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How to gamble and make money?
#1
Maybe I've got something wrong, but couldn't you practically insure that you make some money gambling if you continuously bet more than you did the last time, and start over from a low amount each time you win?

Say you're going on 1 in 2 odds of either doubling your bet or losing it. Each time you lose, you withdraw and gamble twice the amount you've lost on your fail chain. Once you win, you start over at $1.


balancewithdrawal & betresultfail debt
01fail1
-12passx
11passx
21passx
31fail1
22fail3
06passx
6


Even when the odds are against you (though not too extreme) you can still make money. Say you had a 1 in 7 chance of only doubling your money. If you always gambled the same amount, you'd normally lose money, but now it actually speeds up your gainings, even if it does make it more unstable.


balancewithdrawal & betresultfail debt
01fail1
-12fail3
-36fail9
-918fail27
-2754fail81
-81162fail243
-243486passx
243



By using this method, you'd always turn positive the total amount you've withdrawal'd since your last win. The only holes in this method is that it will either take quite a few tries to make any money or you'll just fail so many times in a row that you deplete your very real and finite wallet.

So, uh, does this actually work, or did I just do the math wrong somewhere?
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#2
Except, Gambling is all luck.
You can use all the math in the world, but, it's really all comes down to how lucky you are.
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#3
Plus the fact that most dealers don't offer good odds such as 1 in 2 odds of either doubling your bet or losing it (they lose money in the long run, regardless of any method).

With higher odds...I guess your theory would apply (although I think I'm missing something)...although as you said, one needs a lot of pocket change.

Plus the fact that gambling is more dependent on skill now.
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#4
EmuAlert Wrote:
balancewithdrawal & betresultfail debt
01fail1
-12fail3
-36fail9
-918fail27
-2754fail81
-81162fail243
-243486passx
243


I think it's better to view it this way:

betdebt-resultwin-total-resultprobability of winning
112p
231p
693p
18279p
548127p
16224381p
486729243p


Problem being, it's a higher chance to lose money than earn. That means that you will on average lose money instead of earning it.

It's easier than the "If you flip a coin until you get heads, I will give you [Image: yep7xmn.png] dollars, where n is the amount of tails you got in a row" though. You could try to find the expectation-value of that one! Wink

Noah
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#5
Gambling is not worth it in my books. In my honors class we're learning probablity and the chances of me getting profit from a 8% hand isn't very comforting- even if I played 12 times....
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#6
Noah Wrote:I think it's better to view it this way:

betdebt-resultwin-total-resultprobability of winning
112p
231p
693p
18279p
548127p
16224381p
486729243p


Problem being, it's a higher chance to lose money than earn. That means that you will on average lose money instead of earning it.

It's easier than the "If you flip a coin until you get heads, I will give you [Image: yep7xmn.png] dollars, where n is the amount of tails you got in a row" though. You could try to find the expectation-value of that one! Wink

Noah

Not completely sure what you mean there, but you're Noah, so you're probably right. Also, I'm assuming that I have an infinite money supply. Of course, with that, you can gamble whatever huge amount you want and just quit whenever you make a profit. Of course I can make a huge finite profit out of finite money, but both doubling my money and using it all up have a theoretical 50% success rate if I repeated it infinite times.

Since the amount of money I gamble will always increase exponentially with this method, with a finite wallet I'll be only a handful of failures in a row from emptying it.

Basically, you have an incredibly high chance of making a small portion of your money supply back using this method (which can be as enormous as you want when you have an infinite supply) but your chance of making a large proportion of your supply isn't helped at all. If you inch your way up, you're gambling enough times to put your money at risk, and if you're gaining in leaps and bounds, you're getting dangerously close to losing it all.

IT MAKES SENSE IN MY HEAD.
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#7
its true this strategy will net you profit guaranteed
assuming 2 very large assumptions

1) the game is 50/50
i think some casinos do highlow dice and its 50/50
2) you have infinite money

the second assumption is false, so its not possible

also many games have table limits.
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#8
One of my rl friends actually makes a living off of online gambling. He says that 90% of the people that play end up losing, while the rest of the 10% milk the crap out of those 90%. I have not seen him do anything that the OP figures, and it kinda doesn't make sense to do that when you see the kind of tactics and mind games that go on.

Let's take Texas Hold'em for example. Playing conservative? He'll bet occasionally and take your money (usually the ante) when he's not big blind, based on the fact that you'll fold when you see something large go into the pot. Playing aggressive? He'll just consult the table of odds in his head and just base it on how likely he's going to win, knowing that you're probably exaggerating your hand. And when a potential high hand appears on the table, and it doesn't complete his hand, he'll just back away, far far away. Basically, he WILL tilt the odds in his favor by at most 30%.
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#9
This is interesting.

I feel like simulating this with a finite amount of money and seeing what happens.
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#10
Most gamble games doesn't work as the OP says.
In games such as texas hold'em everyone can bet. Say if you bet $1, and your opponent raises, what would you do now? forced to fold cause you can only bet $1 for the first round. And when you bet big amount, like $1000, your opponent will just fold unless they are 99% sure they can win (in which case u lose for sure), so you don't make $1000 profit.
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#11
Horusmaster Wrote:In games such as texas hold'em everyone can bet. Say if you bet $1, and your opponent raises, what would you do now? forced to fold cause you can only bet $1 for the first round.

Horusmaster sez "I found a new exploit!"

No.
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#12
As I expected... assuming you have a finite amount of money, you'll eventually lose a large bet and be forced to stop.

Starting with 100 money
 Spoiler

Starting with 500 money
 Spoiler

However, the more money you allow yourself to spend (compared to whatever '1' is), the higher chance you have of gaining money. I tried starting with 20000, and got to 120000 on the first trial and 50000 on the second. The flip side to this, though, is that it takes a long time; it takes about 60000 bets to get from 20000 to 50000.

Edit: Actually... I've done it wrong, I misinterpreted him. It's still much the same though, just takes less time and requires even larger initial values to become 'safe'.

Starting with 500 money
 Spoiler
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#13
KajitiSouls Wrote:Horusmaster sez "I found a new exploit!"

No.

what:f6:
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#14
Horusmaster Wrote:what:f6:

You paralyzed everyone except the last guy (and maybe big blind, depending on parameters) from betting, because all it takes to guarantee that you'll win over the guy before you is to bet. Why bet if the next guy can just 1up you and force you to lose? Also, by liberty of your logic, the guy that raised is breaking the rules anyways since he'd have to bet more than $1 to raise someone that bet $1.

I don't know what kind of game you play, but the versions I'm familiar with, when someone makes a bet and another person raises, everyone else (except the person who originally raised) has the option to either "call", "fold", or "re-raise". Sure, there can be a raise limit, which I suspect is what you're confusing yourself over. But there would be no Texas Hold'em or Poker if you went with your example.


(in the case of misinterpretation, no one's stupid enough to only bring $1 to the table or even just $10)
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#15
KajitiSouls Wrote:You paralyzed everyone except the last guy (and maybe big blind, depending on parameters) from betting, because all it takes to guarantee that you'll win over the guy before you is to bet. Why bet if the next guy can just 1up you and force you to lose? Also, by liberty of your logic, the guy that raised is breaking the rules anyways since he'd have to bet more than $1 to raise someone that bet $1.

I don't know what kind of game you play, but the versions I'm familiar with, when someone makes a bet and another person raises, everyone else (except the person who originally raised) has the option to either "call", "fold", or "re-raise". Sure, there can be a raise limit, which I suspect is what you're confusing yourself over. But there would be no Texas Hold'em or Poker if you went with your example.


(in the case of misinterpretation, no one's stupid enough to only bring $1 to the table or even just $10)
I know how to play texas, the example I gave obviously doesn't work to prove OP wrong.
OP is basically saying u bet $1, then $2, then $4, then $8 and etc...
but in texas the first round ur opponent can raise to the max amount of money you have, in that case you have to fold if ur going by OP's way of betting.
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#16
Thinking about the finite amount of money.

If your goal is to double your original amount. Each round you bet some amount, if you win you double it, if you lose you lose the entire bet. The probability of winning in a given round is p.

If you start at the max bid, you have p probability of winning in the first round, thus attaining your goal.
If you start at half the max bid, you can only bid once... if you win, with p probability, you only have 1.5x your original amount. So you start again. Probability of doubling your money is p^2, which, unless you always win, is strictly less than p.
If you start at 1/3 the max bid, you can bid twice.
This makes it complicated to figure out what happens.
> a) 1, next bid 1/3
-> b) p - 4/3, next bid 1/3
--> c) p - 5/3, next bid 1/3
---> d) p - 6/3, win
---> b) (1-p)
--> a) (1-p)
-> d) 1-p - 2/3, next bid 2/3
--> b) p - 4/3, next bid 1/3
--> x) (1-p) - none left.

Solving this sequence is kinda complicated and I'm lazy. After this it just gets worse.
a = p*b + (1-p)*d
b = p*c+ (1-p)*a
c = p*1 + (1-p)*b
d = p*b

Solving for a, b, c, d, gives us a value of a equivalent to the overall probability of winning.
Eliminate d:
a = p*b + (1-p)*(p*b)
Eliminate c:
b = p*(p+(1-p)*b) + (1-p)*a
b = p*(p+b-p*b) + (1-p)*a
b = p*p+p*b-p^2*b + (1-p)*a
b - p*b + p^2*b = p*p + (1-p)*a
b = (p*p+(1-p)*a)/(1 - p + p^2)
Eliminate b:
a = p*(p*p+(1-p)*a)/(1 - p + p^2) + (1-p)*p*(p*p+(1-p)*a)/(1 - p + p^2)
a = (p*(p*p+(1-p)*a) + (1-p)*p*(p*p+(1-p)*a))/(1 - p + p^2)
a - p*a + p^2*a = 2*p^3+2*a*p-3*a*p^2-p^4+a*p^3 -> I used Maple to simplify this for me, lol.
a - 3*p*a + 4*p^2*a - p^3*a = 2*p^3 - p^4
a = (2*p^3 - p^4) / (1 - 3*p + 4*p^2 - p^3)

This is not getting any simpler. But I can use the power of mathematical tools (Maple again) to plot a.

Turns out, if p < 0.5, a < p. That implies that it's better to just use the original 1-shot plan than to do it this way.


I would guess that the best chance you have of doubling your money is (p), no matter what fraction of the original you start with. Unless p > 0.5, in which case the game is rigged in your favour anyway, and the best strategy is to just bet a small amount repeatedly, and accumulate wealth over time.
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