newb gambling question

Quote from AAAintheBeltway:

Ok, before someone jumps down my throat, I admit I'm no expert on this stuff. But isn't this the famous gambler's paradox? In reality the odds on play number 7 are not 99.25% or whatever, but 50%. The likelihood of red or black is 50/50 each time, as the table does not have a memory. Ok, some tables might have a memory but hopefully you're not playing there. So you're risking 6400 to clear 100 on a 50/50 play. Not smart.
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i want to use the no expert disclaimer, too. :)

you're right, in my example #1, on play #7, the odds are the same as on play #1. the point is, the odds of getting 7 losses in a row without a winner is low. so the 7th play is not a 99% winner, but the series together is. the key is starting with the right bet size and sizing the bets right throughout a losing streak so you can withstand it.

i don't know the actual win rates of roulette or the payoffs. i'm just using simple examples here.

for the record, i'm not claiming this makes money in the long term. i'm just saying, if you gamble twice in your life, you may be able to walk away with money doing something like this--if it is possible. i don't know how casinos operate.
 
Just figured I'd let you know... I was flipping a coin today and tracking the results (for an experiment in chaos, why - what did you think?) and within a hundred flips had a series of 7 consecutive heads & 7 consecutive tails.... just a coincidence, but I thought it humorous.
 
Gordon,

I've struggled with this for a long time. In another thread someone explained it in an elegant way, sorry I forgot who it was. The probability of a series of events happening is the product of their individual probabilities. Thus, 7 heads is .5*.5 seven times. But if you have 6 heads already, the probability of having six heads is now 1, so the total probability is 1*.5=.5.
 
Quote from AAAintheBeltway:

Gordon,

I've struggled with this for a long time. In another thread someone explained it in an elegant way, sorry I forgot who it was. The probability of a series of events happening is the product of their individual probabilities. Thus, 7 heads is .5*.5 seven times. But if you have 6 heads already, the probability of having six heads is now 1, so the total probability is 1*.5=.5.
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with a 50% win rate...

...the chance of getting 6 in a row is 1.56%.

...the chance of getting 7 in a row is .78%.
 
Quote from AAAintheBeltway:

But the point is after you have six in a row, the odds of getting seven are 50/50.
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yes.

the odds of each one is always 50/50. the odds of getting 7 heads in a row is .78%.
 
Quote from Gordon Gekko:

i may have asked this question here before, but i'm not sure.

i very rarely buy lottery tickets or scratch tickets. i've been to a casino once in my life and i only brought like $20 (i actually won, too. i remember i bet on the kentucky derby and i picked the winning horse lol i think the horse was "silver bullet"). the markets are the only casino i need. if i went to a casino tonight with friends, i'd probably just watch them and not gamble myself. however, i was thinking about one idea. i just want to know if this would work or if it is even possible.

play roulette or something and bet on colors. find something close to 50/50 with 1:1 payoff or something. say you bring $2,000. first bet $100, if you win, quit. if you lose, next time increase size. if you win, quit. if you keep losing, keep increasing size until you do win, then quit.

the point would be to bring enough money to increase size and outlast most losing streaks. when you do win, quit.

if you know a win rate or loss rate, you can calculate the % chance different streaks will happen. for example, if you lose 51% of the time, there is less than a 1% chance you will get 7 losers in a row. so play roulette, and start with the right size, preparing to increase size for a possible 7 losers in a row. in other words, how much money you have will determine your very first bet size, if you are prepared to increase size for 7 losers in a row and quit on first win to make money.

would this work at all?! if this wouldn't work, is there a way to make this idea work? thanks
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You're much to smart to fall for any of this....:-)

5% negative edge, period (there is no winning game, except for blackjack or poker). The idea of using the basic "Martingale" system is appealing, but many a poor soul has helped to pay our Nevada taxes by trying it.

You reach table limits quickly, you lose 1 or 2 out of 20 times, which takes all your profits plus everything you risk.

Read the blackjack book, or simply trade, IMHO.

Don

See below:

The Martingale system: (Money management betting systems, gambling systems.)
Negative progression system, the oldest known system.

The Martingale system is a very old and extremely simple system. It is based on the probability of losing infinite times in a row and is usually applied to 'even money' bets.

You start with one bet. If you win, you start again with one bet. If you lose, you double your bet. Each time you lose, you double your last lost bet. Eventually you are bound to win. When you win you would recover all your lost bets plus one unit (or chip) profit against your initial wager.

Although infallible in theory, the Martingale system requires a large bankroll, has a very low return and is a very risky one because of the maximum bet limits imposed by the casinos. If you run out of money or reach the house limit, you can lose a lot with no chance to recover your losses.

* Negative progression betting systems: You increase the bet when you lose. Require more capital and usually employed to force a winning outcome following a losing streak. (Nerve-wracking, very painful when you lose. Avoid these systems if you can.)

:)
 
I think a better strategy is playing the three rows of 12 #'s. Once you win, leave out the row that you won on and bet on the other two. The only problem is you would have to triple your bet if you lost. I actually tried this with pennies on an online casino just to test it out. I ended up losing because of the table limits. The black and red strategy is not even worth trying. Isn't roulette set up in a way that no matter what combinations you use the house edge will always remain the same?
 
Quote from Mr. Bollinger:

I think a better strategy is playing the three rows of 12 #'s. Once you win, leave out the row that you won on and bet on the other two. The only problem is you would have to triple your bet if you lost. I actually tried this with pennies on an online casino just to test it out. I ended up losing because of the table limits. The black and red strategy is not even worth trying. Isn't roulette set up in a way that no matter what combinations you use the house edge will always remain the same?
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Don't worry about those zero's and double zero's ....2 of 38 numbers = negative 5% +, ......either way, still negative ....

Don :cool: (keep em' comin" ....come to Vegas, we need the tourist action). LOL.....
 
Quote from Gordon Gekko:





the point would be to bring enough money to increase size and outlast most losing streaks. when you do win, quit.
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correct. The house always wins in the long run. If you are up short term you have to find an exit point and never go back or you will lose it all and then some.
 
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