Always place your bet on black or red and if you lose double your bet until you win again (so you got your money back) and then place your starting bet again and so on
I think you are wrong. In theory the expected valkue of the strategy zipzapzup mentioned above is infinite. It´s the so called St.Petersburg Paradox. That´s the reason for limits in practise.
But I agree that in practise roulette isn´t worth playing. And yes...I was bored ;)
I was talking about roulette as it's played on casinos, because I have read many times that there's no way to make it profitable
I calculated this theory out a bit, but turns out I can't solve it algebrally
let's say first bet is A and amount of rounds is B
therefore total winnings for 1 round are
(1-(18/38)^B*A - (18/38)^B*A >= 0 we are interested about the limit where it gets positive
K = amount of money in the beginning = (2^B-1)*A
I got it solved to A + (36/38)^B = (A+1)(18/38)^B
but that doesn't give any answer to B algebrally (no matter which A>0 is)
If u got matlab or mathematica or something else, u could try that out and see if we get anything realistic as outcome :)
I didn´t expect such a mathematical answer (is this crossfire.nu??? ^^).
It seems that we did different assumptions and so we get different results.
First I have to admit that I didn´t deal with decision theory and probability calculation for a long time. Maybe I will try to solve your calculus with mathematica if I get bored at work. (To be honest I don´t get the main idea at the first view. You want to solve for which number of rounds the game has a positive payoff?)
I thought about a simple approach: It is a game with 2 expected outcomes: You win the first round or you lose the first round.
-> If you win the first round -> payoff is positive
-> If you lose the first round, you double your bets every round until you win -> expected value is positive (equals your initial bet respectively smaller as the probability of winning is smaller then 1/2).
The problem is that there are limits in reality. The expected value of the second case will be negative as you cant double to infinity.
As conclusion: I don´t think that there is a realistic value with a positive expected payoff because casino operators will do such calculus before they set their limits ;)
Additionally there is a vast number of literature concerning the st.petersburg paradox... so I guess we wont find an answer today :)
As an ending anecdote: A friend of mine plays this strategy every time and most time he gets a positive payoff ...but maybe he´s just a lucky bitch
I got the formula from your link, in wikipedia. And it comes from probability theorem. It's the winning expectancy, if u use the method of doubling the bet always and putting the money on either black or red. So, I used the formula that u suggested basically :)
got it from here link there's a direct link from your link to this site (check the example part)
And I like maths, already passed all the hardest maths at uni :p
btw talking about roulette, a soccer friend of me played it at a casino about 4 weeks ago, he won over 3900 euro first 2000 on slots - in the end he took 2 numbers on roulet each for 50 euro's and 1 of it hit =)
count count count
then set 100 on the color that hasent come for some while
EDiT: ow lol doesent apply to your journal at all. care
i get 30$ for free
But I agree that in practise roulette isn´t worth playing. And yes...I was bored ;)
I calculated this theory out a bit, but turns out I can't solve it algebrally
let's say first bet is A and amount of rounds is B
therefore total winnings for 1 round are
(1-(18/38)^B*A - (18/38)^B*A >= 0 we are interested about the limit where it gets positive
K = amount of money in the beginning = (2^B-1)*A
I got it solved to A + (36/38)^B = (A+1)(18/38)^B
but that doesn't give any answer to B algebrally (no matter which A>0 is)
If u got matlab or mathematica or something else, u could try that out and see if we get anything realistic as outcome :)
It seems that we did different assumptions and so we get different results.
First I have to admit that I didn´t deal with decision theory and probability calculation for a long time. Maybe I will try to solve your calculus with mathematica if I get bored at work. (To be honest I don´t get the main idea at the first view. You want to solve for which number of rounds the game has a positive payoff?)
I thought about a simple approach: It is a game with 2 expected outcomes: You win the first round or you lose the first round.
-> If you win the first round -> payoff is positive
-> If you lose the first round, you double your bets every round until you win -> expected value is positive (equals your initial bet respectively smaller as the probability of winning is smaller then 1/2).
The problem is that there are limits in reality. The expected value of the second case will be negative as you cant double to infinity.
As conclusion: I don´t think that there is a realistic value with a positive expected payoff because casino operators will do such calculus before they set their limits ;)
Additionally there is a vast number of literature concerning the st.petersburg paradox... so I guess we wont find an answer today :)
As an ending anecdote: A friend of mine plays this strategy every time and most time he gets a positive payoff ...but maybe he´s just a lucky bitch
got it from here link there's a direct link from your link to this site (check the example part)
And I like maths, already passed all the hardest maths at uni :p
add the numbers up
divide by 5, and again if its above 36....
choose that number, there after, where u want