6 ms·

you could exploit max bet if they allowed it. you start at $1 and if you lose you double your bet, and when you win you reset your bet back to $1.

maaaats 10y ago
This is called Martingale and doesn't work out as nicely as it first seems.
- pvdebbe 10y ago
  Yeah. It works if you have unlimited budget and the table has no limits either. Not very realistic situation.
sleepychu 10y ago
This works but you don't need to lose all that many hands in a row before your stake is huge compared to what you stand to recover. By the 20th hand you're over $1M for the single bet and you've invested ~$500K to get this far, all to eventually win $1 per hand if you have the bank to cover it.
- jlebrech 10y ago
  yep, that's a downside.
  - pc86 10y ago
    Yes, betting a million dollars to win $1 is what most people would call "a downside."
lolc 10y ago
This strategy plays constant small wins against the occasional big loss. If you start your runs with one dollar, you can only win one dollar in any given run, but you can lose your entire budget on any run.
wehadfun 10y ago
I think the best way play $1 if you win play $1 if louse play $1 and enjoy the free drinks and conversation with your buddies.
- jlebrech 10y ago
  that sounds like a plan
Muted 10y ago
As others have mentioned this is indeed not true and is referred to as the Martingale betting system [0]. In the case of casino games Martingale still has negative EV. Your EV for any round n is (1-q^n)B-q^n(B^n-1) (chance of winning within n rounds times amount won minus chance of losing n rounds times amount lost) [1] which is B(1-(2q)^n) with B the initial betting amount and q the chance of losing. Since in any round your chance of losing is always slightly greater than 1/2, you will have a negative EV in any round you play, there is no number of rounds for which the EV becomes positive. The standard deviation of this is fairly large though so that the chance of this happening is very small but nevertheless it isn't zero and what you lose is more than you can expect to win. As sleepychu mentioned, the amount you bet goes up exponentially ($1M after 20 rounds). Aside: It is sometimes said that in order to do this "double when you lose" strategy, you need to e.g. "stick to red for the entire run". This is false. If I bet red-red-red-...-red-red I am just as likely to win/lose as when I bet red-black-odd-odd-even-...-red [0] https://en.wikipedia.org/wiki/Martingale_%28betting_system%29 https://en.wikipedia.org/wiki/Martingale_%28betting_system%2... [1] https://en.wikipedia.org/wiki/Expected_value https://en.wikipedia.org/wiki/Expected_value

maaaats 10y ago

This is called Martingale and doesn't work out as nicely as it first seems.

pvdebbe 10y ago
Yeah. It works if you have unlimited budget and the table has no limits either. Not very realistic situation.

sleepychu 10y ago

This works but you don't need to lose all that many hands in a row before your stake is huge compared to what you stand to recover. By the 20th hand you're over $1M for the single bet and you've invested ~$500K to get this far, all to eventually win $1 per hand if you have the bank to cover it.

jlebrech 10y ago
yep, that's a downside.
- pc86 10y ago
  Yes, betting a million dollars to win $1 is what most people would call "a downside."

lolc 10y ago

This strategy plays constant small wins against the occasional big loss. If you start your runs with one dollar, you can only win one dollar in any given run, but you can lose your entire budget on any run.

wehadfun 10y ago

I think the best way play $1 if you win play $1 if louse play $1 and enjoy the free drinks and conversation with your buddies.

jlebrech 10y ago
that sounds like a plan

Muted 10y ago

As others have mentioned this is indeed not true and is referred to as the Martingale betting system [0]. In the case of casino games Martingale still has negative EV. Your EV for any round n is (1-q^n)B-q^n(B^n-1) (chance of winning within n rounds times amount won minus chance of losing n rounds times amount lost) [1] which is B(1-(2q)^n) with B the initial betting amount and q the chance of losing. Since in any round your chance of losing is always slightly greater than 1/2, you will have a negative EV in any round you play, there is no number of rounds for which the EV becomes positive. The standard deviation of this is fairly large though so that the chance of this happening is very small but nevertheless it isn't zero and what you lose is more than you can expect to win. As sleepychu mentioned, the amount you bet goes up exponentially ($1M after 20 rounds). Aside: It is sometimes said that in order to do this "double when you lose" strategy, you need to e.g. "stick to red for the entire run". This is false. If I bet red-red-red-...-red-red I am just as likely to win/lose as when I bet red-black-odd-odd-even-...-red [0] https://en.wikipedia.org/wiki/Martingale_%28betting_system%29 https://en.wikipedia.org/wiki/Martingale_%28betting_system%2... [1] https://en.wikipedia.org/wiki/Expected_value https://en.wikipedia.org/wiki/Expected_value