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12 January 2009 @ 09:58 pm
A statistics question  
In the online game Kingdom of Loathing, one can purchase raffle tickets. A raffle is drawn every night.

My friend D WINOLJ asked,
"It's fairly obvious, from a lot of different sources, that the folks that are winning the raffles are (more often than not) buying mass quantities of raffle tickets. Now, it's obvious that being in a postion to afford to do so, and buying large numbers of tickets per drawing, radically improves one's odds of winning. But my question is this: which would have better odds: buying, say, 10 tickets each day, so that one had a (small) chance in each of 10 drawings, or playing once in 10 days and buying 100 tickets?"

To which I said,
"Gah. I have no idea how to even approach that. I suspect, but am probably wrong, that it might depend on how many total entries there are each day, so as to consider whether 100 tickets is a statistically significantly larger proportion of total entries than 10 tickets. But I'm not sure.

However, nearly all knowledge is contained in LJ, and I have a couple of statisticians on my friends list, so if you'd like, I'd be happy to pose the question."

Any comments, ideas, thoughts, jeers?
K. Peaseceruleanst on January 13th, 2009 03:21 am (UTC)
I think the odds of winning at least once are identical, but the first method has a greater return because it adds the possibilities of winning twice to ten times.
Buddha Buckblaisepascal on January 13th, 2009 03:38 am (UTC)
OK, let's work this through...

Let's say that X tickets are bought by other people each day, so the two cases we are worried about are 10 tickets out of X+10 for 10 days, versus 100 tickets out of X+100 for one drawing.

If you buy 100 tickets out of X+100, your chances of winning are 100/(X+100). Your chances of losing are X/(X+100).

If you buy 10 tickets out of X+10, your chances of winning are 10/(X+10), and your chances of losing are X/(X+10).

Your chances of losing 10 times in a row in the second case are (X/(X+10))^10, or X^10/(X+10)^10.

So what we want to know is which is greater, X/(X+100) or X^10/(X+10)^10.

Cross-multiplying, we get (X+10)^10*X compared with (X+100)X^10. Expanding both out, that's (X^11 + 100X^10 + C) compared to (X^11 + 100X^10). C in this case is a polynomial in X beginning with 4500X^9 and ending with 10^10*X, with all terms positive. C is clearly positive, so X^11+100X^10+C is bigger than X^11+100X^10.

So X/(X+100) (the chance of losing after buying 100 tickets) is greater than X^10/(X+10)^10 (the chance of losing after buying 10 tickets 10 times). So the 10 tickets a day for 10 days is a better bet than the 100 tickets on one day. Assuming that the number of tickets bought by other people is constant across both scenarios and all days.
tassie_galtassie_gal on January 13th, 2009 03:44 am (UTC)
While I follow the logic can I just say that made my head hurt?
Buddha Buckblaisepascal on January 13th, 2009 03:49 am (UTC)
It's hard to do maths in ASCII. There's a math symbol which was designed for this situation, essentially a less-than sign with a question-mark above it like an accent. It means "is it less than?". That symbol, plus the ability to do better math notation would have made it much prettier and easier to follow.
tassie_galtassie_gal on January 13th, 2009 03:51 am (UTC)
Agreeded! Text <> numbers set out neatly in most cases. I have always felt sorry for the poor ppl who have to write out the masters for stats exams...the set out must be hell.
Villiersdianavilliers on January 13th, 2009 07:43 am (UTC)
There are typesetting scripting languages that make it a whole lot easier. I used to be pretty good at using LaTeX(not the rubbery substance), and there was sci-word for those who found LaTeX too scary.
Buddha Buckblaisepascal on January 13th, 2009 09:52 am (UTC)
Some blogging software allows embedded LaTeX, but not LJ as far as I'm aware. I'd love it if it did, but I can understand why LJ Inc might consider that low priority.
Villiersdianavilliers on January 14th, 2009 02:41 am (UTC)
I was more talking about the typesetting of maths/physics exam papers, and how it's not necessarily as difficult as tassie_gal might think, but that information might come in handy in the future.
redneckgaijinredneckgaijin on January 13th, 2009 05:04 am (UTC)
OK, evidently my probability maths are very bad indeed. Lemme do a real example.

Say that 5,000 tickets- no more, no less- are sold in every raffle.

Method one: Buy 100 in one raffle, sit out nine.

Chances of winning: 100/50,000 or 1 in 500
Chances of losing: 49,900/50,000

Method two: Buy ten per raffle.

Chances of winning: 100/50,000 or 1 in 500
Chances of losing: 49,900/50,000

Identical? Well... no.

Because each drawing has a top prize, right?

With Method A, you're giving up on nine top prizes to maximize your chances on winning the tenth.

With Method B, you've got the chance, albeit a small one, to win more than one top prize.

That's the difference. Method A does not actually increase your overall chances- it just throws them all into one drawing. Method B has the same actual numbers by my estimation, with the advantage of marginally higher potential return.
Buddha Buckblaisepascal on January 13th, 2009 09:51 am (UTC)
Um, no. It doesn't work that way. You run into Birthday Paradox issues.

I made different assumptions about the raffle ticket buying habits, which change things, so I'll first run your second example again, then run mine with figures as well.

Your Method Two: Buy ten per raffle, assuming a raffle is limited to 50,000 tickets, all of which are sold.

Chance of winning each drawing: 10/50,000, or 1 in 5,000.
Chance of losing each drawing: 4999 in 5000, or 0.9998 (exact)
Chance of losing all drawings is the chance of losing the first drawing, times the chance of losing the second drawing, times... time the chance of losing the tenth drawing, or 0.9998^10 = 0.998001799.

So in this case, your chances of losing all 10 drawing with 10 tickets apiece is greater than your chances of losing one drawing with 100 tickets.

But I assumed that the raffle wasn't fully subscribed, and that if I were to buy more tickets it wouldn't force someone else to buy fewer. I assumed that the ticket buying habits of the other players was unaffected by how many tickets I bought.

For sake of example, let's assume that 50,000 tickets were bought by other players.

Method one: I buy 100 tickets in one raffle, for a total raffle drawing of 50,100 tickets.

Chance of winning: 100/50100, or 1 in 501, or 0.001996008
Chance of losing: 50000/50100, or 500 in 501, or 0.998003992

Overall chance of winning is 0.001996008

Method two: I buy 10 tickets each in 10 games, for ten raffle drawings of 50,010 tickets.

Chance of winning each drawing: 10/50010, or 1 in 5001, or 0.00019996
Chance of losing each drawing: 50000/50010, or 5000 in 5001, or 0.99980004
To lose overall, I have to lose each independent drawing, so the chance of losing overall is (5000/5001)^10, or 0.998002198
Overall chance of winning is 0.001997802

Either my maths are off, or with the assumption of a fully subscribed raffle, buying a lump of 100 tickets is the better deal, but with the assumption that tickets can always be bought and my buying habits don't affect others, buying 10 over 10 raffles is a better deal.

Let's work the numbers with a smaller raffle: 200 tickets sold (either total, or to others).

200 tickets sold total, 100 to you:
Chance of winning: 1/2, chance of losing: 1/2

200 tickets sold total per raffle, 10 to you each raffle:
Chance of winning each game: 1/20, chance of losing each game: 19/20, chance of losing all 10 games: (19/20)^10 = 6,131,066,257,801/10,240,000,000,000 = 0.59873693, chance of winning at least one game = 1- 0.59873693 = 0.401263061

Best strategy for you: lump all your chances into one game, as your chance of winning that way is 0.5 versus 0.4 the other way.

200 tickets sold to others, 100 to me:
chance of winning: 1/3, chance of losing: 2/3

200 tickets sold to others per raffle, 10 to me each raffle
Chance of winning each game: 1/21. Chance of losing each game: 20/21. Chance of losing all 10 games: (20/21)^10 = 10,240,000,000,000 /16,679,880,978,201, or 0.613913254. Chance of winning at least one game = 1 - 0.613913254 = 0.386086746

So my best strategy is to spread my tickets out, and let the birthday paradox work in my favor, as my odds are 0.33333333 for lumping, and 0.386086746 for spreading.

So the answer depends on if the raffle is fully subscribed or not.
Peter Engdornbeast on January 13th, 2009 08:37 pm (UTC)
So, based on your assumptions, what is the point at which the player's percentage of tickets sold tips things over from "spread it out" to "all at once?" For that matter, is there one?
Buddha Buckblaisepascal on January 13th, 2009 10:28 pm (UTC)
Based on my assumptions (demand by others is constant, never run out of tickets) it is always appropriate to "spread it out".

Based on the assumption made by someone else that there is a fixed total number of tickets which always sell out, it is always appropriate to "all at once".

rickvsrickvs on January 13th, 2009 03:51 am (UTC)
I'd want to know more about the rules of the raffle.

If it's possible to (buy up all the tickets / cover every possible lottery number combination), then there would be a difference between pursuing that 100% guaranteed win and covering, say, 10% of the possibilities in each of ten different contests.

And depending on the rules, I dunno which strategy will provide a larger return for each $100 spent.

Or if you *really* want to open a can of worms, we can discuss the three-door game show strategy that got dissected in the "Ask Marilyn" column a while back :>
Peter Engdornbeast on January 13th, 2009 08:31 pm (UTC)
As far as I know, the raffle house has unlimited tickets.
L A: Down and Seriouslouisadkins on January 13th, 2009 04:13 am (UTC)
When I was a kid, I was always told that entering multiple tickets in to a raffle reduced your chance of winning because it raised the total entries in the raffle. (This didn't ever make sense to me, really.) I go on to say that in 3rd grade I entered the school fall raffle 75 times (25 cents per ticket) and walked away with about $600 in prizes. A defining function, though, was that each drawing came from the same pot - every time someone won something (that wasn't me) it increased my chances to win the next item. The risk of loss would have been a lot higher if it had been a single drawing. So, I guess the answer would be "it depends." :)
Buddha Buckblaisepascal on January 13th, 2009 09:57 am (UTC)
It's good it didn't make sense to you, because it's wrong.

Let's take a small-number example:

20 people buying raffle tickets, each buys one: 20 tickets sold total, and you have one. Your chance of winning is 1/20.

Now you buy an extra ticket (and everyone else still buys one): your chance of winning in 2/21, which is slightly less than double your chances if you had bought one ticket.

So it's true that buying two tickets doesn't double your chances, but it increases it over buying one ticket.