i should be a professional gambler

[DeadBabySoup]

aside from poker, i don't gamble. and poker isn't really gambling. so basically i don't gamble. but last night i was out with a girl who does. we had a few drinks and she wanted to play a roulette slot machine. so i put in $20 and had a go at it. my strategy was pretty simple. i would bet $1 on either red or black. if i lost, i would double my bet until i won. after winning, i went back to betting $1. after about 20 minutes, i was up to $100 and got a couple rounds of free drinks. so i cashed out and quit. she made fun of me for betting only red or black. but the $80 covered the drinks from earlier so anyway, i just figured out the odds of winning with this system and it works out like this. with $20, you'd have to lose 5 bets in a row to go bust. the chances of doing that are only 2.7%. so i basically had a 97% chance of winning $100 with my $20 bet. the odds increase in your favor the larger your bankroll gets, as you'd have to lose more consecutive bets. the only thing is you have to be committed to doubling every time.

[C8H10N4O2]

You'll still go broke eventually doing this. I'm going down tomorrow to play in a tournament. Maybe I'll try this and report back... and more than likely that I lost my $20 lol.

[DeadBabySoup]

i think that if you set a goal of $100 or bust, the odds are in your favor doing this. the wheel only had a single zero too, which was nice. i'm going to start doing it whenever i'm at a casino with some time to kill.

[C8H10N4O2]

So I went to play a flash game. You start out with $2,000. So I bet $100, and doubled everytime. While I'm up, I just wanted to show the streak of it missing one color.


I just landed 9 in a row on red as well. I've been exclusively betting on red.
Just saying... it's definitely possible to lost more than 4-5 in a row.

[DeadBabySoup]

cool. i was just playing one too. the only thing about playing $100 at a time is you might hit the limit and not be able to continue doubling your bet. i was betting $25 on the one i played. the max bet was $500, so i figured that was a comfortable ceiling. i started with $1000 and quit when i hit $3000

but yes, it's possible to lose more than 5 in a row. but the chances of losing 5 in a row are less than 3%. i'll take those odds

[C8H10N4O2]

Actually... if you figure (20/38)^5, you'll see it's 4.0386%, and you have to realize that this is for 5 consecutive spins. You gain $1 each time you hit, but you attempt this numerous times. So statistically I still say you are going to go bust. I'd like for someone to chime in that can run the stastical numbers, but anyway. Not trying to downplay your winning $80 at all, it's great! I'm just trying to say that it's not as mathematically easy to do as you think.

I don't know. In my thinking, you have to have 80 "successful avoidances" of a 4% chance. That's also assuming that you miss the (20/38)^4 you also have to avoid while your "bankroll" is on the shorter amount, since you can't bet $1, $2, $4, $8, and then $16 with only $20 total. Anyway, 4/100 would be 1 in 25. So with this method 1 in 25 times of spinning 5 times consecutively you are going to go bust, but you must successfully win 80 times. However, I'm not sure exactly how it works since you don't have to spin 5 consecutive times every time to avoid the 4% chance, which is where someone with better statistical math skills could chime in.

[travz21]

It seems foolproof at first, but every gambler has thought of this. If you want to look at odds, you're getting paid 2 to 1 on a bet you're going to win less than 50% of the time. No matter what, this is a losing bet in the long run.

People also exercise this style of betting with blackjack, which is also a losing battle unless you are a top notch card counter.

[C8H10N4O2]

Agreed^
I still want to know what the odds are in what I presented above... but here is an article in the meantime on the "Martingale" system:
"Most players in the casino are looking for a system to help them get the edge. Many of them stumble upon or hear about a surefire system that involves doubling their bets after each loss. They figure that sooner or later they have to win. This system that is called the Martingale system and it has been around for years. It has also been the ruin of many players. Here is how it works. You make a bet and if you lose you double your bet. If you lose again you double your bet. You keep doing this until you win and then go back to your original bet.

Let’s look at an example of the Martingale. For this example we will assume you are playing on a five-dollar table. First of all most casinos have a maximum bet. Many five-dollar tables have a maximum bet of $500. If you bet using the martingale where you double after every loss the progression would look like this:

You bet $5 and you lose.
Your next bet is $10. If you lose:
Your next bet is $20. If you lose:
Your next bet is $40. If you lose:
Your next bet is $80. If you lose:
Your next bet is $160. If you lose:
Your next bet is $320. If you lose:
Your next bet is $640, which exceeds the maximum bet for the table.

It would only take eight losses to exceed the table maximum bet. It doesn't matter how many times you double up your bet. You will up you will only win five dollars if you eventually hit. This takes into account all your progressions up to that point. Even if you could exceed the $500 maximum bet, on the eighth bet you are risking $640 to win $5. You have already invested $635 for your previous seven bets. If you lose that one you are out $1,275. Can you imagine risking over a thousand dollars for a chance to win five dollars? The casinos know that if someone had unlimited resources they would eventually win. That is why the set maximum bets at the table.

Some people who try the Martingale system have an initial success with it. They might win consistently for many sessions and swear the system is fool proof. Sooner or later reality will kick in and they will discover the flaw in the system. I know because years ago when I was naive about the casino world, I was one of those persons who tried the Martingale.

To use the Martingale system you have to place even money bets. One of the most popular bets for players using this system is to bet black or red on the roulette wheel. (This is also one of the best games for showing you the flaws in the Martingale system.) Shortly after I started playing table games, I read about the Martingale and decided to give it a try on my next visit to Atlantic City. I started with a bankroll of $200 and I won $145 on my first trip using the system. I won on three trips after that building a bankroll of $500. This was too easy. I wondered how the casinos could stay in business if everyone used this system?

On my next trip I took my bankroll to the table and placed my $5 on red. I lost and started the progression. I placed my next bet and lost again and again and again. I lost six bets in a row. When I lost the bet for $160 I was down $315 and only had $185 left of my five hundred dollars. I did not have enough money to bet the $320, which was the next step in the progression. I was nervous and down hearted. I couldn't believe this could happen to me. I was all ready to throw down my remaining money when I came to my senses and decided to take break.

I thought about what I had just done and realized that this system was not fool proof. Anything can happen in a casino including losing several bets in a row. That was many years ago and I am wiser to the realities of the games. There are still many players however that are just hearing about the Martingale system and are convinced that they can't lose with it. Here is an easy way to see for yourself.

If you believe that you can't lose seven bets in a row take a walk through the casino and look at the display board that is connected to most roulette wheels showing you the twenty previous spins. How many times do you see one color come up more than seven times in a row? I once saw black appear 17 times in a row! If you started with a five dollar bet and used the Martingale system, your 17th bet would be $327,680.

No betting system will change the overall casino edge and the Martingale system can be disastrous when things go bad. Please don't be one who is fooled by this system."

[thegrizz1111]

aside from poker, i don't gamble. and poker isn't really gambling. so basically i don't gamble. but last night i was out with a girl who does. we had a few drinks and she wanted to play a roulette slot machine. so i put in $20 and had a go at it. my strategy was pretty simple. i would bet $1 on either red or black. if i lost, i would double my bet until i won. after winning, i went back to betting $1. after about 20 minutes, i was up to $100 and got a couple rounds of free drinks. so i cashed out and quit. she made fun of me for betting only red or black. but the $80 covered the drinks from earlier so anyway, i just figured out the odds of winning with this system and it works out like this. with $20, you'd have to lose 5 bets in a row to go bust. the chances of doing that are only 2.7%. so i basically had a 97% chance of winning $100 with my $20 bet. the odds increase in your favor the larger your bankroll gets, as you'd have to lose more consecutive bets. the only thing is you have to be committed to doubling every time.

the sum of negative expectation bets is itself a negative expectation bet.

since your strategy uses multiple bets which are all negative e.v. the strategy must be a negative e.v. strategy.

[travz21]

From your odds above, you should lose 4% of your initial bet in the long run. If you bet $100, you should lose $4 in the long run every spin, regardless if you won or lost that bet.

Conclusion: Every form of table games is unprofitable in the long run.

[DeadBabySoup]

it's not foolproof by any means. but i think it's a good system for a casual gambler playing low limits. the way i played it, the most i could possibly lose was $20. which is less than the 4 free drinks i got. anywhere else in the casino, they would have cost me $40 i calculated the odds with 5 losing bets, as i'd have to lose 5 to completely go bust from the start. to be completely accurate, it would be somewhere in between. even so, you have a 95%+ chance of winning.

[C8H10N4O2]

I think your odds are more in the realm, you will have a total return of 95% of what you wager. Again, it's important to note your system. Are you literally only going to play until you have that 5 bet situation and then stop? If not, you no longer have "a 95% chance of winning." If you are going to look collectively at this entire thing, you have to realize, if I tell you that I rill roll a dice with 100 sides, and you win with 95 of them, but lose with 5 of time, and we repeat this process 100 times, you will have lost an entire 5 times statistically. You are repeating your "method" over and over and over again, so it's no longer a 95%+ chance of winning. You will get a return of 94.7% ((18/38)*2) (*2 because you get 2:1 odds) on all your wagers placed. Therefore your "roi" is -5.3%.

The free drinks are great, and in this situation, if you place a total of $200 worth of bets, you are going to statistically lose $10.60 (if my calculations are at all near correct), which pays for your drinks. Just realize that you can't beat the odds and you are still statistically going to lose money in the long run.

Also, you only need to lose 4 times until your method no longer works. You can only assume 5 times if you are starting out with $31 based completely on your method. I know you are left with $5, but this no longer can use the method as you need another $11 to be able to make your 5th bet.

[DeadBabySoup]

right, you'd have to take into account the number of times that a series of events occurs. i don't think either of us have hit on the probability of hitting $100 with an initial investment of $20. the formula would be a bit more complicated. until someone figures it out, i'll go with real world observation and see how often i win and how often i go bust. either way, it's certainly more interesting than mashing buttons on a slot machine and hoping for the best

[thegrizz1111]

right, you'd have to take into account the number of times that a series of events occurs. i don't think either of us have hit on the probability of hitting $100 with an initial investment of $20. the formula would be a bit more complicated. until someone figures it out, i'll go with real world observation and see how often i win and how often i go bust. either way, it's certainly more interesting than mashing buttons on a slot machine and hoping for the best

if i had the time to type in the numbers to an 82X82 matrix i could give u this number. maybe if i am really bored one day i will do it.

if anyone knows how to calculate the steady state visitation frequency of a markov chain this number can be obtained by setting up an 82 state markov chain.


here is how you would populate the 82X82 transition matrix


states are start, loose, win21, win22, win23,...win100.

from start you move to loose with p=(19/37)^5, move to win21 with p=(18/37), stay in win21 with p=1-(19/37)^5-(18.37) and move to all other states with p=0.

from winX you move to loose with p=(19/37)^y where y is how many bets u can make based on theX, you move to winX+1 with p=(18/37), stay in winX with p=1-(19/37)^y-(18/37) and move to all other states with p=0.

this is true for winX where X is between 21 and 99

for win100 and loose you stay in these states with p=1



using this matrix you could calculate the steady state visitation frequency to states win100 and loose and that would be the probabilities that you are looking for.

[DeadBabySoup]

i googled it and dug this up. for one hour of play on a european roulette table (single zero), you have an 84% chance of a winning session. that's assuming a $1 starting bet, $200 bankroll and $200 table limit. that's not exactly the calculations i'm looking for. but it does show how it can improve your odds in the short run at least.

"Data derived from computer simulations of 100,000 to 1,000,000 sessions for each cell of data. Strategies: Baccarat=banker; Craps=pass line, no odds; Roulette=red or black. Rounds per hour per game: Baccarat=150 (typical for Mini-Baccarat), Craps & Roulette=30. Baccarat is eight decks. Session is over when all rounds have been completed or player lacks sufficient funds to double a bet as required. When table limits prevent the player from doubling the player makes the maximum bet allowed if s/he has funds to do so. Odds used in calculations from WizardOfOdds.com."

[thegrizz1111]

i just did the math with getting to $24 dollars or bust

here is the transition matrix i was talking about


State start loose w21 w22 w23 w24
start 0.477806 0.035707 0.486486 0 0 0
loose 0 1 0 0 0 0
w21 0 0.035707 0.477806 0.486486 0 0
w22 0 0.035707 0 0.477806 0.486486 0
w23 0 0.035707 0 0 0.477806 0.486486
w24 0 0 0 0 0 1

results are you go bust with p=.246721 and you get to $24 with p=.753279

so you with win $4 dollars approximately 75.33 percent of the time and you will go "bust" approximately 24.67% of the time.

since going bust here does not mean having no money this cant be used to give your ev. when you go bust you will have somewhere between 4 and 7 dollars so your ev is somewhere between (24*.753 +4*.246) and (24*.753+7*.246).

you ev on your $20 investment is between 19.07 and 19.81. which is negative. calculating this exact value of ev would require a much more complicated Markov chain.

[DeadBabySoup]

i just did the math with getting to $24 dollars or bust

here is the transition matrix i was talking about


State start loose w21 w22 w23 w24
start 0.477806 0.035707 0.486486 0 0 0
loose 0 1 0 0 0 0
w21 0 0.035707 0.477806 0.486486 0 0
w22 0 0.035707 0 0.477806 0.486486 0
w23 0 0.035707 0 0 0.477806 0.486486
w24 0 0 0 0 0 1

results are you go bust with p=.246721 and you get to $24 with p=.753279

so you with win $4 dollars approximately 75.33 percent of the time and you will go "bust" approximately 24.67% of the time.

since going bust here does not mean having no money this cant be used to give your ev. when you go bust you will have somewhere between 4 and 7 dollars so your ev is somewhere between (24*.753 +4*.246) and (24*.753+7*.246).

you ev on your $20 investment is between 19.07 and 19.81. which is negative. calculating this exact value of ev would require a much more complicated Markov chain.

a 25% chance of losing 4 bets in a row over such a short span seems high. this site claims you have a 82% chance of doubling your money. which is in line with the other site i saw.

http://www.systemsforroulette.com/co...artingale.html

There are roulette players who believe that the Anti-Martingale system works better than the Martingale roulette system, but if we compare the two systems we can see that the old system is more beneficial. It is true that the Anti-Martingale system is less risky but it offers less opportunity for players to get back their losses. As compared with the Martingale roulette system, every player is assured to have an 82 per cent chance of doubling their winnings.

[thegrizz1111]

the example i posted assumes that you start with $20 and bet one dollar, if you loose you double your bet, if you win you bet one dollar again. you continue to do this until you either have $24 dollars or do not have enough money to place a bet.

the most you ever risk is $16 dollars so lets say you start with 16 and play until you have 20 or bust. this is the exact same thing as the previous example mathematically.

now if you win that 4 dollars 82% percent of the time then you ev is greater than 20*0.82.

20*0.82=16.4 which would mean that your expectation on your 16 dollar bet is greater than 16.4. This cannot be the case or roulette would not be offered in any of the casions.





ive edited this post about 10 times but this is what im sticking with. so if you read some of the intermediate versions im sorry for any confusion they may have caused

[DeadBabySoup]

another way of looking at it is this. every time you begin a sequence of bets, you have a 2.7% chance of going bust (until you have a higher bankroll anyway and can afford additional double bets). you can look at that as being cumulative. in order to double your money, you must win 20 series. each time, you have a 2.7% chance of losing, so you add them up for your total risk. 2.7+2.7+2.7..... or 2.7% times 20 for short. this would give you a 54% chance of busting your bankroll. so a 46% chance of success. which would makes sense. in the long run, the odds will always be slightly against you. but in the short run, you will improve your odds. if your goal were to make a $10 profit, you would have a 73% chance of success. but you would be risking $20 for the chance of making $10. i'm not a mathematician, but i'm pretty good at logic. and i think this is correct.

[C8H10N4O2]

I don't think you can add those percents up like that. If that's the case, you can get over 100% given enough tries, and statistically, you can't have over 100% chance.