keep from going to jail :D

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Re: Powerball Odds and Statistics
« Reply #117 on: January 25, 2016, 01:11:58 AM »
Well, that depends on if you know what's being drawn or not.  :P And actually the odds do change. The purpose of the drawing isn't to generate the winning combination but rather to pick one of the tickets already bought by the players as a winner

That's not how Powerball works.  Powerball draws a sequence of numbers completely at random, and then checks to see if any of the entries matches.  If no ticket matches, the the lottery is declared to have no winner.  All tickets are then discarded (if they win secondary prizes they are paid off, but we're only considering winning the jackpot) and the lottery starts over completely from scratch, except the money allocated to the jackpot prize is "rolled over" to the next drawing.  That's how Powerball can end up with a billion dollar prize.  It starts small, and every time there is no winner the pot rolls over to the next lottery.  But you have to buy new tickets to enter that contest.  As Vee mentions above, you are talking about a raffle where the winner is chosen from a set of entries.  Powerball doesn't work that way.  Powerball takes a set of 69 balls each with the numbers one through 69 and draws five of them randomly.  That generates a combination of five numbers from 69.  Then it draws another ball from a different set of balls numbered one through 26.  The powerball number can theoretically be identical in number to one of the other previously drawn balls, although the first five must obviously be different.

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Re: Powerball Odds and Statistics
« Reply #118 on: January 25, 2016, 01:27:44 AM »
It was a quick Google search. I was curious where she was getting her numbers. But her saying that only a single number was being picked was definitely not very accurate.  ;D

I never said that.  I said the powerball lottery drawing is mathematically equivalent to choosing a single integer from one to 292,201,338.  The process of picking balls and having people choose six numbers rather than one is just a human convenience.  Much like IP addresses.  This is not an IP address: 10.1.1.1.  That's a human convenience for expressing IP addresses aka dotted decimal.  The actual IP address that corresponds to what a human being would write as 10.1.1.1 is 167837953 in decimal, or 00001010000000010000000100000001 in binary.  In other words, an IP address is an unsigned 32-bit integer.  A single number.  For anyone who hasn't done this before, I suggest trying this: ping 167837953.  See what happens.  Your computer knows what I do: an IP address is a single number, and when human beings type it using four numbers it has to be converted into a single number first.  When you give it a single number, it says "oh good" and uses that directly without conversion.

The lottery operators know what they are doing.  They engineer the number of balls and the way they are chosen to create a very specific chance of winning per ticket entered.  And the lottery operators both know what that chance is, and are required by law to publish it.  The odds of winning the powerball (given the current drawing rules) is one in 292,201,338.  If those odds varied based on the number of people who entered the lottery, they would be required by law to state that.  They do not.  The reason why they don't require people to pick a number between one and 292,201,338 is gamesmanship, and nothing more.  Mathematically the lottery would operate the same if that were the case.

Finding and recognizing mathematically equivalent or mathematically congruent situations is a foundational principle of analytical mathematics.

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Re: Powerball Odds and Statistics
« Reply #119 on: January 25, 2016, 01:33:21 AM »