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Roulette System

Lose at roulette
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An easy system to lose at roulette:

Budget $100. Bet $5 on any number until you win. When you win then bet the winnings plus what is left of the $100 on any number.

Repeat.

Claim 1: With American roulette the average time to win back to back on the same number is 38*38=1444 attempts.

Demonstration Program: http://ideone.com/OVjjqW

zelah, Oct 06 2016

My version of code http://ideone.com/u1ev6T
[caspian, Oct 09 2016]

Second try http://ideone.com/4vbcMj
[caspian, Oct 09 2016]

[link]






       Welcome to the Halfbakery!   

       Unfortunately, this system will not work. Nor will "systems" like doubling up, and things like that. In fact, on any "fair" roulette wheel, you will lose slowly, on average. You'll lose faster on an American wheel than a European wheel, because the American wheel has two zeroes rather than one.   

       The only way to beat roulette is either to find an unbalanced wheel (in which case you can get a slight edge - but probably not enough to outweigh the house percentage); or to use Newtonian physics and some fancy maths to predict where the ball will land, just before betting closes. The latter has been tried (and works), using computers concealed in shoes etc, but most casinos are wise to this and you will not get very far.
MaxwellBuchanan, Oct 06 2016
  

       //Budget $100. Bet $5 on any number until you win. When you win then bet the winnings plus what is left of the $100 on any number.//   

       So what do you do when you've spent all the money and haven't won? There are 37 (or 38) numbers, and you've only got 20 goes in your $100, so the chances are you'll go bust. For the french table, with 37 slots:
Chance of zero successes : (36/37)^20=0.538
So over half the time you fail at the first hurdle.
Obviously, you've a 36 in 37 (or 37 in 38) chance of failure at the 'all in' phase.
  

       Probability of success : (1-0.538)*1/37=0.0114
At just over 1%, these arn't great odds.
  

       Way to win at roulette: be the house.
Loris, Oct 06 2016
  

       Only time I won on roulette was when I was showing someone how pointless gambling is. Such is life.   

       [+] to Loris, owning the casino is the best way to make money from gambling. Or being in government, or a banker...
not_morrison_rm, Oct 06 2016
  

       Methinks, zelah has a roulette wheel license.
wjt, Oct 07 2016
  

       If you calculate the mathematical expectation correctly, you might be very displeased.
MaxwellBuchanan, Oct 07 2016
  

       So then, what is the correct mathematical expectation? I think I will bet an average of $7220 before winning back an average of $8280.
zelah, Oct 07 2016
  

       //I think I will bet an average of $7220 before winning back an average of $8280//   

       I think "loooong before" would be a more appropriate time frame.   

       But hey, feel free to post a math proof, or maybe a bit of reasoning.
FlyingToaster, Oct 07 2016
  

      
zelah, Oct 07 2016
  

       Zelah, your maths is screwy.   

       I expect you are thinking of probabilities as additive where they should be multiplicative.   

       The odds of your losing 20 times in a row (i.e., blowing your whole stake) is (1-[1/37])^20. Start from there and work forward.   

       I wrote a small program that simulates this, using a European (single-zero) model. On average (averaged over 1,000,000 games), you lose. You'd lose more on a US (double-zero) wheel.   

       Also, did you not notice that (a) lots of people play roulette (b) this is a fairly simple scheme and yet (c) nobody has consistently got rich playing roulette without actually cheating?   

       It's the same basic idea as "doubling up" on an evens bet. Yes, if you continue for an infinite number of goes, you will win - but your winning becomes infinitesimal as you approach infinity. It's more likely that you will run out of stake money before winning.
MaxwellBuchanan, Oct 07 2016
  

       //It's the same basic idea as "doubling up" on an evens bet. Yes, if you continue for an infinite number of goes, you will win - but your winning becomes infinitesimal as you approach infinity. It's more likely that you will run out of stake money before winning.//   

       The difference here seems to be that I never need to bet more than I have budgeted.
zelah, Oct 07 2016
  

       There's two different plays here, the "before" winning strategy ($5) and the "after" winning strategy ($all-in) But assuming you win once, before running out of your $100, that means putting your entire stake on a single spin - since the odds are still 37/1 (or 38/1) - can you explain the reasoning that suggests changing strategies after the initial win? i.e. why not play all-in on your initial spin?
zen_tom, Oct 07 2016
  

       Yes, but you will lose most of the time. If you play once, starting with $100, there is a roughly 57% chance that you will go through it all without a win.   

       But suppose you win on your 10th spin. You now have $180 in winnings, plus $50 left. You bet $185 on the next spin. You will probably (36/37) lose, leaving you with $45. And you will probably lose that $45 (in 9x $5 bets) without winning again.   

       Even if you win on that $185 bet (winning $185 x 36), you will then bet all your winnings plus $5 on your next spin, and will probably (36/37) lose, leaving you with $40...   

       Etc. etc. etc.   

       You can consider other permutations but, statistically, you lose overall. Seriously - I'm sure you can program in some language or another, so write a script and test it yourself.
MaxwellBuchanan, Oct 07 2016
  

       //But suppose you win on your 10th spin. You now have $180 in winnings, plus $50 left. You bet $185 on the next spin. You will probably (36/37) lose, leaving you with $45. And you will probably lose that $45 (in 9x $5 bets) without winning again.   

       Even if you win on that $185 bet (winning $185 x 36), you will then bet all your winnings plus $5 on your next spin, and will probably (36/37) lose.//   

       Yes - but this is not my system. You would bet $185+$50 on one number.
zelah, Oct 07 2016
  

       Would anyone care to challenge Claim 1?
zelah, Oct 07 2016
  

       //You would bet $185+$50 on one number.//   

       In that case, you are probably (36/37) going to lose, in which case you're out of cash, end of.   

       Seriously, it is much easier to just model this in software if you want to see for yourself.
MaxwellBuchanan, Oct 07 2016
  

       what's the payout on roulette wheels ?
FlyingToaster, Oct 07 2016
  

       Excellent, if you're the casino ...
8th of 7, Oct 07 2016
  

       //what's the payout on roulette wheels ?//   

       The payout is directly equivalent to the odds - but only after you disregard the zero (or zeroes). So, if you bet on red (or on black) or on odd (or even) the payout is 2:1; if you bet on one of the 36 numbers, it's 36:1; etc.   

       The house makes its share from the zero (or zeroes): if you bet on red, or on a number, and it comes up zero, the house wins. Because American roulette has two zeroes, the house take is twice as high as in European roulette, so that people can be in and out of the casino as fast as possible.
MaxwellBuchanan, Oct 07 2016
  

       Roulette is a school in applied probability; its cost depends on how you play and how you learn. If you pretend to play, and really pay attention, you can learn very cheaply. If you really play, but only pretend to pay attention, the tuition will take everything you have.
lurch, Oct 07 2016
  

       I have shown computer code demonstrating the effectiveness of the method. Would anyone care to critique my code or at least share their own?
zelah, Oct 08 2016
  

       Yeah, I'm having a look. Thanks for putting up the code
caspian, Oct 09 2016
  

       Your original code didn't make you lose the $100 when you win a bet, and bet the winnings and remains of the $100, and lose it.   

       Modified version: http://ideone.com/u1ev6T
caspian, Oct 09 2016
  

       Why the two random generation if tests? Surely the real world example is only one if. The test either being a random against a const choice or matching two random numbers.
wjt, Oct 09 2016
  

       //Your original code didn't make you lose the $100 when you win a bet, and bet the winnings and remains of the $100, and lose it.//   

       The code you post is incorrect since there is a path through your program which has you lose the 100 dollars twice however my code does necessarily take away the 100 dollars.
zelah, Oct 09 2016
  

       Hey, can I get the Fields Medal already for destroying mathematics? Just putting it out there.
zelah, Oct 09 2016
  

       Oops. Does my second try look right?
caspian, Oct 09 2016
  

       New critique: Your original code continued to bet the $100 when you win a bet, and bet the winnings and remains of the $100, and lose it. It didn't distinguish that case from when you simply hasn't won yet.   

       [wjt] not sure if that's addressed to me, But the reason I have two "if" conditions is one is for the $5 bet, and the other one is for being the winnings plus what's left of the $100
caspian, Oct 09 2016
  

       Caspian is correct. The betting scheme is DOA. Too bad it could not work but perhaps there is something to this "mathematics" thing after all. I am eternally grateful for this since I was almost ready to go to the casino. Very disappointed!!
zelah, Oct 09 2016
  

       [either caspian or zelah] I'm still not understanding the code. Don't you have to take the second $5 off your $100 if you are only doing 20 bets. This is probably just my stupidity.
wjt, Oct 09 2016
  

       If you mean a second $5 for the next bet after a winning $5 bet, no, the next bet isn't for $5, it's for all that remains of the $100. If you lose you jump out of the loop that counts down the $100, because there's none left.   

       If you mean a second $5 for the next $5 bet after losing the $5, that happens next time around the loop, there isn't another bet until then.
caspian, Oct 09 2016
  

       Welcome -zelah. Better luck next time!
xenzag, Oct 09 2016
  

       From the code It looks like the betting is stopped if there is the 'rest in' bet win even though the rest amount is returned and the $5 loops can carry on.   

       The key to the method, I thought, was repeat until $100 no more.
wjt, Oct 09 2016
  

       Enter with 100$. Worst senario leave with 0$ (usual). Best senario 10 211 280 550 505 056 301 907 102 295 654 400$ (heart of gold with fresh cuppa). This assumes the casino doesn't forcibly remove you.
wjt, Oct 10 2016
  
      
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