How do you find the Probability of something like this?
October 20th, 2010 at 10:30:11 AM
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I'm not touting this betting progression as a winner, just simply asking what is the probability of it losing? I know in the long run, it will lose more or less the house edge of whatever game it's applied to, however I just want to know.
This progression is called the "Carsch Progression" and is named from the user who created it, and first posted it on Gamblers Glen. It's a Star System/Fibo variant.
For every loss, you move down the progression, for every win, you repeat the last bet. Win two in a row and start over. WLWL doesn't hurt you.. So how do you calculate the probability of losing the progression when ostensibly, you could play 1000s of decisions and neither win nor lose?
Progression: 1,1,2,3,4,6,9,13,21,32
-TheArchitect
This progression is called the "Carsch Progression" and is named from the user who created it, and first posted it on Gamblers Glen. It's a Star System/Fibo variant.
For every loss, you move down the progression, for every win, you repeat the last bet. Win two in a row and start over. WLWL doesn't hurt you.. So how do you calculate the probability of losing the progression when ostensibly, you could play 1000s of decisions and neither win nor lose?
Progression: 1,1,2,3,4,6,9,13,21,32
-TheArchitect
October 20th, 2010 at 10:40:00 AM
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It would depend on how "losing" is defined. Whatever the answer, the answer is probably best arrived at by random simulation.
"For with much wisdom comes much sorrow." -- Ecclesiastes 1:18 (NIV)
October 20th, 2010 at 10:44:17 AM
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Thank you for responding Wizard.
Losing would be defined by losing at the 32unit bet.
I can calculate the probability of losing all 10 bets in a row, however it gets tricky when you add in non progression clearing wins, and double wins to clear. I agree, a simulation would likely solve it rather easily.. Of course I'm one of the simulation challenged members of your great forum.
-TheArchitect
Losing would be defined by losing at the 32unit bet.
I can calculate the probability of losing all 10 bets in a row, however it gets tricky when you add in non progression clearing wins, and double wins to clear. I agree, a simulation would likely solve it rather easily.. Of course I'm one of the simulation challenged members of your great forum.
-TheArchitect
October 20th, 2010 at 11:27:02 AM
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Does anyone out there have an easy simulation for this? -TheArchitect
October 20th, 2010 at 12:08:31 PM
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Quote: TheArchitectI'm not touting this betting progression as a winner, just simply asking what is the probability of it losing? I know in the long run, it will lose more or less the house edge of whatever game it's applied to, however I just want to know.
This progression is called the "Carsch Progression" and is named from the user who created it, and first posted it on Gamblers Glen. It's a Star System/Fibo variant.
For every loss, you move down the progression, for every win, you repeat the last bet. Win two in a row and start over. WLWL doesn't hurt you.. So how do you calculate the probability of losing the progression when ostensibly, you could play 1000s of decisions and neither win nor lose?
Progression: 1,1,2,3,4,6,9,13,21,32
-TheArchitect
Well, I am sure I have made a bunch of technical mistakes below, so, the final answer is probably off, but here is the idea.
Let p be the probability of a single loss, and f(k) the probability of a loss given that you are k steps from the end, and N be the total number of steps (10 in this case).
Then:
f(1) = p + (1-p)^2*f(N)
f(2) = p*f(1) + (1-p)^2*f(N)
...
f(n) = p*f(n-1) + (1-p)^2*f(N)
...
Let F=(1-p)^2*f(N) just for brevity:
f(n) = p*f(n-1) + F = p^2*f(n-2)+pF + F = p^3*f(n-3) + p^2*F + p*F + F = ... = p^(n-1)*f(1) + F*sum([k=0;n-2] p^k)
f(n) = p^n + F*(1-p^(n-2))/(1-p) = p^n + f(N)*(1-p)*(1-p^(n-2))
So F(N) = p^N/(1 - (1-p)*(1-p^(N-2)))
In particular F(10) = p^10/(1-(1-p)*(1-p^8))
For example, if p=0.51, then F(10) = 0.002324 (take this with a grain of salt, because, like I said, I dunno how many errors I made along the way, but the basic idea is, I think, correct).
"When two people always agree one of them is unnecessary"
October 20th, 2010 at 12:10:21 PM
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Quote: TheArchitectI'm not touting this betting progression as a winner, just simply asking what is the probability of it losing?
Certainty.
The fact that a believer is happier than a skeptic is no more to the point than the fact that a drunken man is happier than a sober one. The happiness of credulity is a cheap and dangerous quality.---George Bernard Shaw
October 20th, 2010 at 2:32:44 PM
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Quote: TheArchitectFor every loss, you move down the progression, for every win, you repeat the last bet.
I'm not sure I understand how you move up the progression, if you repeat the last bet with every win.
Simplicity is the ultimate sophistication - Leonardo da Vinci
October 20th, 2010 at 3:57:52 PM
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Quote: AyecarumbaI'm not sure I understand how you move up the progression, if you repeat the last bet with every win.
That was going to be my question. On the other hand, the sequence itself seems very similar to the Fibonacci sequence, and I've had some very, very big wins playing a straight positive progression on the Fibonacci.
One of the interesting things about the Fibonacci is that, when applied to a streak of winning bets, the rate of wager increase and bankroll increase are equal. Contrast that with a 100% parlay where your wager increase is maximal but you always lose the unit you started with unless you hit the table max.
"In my own case, when it seemed to me after a long illness that death was close at hand, I found no little solace in playing constantly at dice."
-- Girolamo Cardano, 1563
October 20th, 2010 at 7:44:13 PM
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I should be more clear with the rules.
When you play this progression and lose a wager, you move to the next level. When you win at any level (except the first 1unit bet) you will have to repeat that beat and win to clear the progression (similar to the fibo where you have to win two consecutive bets to clear). If you lose the "repeat" bet, you do not progress.. So say you make it to the 3 unit bet and win, your next bet is 3 units and lose, you do not progress down, you repeat the 3 unit bet again.. This is where I'm having a difficult time simulating this progression..
In theory, you could find yourself in a WLWL situation for a large number of bets without winning or losing..
Any suggestions?
When you play this progression and lose a wager, you move to the next level. When you win at any level (except the first 1unit bet) you will have to repeat that beat and win to clear the progression (similar to the fibo where you have to win two consecutive bets to clear). If you lose the "repeat" bet, you do not progress.. So say you make it to the 3 unit bet and win, your next bet is 3 units and lose, you do not progress down, you repeat the 3 unit bet again.. This is where I'm having a difficult time simulating this progression..
In theory, you could find yourself in a WLWL situation for a large number of bets without winning or losing..
Any suggestions?
October 21st, 2010 at 5:22:56 AM
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Quote: TheArchitectI should be more clear with the rules.
When you play this progression and lose a wager, you move to the next level. When you win at any level (except the first 1unit bet) you will have to repeat that beat and win to clear the progression (similar to the fibo where you have to win two consecutive bets to clear). If you lose the "repeat" bet, you do not progress.. So say you make it to the 3 unit bet and win, your next bet is 3 units and lose, you do not progress down, you repeat the 3 unit bet again.. This is where I'm having a difficult time simulating this progression..
In theory, you could find yourself in a WLWL situation for a large number of bets without winning or losing..
Any suggestions?
So, what is a "win" then? You got it that you "lose" when you get to the end of the sequence, but if you don't get there, when do you stop, and say that you won? After any two wins in a row?
If so, the probability of losing one bet is:
f1 = p + (1-p)*p*f1
(either you lose straight away (p) or you win (1-p), then lose (p), and then start over (f1)).
Solve it for f1:
f1 = p/(1-p(1-p))
Your sequence is 10 steps long, so the probability of getting to the end of it before returning to the start f1^10, or (p/(1-p(1-p)))^10 (about 2.1% if p =0.51)
"When two people always agree one of them is unnecessary"
October 21st, 2010 at 5:29:19 AM
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A Win is any two wins in a row.
October 21st, 2010 at 5:44:05 AM
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Quote: TheArchitectA Win is any two wins in a row.
I added the formula to my previous post, that I think should give you the probability of getting to the end of the sequence before winning twice in a row:
P = (p/(1-p(1-p)))^10, where p is the probability of a single loss.
"When two people always agree one of them is unnecessary"
October 21st, 2010 at 7:29:37 AM
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And here is the simulation. I wrote it in perl, because it's the best language ever, but it's fairly simple, so it should not be hard to convert to whatever you prefer. You can adjust the length of the sequence ($n) and the probability of the loss ($p) at the top of the script. It runs infinitely, and prints out the current result every 1 million experiments.
The result matches the numbers I get from my formula, which makes me think both are correct this time :)
The result matches the numbers I get from my formula, which makes me think both are correct this time :)
#!/usr/bin/perl
$num=10;
$p = 0.51;
$total=$lost=$res=0;
@seq = (1,1,2,3,4,6,9,13,21,32);
EXP: while(1)
{
for($n = 0; $n < $num;)
{
($res-=$seq[$n++],next) if rand() <= $p;
$res+=$seq[$n];
($res+=$seq[$n], next EXP) if rand() > $p;
$res-=$seq[$n];
}
$lost++;
}
continue
{
$total++;
printf("%d: %8.6f%%, %5.2f\n", $total, $lost*100/$total, $res) unless $total%1000000;
}
"When two people always agree one of them is unnecessary"
October 21st, 2010 at 9:17:50 AM
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Being a mac user, where can I use your perl code to simulate it for myself?
October 21st, 2010 at 10:11:00 AM
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perl comes with Macs. Open a terminal and run perl -v to confirm.
"In my own case, when it seemed to me after a long illness that death was close at hand, I found no little solace in playing constantly at dice."
-- Girolamo Cardano, 1563
October 22nd, 2010 at 8:31:25 PM
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I normally count at the mohegansun and usually hold my own, still up for this year. Tonite I started playing and couldn't get any advantage counts. I decided what the hell I was going to try this betting progression. One unit was $ 100 and playing two hands. I know all the stats about any kind of system beating the house advantage, and I still believe that the only way to win in the long run is by being an advantage player, but if anyone is interested, I started down the nite $1200 while counting, then went completely into this betting mode and in three shoes won $4000. I will still count but this betting progression does have some legs. By the way I never lost more than two hands in a row thru 3 shoes. I left at the high point.
IF YOU PLAY "PLAY TO WIN"
October 23rd, 2010 at 6:50:39 AM
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Quote: MathExtremistperl comes with Macs. Open a terminal and run perl -v to confirm.
Hey guys,
Thank you for taking the time to share your programming knowledge with us mortals. I tried running weaselman's perl script in my terminal, however I keep getting errors. they all look like this:
"-bash: =10: command not found"
Again, I'm, not touting this as a winning betting system, the holy grail, or anything.. I am just looking for why it will fail.
Thanks again for everyone's help and input on this.
-TheArchitect
October 23rd, 2010 at 9:21:31 AM
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You need to save it in a file, not just paste into the terminal.
Put the text into the file, make the file executable (chmod +x filename) and run (by typing ./filename in the command line)
Put the text into the file, make the file executable (chmod +x filename) and run (by typing ./filename in the command line)
"When two people always agree one of them is unnecessary"
October 23rd, 2010 at 9:43:00 AM
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Weaselman,
Thank you for all of your great help. I got it to work (rather easily once I knew what in the world I was doing..)
It's printed out the P Probability of it losing which hovered around 2.1%..
Is there perhaps a way to modify the script to simulate actual bankroll totals and/or bust outs? (as more of a wagering simulation and closer to what the Wiz or BlueJay would for their challenge)
Sorry, you just seem to know how to do this rather easily, as where I look at it like a monkey doing arithmetic..
Thank you again,
-TheArchitect
Thank you for all of your great help. I got it to work (rather easily once I knew what in the world I was doing..)
It's printed out the P Probability of it losing which hovered around 2.1%..
Is there perhaps a way to modify the script to simulate actual bankroll totals and/or bust outs? (as more of a wagering simulation and closer to what the Wiz or BlueJay would for their challenge)
Sorry, you just seem to know how to do this rather easily, as where I look at it like a monkey doing arithmetic..
Thank you again,
-TheArchitect
October 23rd, 2010 at 12:33:17 PM
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I modified the script in my original post to keep track of the bankroll, and print out the running total (assuming the win pays even money).
"When two people always agree one of them is unnecessary"
October 23rd, 2010 at 1:28:29 PM
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Quote: weaselmanI modified the script in my original post to keep track of the bankroll, and print out the running total (assuming the win pays even money).
I cannot thank you enough for taking you time and helping me with this. I really appreciate it.
I'm looking at your code and looking a basic perl commands on a tutorial page to try and learn enough to make modifications to this and to be able to write new ones as needed. One question I do have however is, how would I adjust the code for less than 1:1 payouts? ie. the banker bet in baccarat?
Again, thank you very much
-TheArchitect
October 23rd, 2010 at 1:58:40 PM
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look for commands that look like $res+=$seq[$n] (there are two of them). Insert a multiplier before each $seq to change the payout to your liking. For example, if the bet pays 3:2, change both those commands to look like $res+=1.5*$seq[$n], if it pays 1:2, make it $res+=0.5*$seq[$n] etc.
"When two people always agree one of them is unnecessary"
October 23rd, 2010 at 2:24:50 PM
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Quote: weaselmanlook for commands that look like $res+=$seq[$n] (there are two of them). Insert a multiplier before each $seq to change the payout to your liking. For example, if the bet pays 3:2, change both those commands to look like $res+=1.5*$seq[$n], if it pays 1:2, make it $res+=0.5*$seq[$n] etc.
Again, my hat's off to you weaselman. Thank you for all of your patience and help.
-TheArchitect
