50 blacks in a row
Quote: tehepidemickI wouldnt consider losing 2,000 going broke. It started as just an idea hat I was curious of the outcome, but now I am curious if it would work or not.
I will say this:
If you are only risking $2000 and want to make $30 per day, that is a 1.5% return per day. That means that you will have to have at least a 1.5% chance of losing the $2000 every single day.
In other words, you're likely to lose it all within a few months.
Essentially In order to lose my entire bankroll I have to lose 50 times in a row on a 49.5% flip.
Now, I propose that this will happen once every 38million years(i dont know if my maths right, no one will help me)
I do know that once in that 38million years I will lose more then I have won however I am proposing it is worth risking that for $30 a day, and counting on the losing sequence occurring sometime outside of my life span.
Quote: tehepidemickSo with a 49.5% chance of winning even money, using the martingale system and being able to double 50 times is a -EV.
Can you explain why.
Yes.
Look at it this way.
If you have a 49.5% of winning even money and can double 2 times, your chances of losing are
.505^2= 25.5025%
So you will win 1 unit exactly 74.4975% of the time.
You will lose 25.5025% of the time. And when you lose, you will have bet 1 + 2 = 3 units.
($1 * 74.4975%) + (-$3 * 25.5025% ) = -$0.0201.
Now just do the exact same thing for 50 double ups.
If you have a 49.5% of winning even money and can double 50 times, your chances of losing are
.505^50 = 1.460726492513837× 10^-15
So you will win 1 unit 99.9999999999998539273507486163% of the time.
You will lose .0000000000001460726492513837% of the time. And when you lose, you will have bet (2^50) -1 units. That's 1125899906842623 units.
($1 * 99.9999999999998539273507486163%) + (-$1125899906842623 * .0000000000001460726492513837%) = -$0.64.
so you see no matter how many times you can double up, it stays negative. and in fact the more times you can double up, the more money you can expect to lose, because you are betting more.
Quote: tehepidemickI can tell you im being civil. Your points were addressed. Ill do so again.
For you to say that it is impossible to figure out the answer to "what are the chances a coin flip land on heads 50 times in a row." can not be answered in this or any other universe is very odd to say the least.
Im not betting red-black, its a 49.5% chance of winning bet.
u dont care what the odds are but the system doesnt work, can you explain why?
Sorry if i seemed uncivil
Go back and read what I posted. I never once said it was impossible to figure out the probability of hitting 50 heads in a row, I said it was impossible to solve your problem because you cannot overcome the HE on Roulette if you bet on RED or BLACK an infinite number of times.
Ok so I know very little about math but let me try this......
In order to get a fair coin to flip 50 times on the same outcome, IN A ROW, would be something like 0.5^50 which equals something like 8.9 x 10(to the power of -16).
It would take you approximately 36 million years to approach a 100% outcome (even then it won't be 100% in my opinion)
I can't do any better than that. It still seems absolutely impossible to dow ith roulette as you have a - Expectency game which can never actually make you a profit if the sample size is big enough.
I know I probably did something wrong in my calculations but I'm no mathematician, in fact I barely passed maths in school.
I had to triple check my numbers firstQuote: tehepidemickDid you read any or the posts? Its not impossible.
My goal.
To make roughly $30 a day with absolutely no effort.
here goes my attempt
you want to win 30units a day
and for how long? how many days?
that determines your chances of hitting a win goal before the first loss.
lets see if it is practical and how many bets would be needed per day winning
0.00000001 for each progression win. (looks like a bitcoin casino)
00Roulette example
(I can change these for 0 Roulette. maybe you want to win more at each winning step?)
1-(20/38)^50 = 0.99999999999998845696659845540232 =
probability of winning 0.00000001 one time
starting at 0.00000001
a 50 step Marty requires a total bankroll of only
11,258,999.0684262
and a max bet of 5,629,500
in order to win 1,000,000 units before the first loss
we would need to win how many times in a row before 50 in a row?
1,000,000 / 0.00000001 = 100,000,000,000,000
0.99999999999998845696659845540232^100,000,000,000,000 =
0.315173516555021000
not great
in order to win 5,000,000 units before the first loss
we would need to win how many times in a row before 50 in a row?
5,000,000 / 0.00000001 = 500,000,000,000,000
0.99999999999998845696659845540232^500,000,000,000,000 =
0.003109915486090210
got worse
attempting to win 600,000 units looks to be a coin flip 50/50
attempting to win 100,000 units looks to be .89095336
I like that one much better.
==================================================
in order to win 30 units before the first loss
we would need to win how many times in a row before 50 in a row?
30 / 0.00000001 = 3,000,000,000
can one make this many bets in a day?
0.99999999999998845696659845540232^3,000,000,000 = 0.999965361641256000
probability of a 30 unit win for one day only
how many days in a row do you want to win?
3,334?
That would be 100,000 total units
we know that is .89095336
not close to 100% any more.
you should be able to play with the numbers or change them to 0Roulette
that would make winning 0.00000001 = 0.99999999999999663027977705100943
1-(19/37)^50
Good Luck
I need it myself
Quote: TomspurGo back and read what I posted. I never once said it was impossible to figure out the probability of hitting 50 heads in a row, I said it was impossible to solve your problem because you cannot overcome the HE on Roulette if you bet on RED or BLACK an infinite number of times.
Ok so I know very little about math but let me try this......
In order to get a fair coin to flip 50 times on the same outcome, IN A ROW, would be something like 0.5^50 which equals something like 8.9 x 10(to the power of -16).
It would take you approximately 36 million years to approach a 100% outcome (even then it won't be 100% in my opinion)
I can't do any better than that. It still seems absolutely impossible to dow ith roulette as you have a - Expectency game which can never actually make you a profit if the sample size is big enough.
I know I probably did something wrong in my calculations but I'm no mathematician, in fact I barely passed maths in school.
See, this is the kind of post I was looking for. Now going from the numbers.
This sequence of losing 50 times in a row happens once every 38million years.(in theory)
The key here is you said you can never make a profit in a -EV game if the sample size is big enough. The sample size here is probably a year. What are the chances that the 50 lose streak occurs within the next year.
And the opinion quesiton-
Is it worth risking $10,000 to make $30 a day, considering the chances of the 50 in a row happening within the time span that you are playing.
I completely understand that if you play for 38million years you will lose. However the time frame here is between 1-5 years(or hopefully my entire life)
Quote: tehepidemick1
And the opinion quesiton-
Is it worth risking
No. That's how house edges work. The potential gains never justify the risks. That's the entire point.
Also, read my post above. I went through the trouble to calculate your exact EV for doubling 50 times.
Quote: tehepidemick1See, this is the kind of post I was looking for. Now going from the numbers.
This sequence of losing 50 times in a row happens once every 38million years.(in theory)
The key here is you said you can never make a profit in a -EV game if the sample size is big enough. The sample size here is probably a year. What are the chances that the 50 lose streak occurs within the next year.
And the opinion quesiton-
Is it worth risking $10,000 to make $30 a day, considering the chances of the 50 in a row happening within the time span that you are playing.
I completely understand that if you play for 38million years you will lose. However the time frame here is between 1-5 years(or hopefully my entire life)
Ok so now all I can add is that you have a random variable at work here. You have no idea when you may lose a certain amount of hands in a row. It could happen in 38 millino years or it can happen tomorrow.
In my opinion nobody has the ability to figure out when that may happen. Are you willing to risk your entire bankroll, whether it be $9000 or some crazy unquantifyable figure tomorrow in order to eek out a $30 profit?
are you a twin?Quote: tehepidemick1See, this is the kind of post I was looking for. Now going from the numbers.
more than one account on day one
OK
tehepidemick1
tehepidemick
Quote: TomspurOk so now all I can add is that you have a random variable at work here. You have no idea when you may lose a certain amount of hands in a row. It could happen in 38 millino years or it can happen tomorrow.
In my opinion nobody has the ability to figure out when that may happen. Are you willing to risk your entire bankroll, whether it be $9000 or some crazy unquantifyable figure tomorrow in order to eek out a $30 profit?
The real question is are you willing to bet $10,000 for a 99.99999999999998845696659845540232 chance of winning $30
Quote: 7crapsare you a twin?
more than one account on day one
OK
unfortunatly I had to circumvent the rules to keep the conversation going. I will most likely get banned. Im reading over your numbers you posted.
Quote: tehepidemick1The real question is are you willing to bet $10,000 for a 99.99999999999998845696659845540232 chance of winning $30
That's not what is happening. You would be betting $1125899906842623 for a 99.99999999999998845696659845540232% chance of winning $1.
Quote: sodawaterThat's not what is happening. You would be betting $1125899906842623 for a 99.99999999999998845696659845540232% chance of winning $1.
Thank you!!!!
Quote: tehepidemick1unfortunatly I had to circumvent the rules to keep the conversation going. I will most likely get banned. Im reading over your numbers you posted.
Be kind of hypocritical of me, but at least I got permission first. I'm banning the first account, please don't circumvent the Rule with this second account you have. I, at least, tried to respond to multiple posters right off the bat within one post.
Also, the question has been answered in many ways. You either accept the answer, which is mathematically infallible, or you do not accept the answer and/or want to take the risk.
If you want to take the risk, just do it, nobody here will give you the answer you want, but they'll provide you with a mathematical truth and prove it. People risk much larger sums on -EV propositions every day, and nobody is denying your tremendous probability of winning a single trial, or perhaps, even of lifetime profit before your system fails.
An Internet gambling site, though?
There's the real gamble.
Positive Variance to you.
Quote: sodawaterThat's not what is happening. You would be betting $1125899906842623 for a 99.99999999999998845696659845540232% chance of winning $1.
eh to put it in perspective its more like betting $700 to win .01. So is there something wrong with making this bet 1,000 times?
Quote: Mission146Be kind of hypocritical of me, but at least I got permission first. I'm banning the first account, please don't circumvent the Rule with this second account you have. I, at least, tried to respond to multiple posters right off the bat within one post.
Also, the question has been answered in many ways. You either accept the answer, which is mathematically infallible, or you do not accept the answer and/or want to take the risk.
If you want to take the risk, just do it, nobody here will give you the answer you want, but they'll provide you with a mathematical truth and prove it. People risk much larger sums on -EV propositions every day, and nobody is denying your tremendous probability of winning a single trial, or perhaps, even of lifetime profit before your system fails.
An Internet gambling site, though?
There's the real gamble.
Positive Variance to you.
Thank you, I will now try to respond to multiple posters in one post. And the first 3 pages were people telling me there isn't enough money in the world. I accept the overall loss of the model, I question the reasoning behind making it a losing proposition throughout 1 lifetime. My question was how high was the "temendous probability" of winning each single trial, and if the probability of losing was close enough to 0% to make the risk worth it.
The internet gambling aspect is definatly the biggest gamble.
Quote: tehepidemick1eh to put it in perspective its more like betting $700 to win .01. So is there something wrong with making this bet 1,000 times?
No. But it doesn't make it a winning system. it just means it will win a little a lot more often than it loses a massive amount. As someone already worked out what each bet you are making actually is, you can then work it downwards into the chance -per 1000 iterations- of losing X and gaining Y. The work out the average length of each sequence. And work out the winning per hour, with the risk of losing the 50th bet every hour and make your own decision.
Just understand, the more time you roll those dice, the more chance you will come up a loser.
Each bet is still a negative EV. There are no guarantees. You might decide a 1 in million chance per hour is fine. It's your money.
Semi-interesting effort to work it out, and go play with bitcoins, I guess.
Quote: thecesspitNo. But it doesn't make it a winning system. it just means it will win a little a lot more often than it loses a massive amount. As someone already worked out what each bet you are making actually is, you can then work it downwards into the chance -per 1000 iterations- of losing X and gaining Y. The work out the average length of each sequence. And work out the winning per hour, with the risk of losing the 50th bet every hour and make your own decision.
Just understand, the more time you roll those dice, the more chance you will come up a loser.
Each bet is still a negative EV. There are no guarantees. You might decide a 1 in million chance per hour is fine. It's your money.
I never claimed it was a winning system. Simply a profitable one if done throughout one lifetime.
Quote: tehepidemick1The real question is are you willing to bet $10,000 for a 99.99999999999998845696659845540232 chance of winning $30
You are not getting odds anywhere near that good.
If you risk $10,000 and want to win $30/day, your chances of losing the $10k are larger than 1 in 333 (they would be 1 in 333 if the game had no edge).
So, more often than once per year, you can expect to lose that $10k.
If you want to go broke only once every 38 million years, then you need a bankroll of more than $416 trillion.
And, really, $30/day is not a good return for $416 trillion.
Quote: 7crapsI had to triple check my numbers first
here goes my attempt
you want to win 30units a day
and for how long? how many days?
that determines your chances of hitting a win goal before the first loss.
lets see if it is practical and how many bets would be needed per day winning
0.00000001 for each progression win. (looks like a bitcoin casino)
00Roulette example
(I can change these for 0 Roulette. maybe you want to win more at each winning step?)
1-(20/38)^50 = 0.99999999999998845696659845540232 =
probability of winning 0.00000001 one time
starting at 0.00000001
a 50 step Marty requires a total bankroll of only
11,258,999.0684262
and a max bet of 5,629,500
in order to win 1,000,000 units before the first loss
we would need to win how many times in a row before 50 in a row?
1,000,000 / 0.00000001 = 100,000,000,000,000
0.99999999999998845696659845540232^100,000,000,000,000 =
0.315173516555021000
not great
in order to win 5,000,000 units before the first loss
we would need to win how many times in a row before 50 in a row?
5,000,000 / 0.00000001 = 500,000,000,000,000
0.99999999999998845696659845540232^500,000,000,000,000 =
0.003109915486090210
got worse
attempting to win 600,000 units looks to be a coin flip 50/50
attempting to win 100,000 units looks to be .89095336
I like that one much better.
==================================================
in order to win 30 units before the first loss
we would need to win how many times in a row before 50 in a row?
30 / 0.00000001 = 3,000,000,000
can one make this many bets in a day?
0.99999999999998845696659845540232^3,000,000,000 = 0.999965361641256000
probability of a 30 unit win for one day only
how many days in a row do you want to win?
3,334?
That would be 100,000 total units
we know that is .89095336
not close to 100% any more.
you should be able to play with the numbers or change them to 0Roulette
that would make winning 0.00000001 = 0.99999999999999663027977705100943
1-(19/37)^50
Good Luck
I need it myself
Thank you very much for this, however, the first equation is not correct, for me at least.
1-(20/38)^50 = 0.99999999999998845696659845540232 =
probability of winning 0.00000001 one time
the bet has a 49.5% chance of winning
Quote: AxiomOfChoiceYou are not getting odds anywhere near that good.
If you risk $10,000 and want to win $30/day, your chances of losing the $10k are larger than 1 in 333 (they would be 1 in 333 if the game had no edge).
So, more often than once per year, you can expect to lose that $10k.
If you want to go broke only once every 38 million years, then you need a bankroll of more than $416 trillion.
And, really, $30/day is not a good return for $416 trillion.
I will agree if i had $416 trillion we wouldn't be having this conversation.
Still don't know what the HE is; did we establish it's single 0 Roulette, getting paid 18 of 37 for win ratio ~48.6%? (Not sure how that fits with 49.5% but maybe that was a hypothetical example to demonstrate HE compounding).
And the tiny increments add up, while the Martigale keeps you from losing that side. And you're getting 300 bets/minute. So you calculate it'd be worth roughly $30/day with a possible loss of $10000 if you hit the 50x streak, just because the bet turns over so fast that a 24/7 play would add up to that much. But in a year, if you never hit that 50x loss point, you'd double your money, because you have enough on account to chase the tiny amount multiplied into big money.
I'm no mathematician, call me silly (if you do so with affection), but with those parameters, I don't see why it couldn't work. I just don't know about the 38 million years; the bet cycle is so accelerated that maybe that number is compressed as well? So...18,000 bets/hour; 432,000 bets/day; 157,680,000 bets/year. Take that times (.514^50), I get 1 in 1,795,012 chance that you'll have a 50x loss each year. Seems like a good bet. Or bad math on my part.
Its not single 0 roulette, its simply you roll a dice that gives you a random number between 0.00 and 99.99 you win if the number is lower than 49.50 giving you a 49.5% chance of winning.
Right, the 38 million years is what my math generated. im not sure either.
Im trying to decide if its a good bet to do for probably six months.
Quote: tehepidemick1
Im trying to decide if its a good bet to do for probably six months.
Why only six months?
If it is a good bet for six months, then it is a good bet for your whole life, seventy-eight billion years, or two minutes.
Same goes with a bad bet.
You have a very high probability of going six months without a system failure of fifty consecutive losses, I don't believe anyone ever disputed that. Your expected return will still be negative, though.
each bet, the risk doesn't justify the reward. that holds true no matter where you are in your progression.
Say you lost 49 in a row. Now you have 1 more bet you can make to make all your losses back and go ahead 1 unit.
You still shouldn't make that last bet.
By recursion, you can show you shouldn't make the 48th bet, or the 47th, or the 46th, or even the 1st.
People play negative games for entertainment value. I don't see what is entertaining about trying, futilely, to grind out a tiny profit on a bitcoin casino by risking a lot in a negative game.
You'd be better off playing roulette in a casino because at least some people find that fun.
Quote: Mission146Why only six months?
If it is a good bet for six months, then it is a good bet for your whole life, seventy-eight billion years, or two minutes.
Same goes with a bad bet.
You have a very high probability of going six months without a system failure of fifty consecutive losses, I don't believe anyone ever disputed that. Your expected return will still be negative, though.
Your saying the expected return for those 6 months will be negative?
Quote: Mission146Why only six months?
If it is a good bet for six months, then it is a good bet for your whole life, seventy-eight billion years, or two minutes.
Same goes with a bad bet.
You have a very high probability of going six months without a system failure of fifty consecutive losses, I don't believe anyone ever disputed that. Your expected return will still be negative, though.
Ok, Mission, now I'm missing something. If your losses are zeroed out, and you're accruing your wins, why are you still losing overall?
Quote: tehepidemick1Your saying the expected return for those 6 months will be negative?
the expected return in a negative game is negative for any number of trials, any number of time periods, with any and every betting system, forever and ever, till the end of time.
Quote: beachbumbabsOk, Mission, now I'm missing something. If your losses are zeroed out, and you're accruing your wins, why are you still losing overall?
See, I guess im missing something to. people keep saying it will fail, of course it will fail, but what are the chances of it failing in those 6 months. 7Craps made a good post a few pages back but used the slightly wrong figures, figuring HE at a little over 2% instead of 1%, i think that would double the profit...
Quote: beachbumbabsOk, Mission, now I'm missing something. If your losses are zeroed out, and you're accruing your wins, why are you still losing overall?
Because there are 505 losing numbers and only 495 winning numbers and they pay even money. No accounting method can get past that.
Quote: sodawaterthe expected return in a negative game is negative for any number of trials, any number of time periods, with any and every betting system, forever and ever, till the end of time.
So using this system 10,000 trials would fail?
Quote: tehepidemick1So using this system 10,000 trials would fail?
Most of the time, it would show a profit. However, the reward for the times it works doesn't justify the risks for the times it fails. That's a fancy way of saying it's a negative game. And if it's a negative game, you can expect it to be negative for 1 trial or 10,000 or any number you like.
that makes it even easierQuote: tehepidemick1Thank you very much for this, however, the first equation is not correct, for me at least.
1-(20/38)^50 = 0.99999999999998845696659845540232 =
probability of winning 0.00000001 one time
the bet has a 49.5% chance of winning
1-(.505)^50 = 0.99999999999999853927350748616307 =
probability of winning 0.00000001 one time (you net only 0.00000001 no matter what step you finally win on)
in order to win 30 units before the first loss on the first day
we would need to win how many times in a row?
30 / 0.00000001 = 3,000,000,000 (3 Billion)
can one make this many bets in a day?
0.99999999999999853927350748616307^3,000,000,000 = 0.999995670139541000
probability of a 30 unit win for one day only IF you
can one make this many bets in a day? 3 Billion
IF not, you will win way less than 30 and it may not be worth it
20units = 2 Billion bets each day
10units = 1 Billion bets each day
and so on
Good Luck to you
Quote: sodawaterMost of the time, it would show a profit. However, the reward for the times it works doesn't justify the risks for the times it does fail. That's a fancy way of saying it's a negative game. And if it's a negative game, you can expect it to be negative for 1 trial or 10,000 or any number you like.
See thats the catch there. Most of the time it will show a profit. If it shows a profit 95,000/100,000 times and I only do it 5,000 times, I can expect a Positive EV from a -EV game
Quote: sodawaterBecause there are 505 losing numbers and only 495 winning numbers and they pay even money. No accounting method can get past that.
The 505 losing numbers would have to lose in a 50 unit increment for them to remain losing numbers, with none of the 495 winning numbers in the sequence. The 505-495 assumes flat betting. Save for an infintesmally small chance of 50 in a row, the losses will not be counted against you; only the winnings will count for you. Even 1 winning number wipes out all your losses, and that keeps happening; your losses reset to zero each time, but your winnings don't.
Quote: tehepidemick1See thats the catch there. Most of the time it will show a profit. If it shows a profit 95,000/100,000 times and I only do it 5,000 times, I can expect a Positive EV from a -EV game
That's just not how probability works. When calculating your expectation, you can't just count the times you win. You need to multiply your winning percentage by your win, and then add it to your losing percentage times your loss. If you do that you will see that the game is negative no matter how many times you play it.
Quote: beachbumbabsThe 505 losing numbers would have to lose in a 50 unit increment for them to remain losing numbers, with none of the 495 winning numbers in the sequence. The 505-495 assumes flat betting. Save for an infinitesimally small chance of 50 in a row, the losses will not be counted against you; only the winnings will count for you. Even 1 winning number wipes out all your losses, and that keeps happening; your losses reset to zero each time, but your winnings don't.
Yes, looking backwards in time, if you play this system and you win with it, you have won.
Just like if you play roulette and you win, you have won.
But that's not how probability works. Expectation is about what you can expect going forward, before you play. And even if you can double up 50 times or 500 times, that "infinitesimally small chance" of losing will make the entire proposition a loser.
Quote: sodawaterThat's just not how probability works. When calculating your expectation, you can't just count the times you win. You need to multiply your winning percentage by your win, and then add it to your losing percentage times your loss. If you do that you will see that the game is negative no matter how many times you play it.
I simply do not get how, if my chances of losing are 1-100000 and I play 5,000 your considering it -EV.
There would hypothetically be no loses.
Quote: beachbumbabsOk, Mission, now I'm missing something. If your losses are zeroed out, and you're accruing your wins, why are you still losing overall?
Because it is fundamentally a negative expectation bet, you're not being paid fair odds.
The only difference in this scenario, with the BitCoin thing, is that it enables a Martingale that starts with a base bet so low that it could not have ever been previously considered.
Aside from that, it's no different than a two-step or ten-step Martingale or any number in-between, there are just more steps. We can look at a Martingale that starts with something like a base bet of $5, figure out the probability of outright failure for a five-step with a .505 chance of losing = 0.03284406406 and then we figure that the loss when that happens is $155.
(0.03284406406 * -155) + (.495 * 5) + (.505 * .495 * 5) + (.505 * .505 * .495 * 5) + (.505 * .505 * .505 * .495 * 5) + (.505 * .505 * .505 * .505 * .495 * 5) = -0.25505024962
Okay, average total bet:
(5 * .495) + (15 * .505 * .495) + (35 * .505 * .505 * .495) + (75 * .505 * .505 * .505 * .495) + (155 * .505 * .505 * .505 * .505) = 25.50502505
Okay, now take that average total bet and apply it to the base probabilities:
(.505 * -25.50502505) + (.495 * 25.50502505) = -0.2550502505---Errors, Rounding
We see that the Expected Loss is the same based on the average amount bet and the probabilities, and the question just becomes one of distribution.
The reason is because something either pays fair odds or it doesn't, and if it doesn't, either the player or the casino have the edge. In this case, the casino has the edge.
It all boils down to a simple mathematical principle that adding negative numbers will always result in a negative number, regardless of how slowly you add them. The Martingale can have as many steps as it wants to and the fact remains unchanged.
The Expected Return is the same, it just takes longer to play out. If the player loses the $155 on the first attempt, then it still takes longer to play out (if he keeps going) because he will eventually approach expectations and eventually result in something that reflects the House Edge of the game.
Again, (.505)^50 probably won't happen anytime soon and may not happen in his life, nobody disputes that, but the House Edge exists without concern for when it manifests.
Quote: tehepidemick1I simply do not get how, if my chances of losing are 1-100000 and I play 5,000 your considering it -EV.
There would hypothetically be no loses.
"EV" refers to the average.
If you lose 1 time in 1000, but the time you lose you lose more than you won with the 999 wins, the EV is negative.
Yes, you will probably win. But you are risking a lot more than you stand to win. You are risking too much to justify the odds of you winning. That's all there is to it.
Quote: tehepidemick1I simply do not get how, if my chances of losing are 1-100000 and I play 5,000 your considering it -EV.
There would hypothetically be no loses.
Let me simplify it for you.
Say there's a drum with 9 red balls and 1 black ball. You get to reach in blindfolded and pick a ball. If you get a red ball, I give you $1. If you get a black ball, you give me $20.
You will win 90% of the time. If you do it 1 time, there is a 90% chance I will have to pay you.
Doesn't change the fact that it's not worth doing at all.
Quote: sodawaterLet me simplify it for you.
Say there's a drum with 9 red balls and 1 black ball. You get to reach in blindfolded and pick a ball. If you get a red ball, I give you $1. If you get a black ball, you give me $20.
You will win 90% of the time. If you do it 1 time, there is a 90% chance I will have to pay you.
Doesn't change the fact that it's not worth doing at all.
Sure but lets change it up a little. Lets say there are 100,000 balls, 999,999 red and 1 black. If i pick the red ball I get $1. If i pick the black ball I give you $100,000. Would you do this, 20 times.
With the ball example, now imagine there are 10 trillion red balls and 1 black ball. If i pick the red ball I get $1. If i pick the black ball I give u $1 trillion dollars. Would you make this bet 10 times.
Quote: tehepidemick1Sure but lets change it up a little. Lets say there are 1,000 balls, 999,999 red and 1 black. If i pick the red ball I get $1. If i pick the black ball I give you $2,000. Would you do this, 20 times.
haha, OK, I think we are just feeding the troll here.
Quote: sodawaterhaha, OK, I think we are just feeding the troll here.
why is that? its how the math works out, essentially.
its basically 10 trillion red balls 1 black ball. picking the red ball i win $.00000001 and picking the black ball I lose $500. If I play 1 trillion times, i think I will profit.
Quote: beachbumbabsSo, if you bet 432000x/day, you are expecting to win 213840 of those bets at 49.5%. Your losses don't count against you unless you lose it all with a 50x streak. To win $30/day, you need to bet $0.00014029/bet. So what does that amount Martingale 50x to? At 26x, the bet has to be 9414.82, but you've already spent most of the 10K bankroll and can't make that bet. So your bankroll fails at 25x, not 50x.
theres no daily win goal, the goal is to not have the martingale fail. so the winning per day can be $5 for all i care as long as the martingale can go to 50x
Quote: tehepidemick1See thats the catch there. Most of the time it will show a profit. If it shows a profit 95,000/100,000 times and I only do it 5,000 times, I can expect a Positive EV from a -EV game
Ick.
Your terms are erroneous now, the Expected Value of the game is still negative and always will be. What you are saying is that you can expect a positive Actual Return, and I would suggest that you can, with respect to probability in a limited sample size of that nature.
However, you can't use actual return to say you extracted Positive EV from a Negative EV game because that's retroactive. There's also still Variance, as in, you could lose, so as long as losing is possible, you can't take a positive actual return and make the statement that you had positive expected value because of it.
How could you do that?
If the possibility of losing exists prior to attempting the system, then you did not know that you would have a positive actual return until you conducted the system for x trials and ended up with a positive actual return. You could have lost the first one, and you see to acknowledge that you expect the system to fail once every x million years.
Funny thing about Roulette. Double-Zero. You expect every ball to come up 1 in 38 spins, right?
Every time you make a spin the ball lands somewhere, so if you buy a new Roulette wheel, the first spin will result in a 1/38 shot going 1 for 1 and 37 1/38 shots going 0 for 1.
And, you know it's going to happen, unless some cosmic force prevents you from spinning the wheel and ball.
If you know it could lose sometime, then it could lose anytime, and if the Odds are against you, it's -EV and always will be.
Actual results may vary.
Quote: Mission146Ick.
Your terms are erroneous now, the Expected Value of the game is still negative and always will be. What you are saying is that you can expect a positive Actual Return, and I would suggest that you can, with respect to probability in a limited sample size of that nature.
However, you can't use actual return to say you extracted Positive EV from a Negative EV game because that's retroactive. There's also still Variance, as in, you could lose, so as long as losing is possible, you can't take a positive actual return and make the statement that you had positive expected value because of it.
How could you do that?
If the possibility of losing exists prior to attempting the system, then you did not know that you would have a positive actual return until you conducted the system for x trials and ended up with a positive actual return. You could have lost the first one, and you see to acknowledge that you expect the system to fail once every x million years.
Funny thing about Roulette. Double-Zero. You expect every ball to come up 1 in 38 spins, right?
Every time you make a spin the ball lands somewhere, so if you buy a new Roulette wheel, the first spin will result in a 1/38 shot going 1 for 1 and 37 1/38 shots going 0 for 1.
And, you know it's going to happen, unless some cosmic force prevents you from spinning the wheel and ball.
If you know it could lose sometime, then it could lose anytime, and if the Odds are against you, it's -EV and always will be.
Actual results may vary.
Your right I misspoke. I cant expect a Positive EV just a Positive Actual Return.
First of all, when you are getting into hyper large amounts of money, you have to consider that the use of money is not linear. At some point, you should turn down a good bet because the odds are too low.
Would I risk $1 to win a billion dollars if my odds were 500 million to 1? definitely. Because a billion dollars holds meaning in the world. There are things you can do with a billion dollars that you can't do with, say, $10 million. The money is meaningful most of the way up.
Would I risk $1 to win a trillion dollars if my odds were 500 billion to 1? Definitely not. The expectation is the same as above, but there's no difference in utility between ~$10 billion and $1 trillion. That makes it a bad bet because the value of money is not linear. It's closer to logarithmic.
And all of the above is with positive expectation bets. From the other end, your end, it would be negative expectation AND lack utility. If you had a billion dollars, you don't need to risk it to win $1, especially if you will lose your billion more than once in a billion. That money can be put to use in any number of ways better than risking it to chase a small profit in a negative game.
The bottom line is risk vs reward. If you can't understand that using a large bankroll to play a negative game to chase a small profit is not good from a risk/reward perspective, you probably should just avoid the game altogether.
Quote: sodawater
The bottom line is risk vs reward. If you can't understand that using a large bankroll to play a negative game to chase a small profit is not good from a risk/reward perspective, you probably should just avoid the game altogether.
even if the risk is .00000000000000000001?
the math says so with a probability ofQuote: tehepidemick1i win $.00000001 and picking the black ball I lose $500. If I play 1 trillion times, i think I will profit.
0.99855775 to make $10,000 (but you would have to risk way more than $500. $11k gets you to 40 losses)
14 out of 10,000 trying this on average would not make it
Go for it!
Oh, it is online casino
and you trust them?
to make 10 trillion bets and win $100,000 = 0.98567075569
for a cool $1 million and 100 trillion bets = 0.8656029237
risk there
maybe this is too much info
