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I wonder if you could game that jackpot. Statistically, with even-odd tickets the net value of the ticket is always less than the ticket cost -- meaning statist

I wonder if you could game that jackpot. Statistically, with even-odd tickets the net value of the ticket is always less than the ticket cost -- meaning statistically you loose money (ex: a $2 ticket, on average, has a return of $1.50). If you could monitor the jackpot and, at the very last minute, more than double the current pot then you'd have over a 50% chance of winning. You'd win some and loose some, but statistically over time I think you'd end up positive. Any mathematicians want to chime in? EDIT: As fela, zck, and thedufer pointed out below, this would not work in fact. Thank you guys for the math lesson!

thedufer 12y ago
If you exactly double the pot right before a drawing, you'll have a 50% of losing 50% of the pot (what you put in) and a 50% chance of winning 48% of the pot, with an expected value of, you guessed it, losing 2% of your money each time. It's not a coincidence that this 2% is equal to the cut taken. That will always be your expected loss. Unless it does not behave as advertised, you can't game it.
- chrisfosterelli 12y ago
  I said "more than double the pot". So your odds would be over 50%. And you win the entire pot if you win, unless I'm misunderstanding? You don't have a 50% chance of winning 48%, you have a >50% chance of winning 98%.
  - zck 12y ago
    The site says "The draw is weighted by payment amount". I'm assuming that's linear -- if I put in 1 stellar, and you put in 2, you're twice as likely to win than you are. If the weighting is different (say, you're four times as likely to win), then that changes the math. Let's say that before you put anything in, the pot holds X stellar. You then put in Y stellar. So I'm assuming that you win with Y/(X+Y) probability. If you win, you win .98(X+Y). If you lose, you lose everything. So your expected value is the chance you win (Y/(X+Y)) times the return (.98(X+Y)). Y --- * .98 * (X+Y) X+Y The (X+Y) parts cancel, and you end up with .98Y, or less than what you put in. And since we didn't use the fact that Y>X, it's independent from the actual values of X and Y, so it holds no matter what they are^1. [1] Assuming they're positive, and X+Y is nonzero.
  - thedufer 12y ago
    I used 50% because it is easy to work with, and it shows that 50% isn't a magical breakeven point - which would have to exist if you agree that the expected return is a continuous function and you believe there is a point where your expected return is greater than 0. > You don't have a 50% chance of winning 48%, you have a >50% chance of winning 98%. You only win 48% in my model because the other 50% is your own money that you're getting back. Anyway, let's do it generically. You have put in x of the money (more than 0.5 in your scenario, but it doesn't matter). Then there are two ways to formulate your expected return. First, what I was doing before - if you lose, you lose x and if you win you win (.98-x). You have x chance of winning, so expected return is: (1-x) chance of losing x + x chance of winning (.98-x) (1-x) * -x + (.98-x) * x (x-1 + .98-x) * x -.02x Alternatively, you initially give up x and then have x chance of winning the whole pot, .98: give up x + x chance of winning .98 -x + .98x -.02x So either way, no matter what portion of the pot is yours, your expected return is to lose 2% of the money you put in.
cissou 12y ago
Well no matter how you look at it, the expected value is .98 of whatever you paid. So in the long run you're certainly going to lose 2% of what you invested.
- chrisfosterelli 12y ago
  But you wouldn't be getting just your money back, you'd be winning the entire pot. Unless I misunderstand how this works.
  - fela 12y ago
    say the pot contains 20, and you pay 80, you will have 80% chance of winning 18 in addition to your 80 and 20% chance of losing everything, if you do this many times you will still on average lose
    - chrisfosterelli 12y ago
      Right, so (.80 * .98(100) + .20 * 0) = 78.4. Which is 2% less than the 80 you paid to entry... This makes more sense when I use an absolute (100), I think my mistake was in trying to calculate 100 as a variable amount, I must've made an error there. Thanks for explaining :)

thedufer 12y ago

If you exactly double the pot right before a drawing, you'll have a 50% of losing 50% of the pot (what you put in) and a 50% chance of winning 48% of the pot, with an expected value of, you guessed it, losing 2% of your money each time. It's not a coincidence that this 2% is equal to the cut taken. That will always be your expected loss. Unless it does not behave as advertised, you can't game it.

chrisfosterelli 12y ago
I said "more than double the pot". So your odds would be over 50%. And you win the entire pot if you win, unless I'm misunderstanding? You don't have a 50% chance of winning 48%, you have a >50% chance of winning 98%.
- zck 12y ago
  The site says "The draw is weighted by payment amount". I'm assuming that's linear -- if I put in 1 stellar, and you put in 2, you're twice as likely to win than you are. If the weighting is different (say, you're four times as likely to win), then that changes the math. Let's say that before you put anything in, the pot holds X stellar. You then put in Y stellar. So I'm assuming that you win with Y/(X+Y) probability. If you win, you win .98(X+Y). If you lose, you lose everything. So your expected value is the chance you win (Y/(X+Y)) times the return (.98(X+Y)). Y --- * .98 * (X+Y) X+Y The (X+Y) parts cancel, and you end up with .98Y, or less than what you put in. And since we didn't use the fact that Y>X, it's independent from the actual values of X and Y, so it holds no matter what they are^1. [1] Assuming they're positive, and X+Y is nonzero.
- thedufer 12y ago
  I used 50% because it is easy to work with, and it shows that 50% isn't a magical breakeven point - which would have to exist if you agree that the expected return is a continuous function and you believe there is a point where your expected return is greater than 0. > You don't have a 50% chance of winning 48%, you have a >50% chance of winning 98%. You only win 48% in my model because the other 50% is your own money that you're getting back. Anyway, let's do it generically. You have put in x of the money (more than 0.5 in your scenario, but it doesn't matter). Then there are two ways to formulate your expected return. First, what I was doing before - if you lose, you lose x and if you win you win (.98-x). You have x chance of winning, so expected return is: (1-x) chance of losing x + x chance of winning (.98-x) (1-x) * -x + (.98-x) * x (x-1 + .98-x) * x -.02x Alternatively, you initially give up x and then have x chance of winning the whole pot, .98: give up x + x chance of winning .98 -x + .98x -.02x So either way, no matter what portion of the pot is yours, your expected return is to lose 2% of the money you put in.

cissou 12y ago

Well no matter how you look at it, the expected value is .98 of whatever you paid. So in the long run you're certainly going to lose 2% of what you invested.

chrisfosterelli 12y ago
But you wouldn't be getting just your money back, you'd be winning the entire pot. Unless I misunderstand how this works.
- fela 12y ago
  say the pot contains 20, and you pay 80, you will have 80% chance of winning 18 in addition to your 80 and 20% chance of losing everything, if you do this many times you will still on average lose
  - chrisfosterelli 12y ago
    Right, so (.80 * .98(100) + .20 * 0) = 78.4. Which is 2% less than the 80 you paid to entry... This makes more sense when I use an absolute (100), I think my mistake was in trying to calculate 100 as a variable amount, I must've made an error there. Thanks for explaining :)