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(Odds of winning * 20 bets * $1) has exactly the same expected outcome as (odds of winning * 1 bet * $20), assuming the odds are the same in the two brackets, b

by delecti 11y ago

(Odds of winning * 20 bets * $1) has exactly the same expected outcome as (odds of winning * 1 bet * $20), assuming the odds are the same in the two brackets, but the difference is in the latter category you get to play 20 times.

Retric 11y ago
I already had another post on this, but try simulating it. 10% chance of winning 9.99x your bet. To make it really obvious make it 10,000 bets not just 20 bets. PS: Now, if the odds are in your favor that's another thing. But, in that case you don't expect to lose your shirt over time.
- delecti 11y ago
  The list of possible outcomes changes, but the expected outcome does not. 2x winnnings * 50% chance of winning * 20 games * $1 per game = expected outcome $20 2x winnings * 50% chance of winning * 1 game * $20 per game = expected outcome $20 Your fuzzy logic around "winning big" "losing big" etc doesn't change the basic math of the expected outcomes, which is simply (chance of a thing happening) * (result if it happens).
  - Retric 11y ago
    These are not even games; the expected value is < 1. That said if you like playing then the actual math is almost meaningless. Nobody complains about not breaking even when going to the movies. With one game you never actually end up with ~20$. It's either 0 or ~40$. Expected value is close to break even, but you get an actual bet going. With 20 games you never end up at 0 or ~40$ it's 'always' (524,287/524,288) something between those extremes. However, with negative expected outcome games the average case is bad, so you odds of winning are less than 50/50 and your average win is tiny. Further, many people play not just a fixed number of games, but keep playing until there stake is gone, which tends very strongly toward zero. Net result, if you play just one game you have a reasonable chance of winning. If you play lot's of games you most likely will keep losing money until you stop. PS: This is why most casinos don't take big bets; they make far more money from people making lots of small bets over time.
    - delecti 11y ago
      If you assume that the odds of winning are the same, the payoff ratio is the same, and that you don't play again with anything you win, then the expected results are exactly the same. If someone is going to keep playing with their winnings, they're likely to do so either way.
      - Retric 11y ago
        EV is not the only number, your describing an average and the standard diviation is important. Ex: if you have a slight negative bias say 50:50 to win 1.999x and play a trillion games your going to flat out lose money with Effectivly zero % chance of winning.
        delecti 11y ago
        But if you play 1 game with a trillion times the bet, you have a 50% chance of walking away with 1.999x your bet, and a 50% chance of walking away with 0. So about a 50% chance to nearly double your bet, and a 50% chance to lose. Whereas with a trillion smaller games you'll probably end up at about 0.999x your starting money, which is the same on average.

by delecti 11y ago

Retric 11y ago

I already had another post on this, but try simulating it. 10% chance of winning 9.99x your bet. To make it really obvious make it 10,000 bets not just 20 bets. PS: Now, if the odds are in your favor that's another thing. But, in that case you don't expect to lose your shirt over time.

delecti 11y ago
The list of possible outcomes changes, but the expected outcome does not. 2x winnnings * 50% chance of winning * 20 games * $1 per game = expected outcome $20 2x winnings * 50% chance of winning * 1 game * $20 per game = expected outcome $20 Your fuzzy logic around "winning big" "losing big" etc doesn't change the basic math of the expected outcomes, which is simply (chance of a thing happening) * (result if it happens).
- Retric 11y ago
  These are not even games; the expected value is < 1. That said if you like playing then the actual math is almost meaningless. Nobody complains about not breaking even when going to the movies. With one game you never actually end up with ~20$. It's either 0 or ~40$. Expected value is close to break even, but you get an actual bet going. With 20 games you never end up at 0 or ~40$ it's 'always' (524,287/524,288) something between those extremes. However, with negative expected outcome games the average case is bad, so you odds of winning are less than 50/50 and your average win is tiny. Further, many people play not just a fixed number of games, but keep playing until there stake is gone, which tends very strongly toward zero. Net result, if you play just one game you have a reasonable chance of winning. If you play lot's of games you most likely will keep losing money until you stop. PS: This is why most casinos don't take big bets; they make far more money from people making lots of small bets over time.
  - delecti 11y ago
    If you assume that the odds of winning are the same, the payoff ratio is the same, and that you don't play again with anything you win, then the expected results are exactly the same. If someone is going to keep playing with their winnings, they're likely to do so either way.
    - Retric 11y ago
      EV is not the only number, your describing an average and the standard diviation is important. Ex: if you have a slight negative bias say 50:50 to win 1.999x and play a trillion games your going to flat out lose money with Effectivly zero % chance of winning.
      - delecti 11y ago
        But if you play 1 game with a trillion times the bet, you have a 50% chance of walking away with 1.999x your bet, and a 50% chance of walking away with 0. So about a 50% chance to nearly double your bet, and a 50% chance to lose. Whereas with a trillion smaller games you'll probably end up at about 0.999x your starting money, which is the same on average.