probability in the casino












2












$begingroup$


I have a question regarding probability.



Lets assume that roulette has just red and black, no green and an equal amount of both.



Lets also assume we start each game with a bet of 1 dollar.
We also have enough dollars to be able to double our bet up to 40 times.



Now an example game is we bet our 1 dollar on black.
If we lose we bet 2 dollars on black, the total expenditure is now 3 dollars.
If we lose again we bet 4 dollars on black, the total expenditure is now 7 dollars.



If we win we get 8 dollars back. 1 of which is a profit.



Remember we can lose 40 times in a row before we cannot double up any longer.



The odds of getting 2 reds in a row is 25% so getting 40 reds in a row would be very low.



My question is what is the probability of making a profit after one entire game ( a game is where you keep doubling until you can or you profit)



and after 1000 games.






probability







share|cite|improve this question









$endgroup$












  • $begingroup$
    Just remember before setting off to the casino with your doubling scheme that although you will win a little very often, when you loose you loose a LOT. Your expected return is still 0 on your even roulette wheel, and still negative on a wheel with one or two zeroes.
    $endgroup$
    – Tom Collinge
    Mar 11 '14 at 14:33






  • 2




    $begingroup$
    In practice this strategy is foiled by casino bet limits.
    $endgroup$
    – MJD
    Mar 11 '14 at 14:51












  • $begingroup$
    Congratulations, you just discovered the en.wikipedia.org/wiki/Martingale_(betting_system) While it will win in the long run, if there are no zer0s, it will win you only one betting unit. E.g. you will lose 1 + 2 + 4 + 8 (= 15) to win 16, or lose 31 to win 32, etc
    $endgroup$
    – Mawg
    Jan 11 '16 at 14:14
















2












$begingroup$


I have a question regarding probability.



Lets assume that roulette has just red and black, no green and an equal amount of both.



Lets also assume we start each game with a bet of 1 dollar.
We also have enough dollars to be able to double our bet up to 40 times.



Now an example game is we bet our 1 dollar on black.
If we lose we bet 2 dollars on black, the total expenditure is now 3 dollars.
If we lose again we bet 4 dollars on black, the total expenditure is now 7 dollars.



If we win we get 8 dollars back. 1 of which is a profit.



Remember we can lose 40 times in a row before we cannot double up any longer.



The odds of getting 2 reds in a row is 25% so getting 40 reds in a row would be very low.



My question is what is the probability of making a profit after one entire game ( a game is where you keep doubling until you can or you profit)



and after 1000 games.






probability







share|cite|improve this question









$endgroup$












  • $begingroup$
    Just remember before setting off to the casino with your doubling scheme that although you will win a little very often, when you loose you loose a LOT. Your expected return is still 0 on your even roulette wheel, and still negative on a wheel with one or two zeroes.
    $endgroup$
    – Tom Collinge
    Mar 11 '14 at 14:33






  • 2




    $begingroup$
    In practice this strategy is foiled by casino bet limits.
    $endgroup$
    – MJD
    Mar 11 '14 at 14:51












  • $begingroup$
    Congratulations, you just discovered the en.wikipedia.org/wiki/Martingale_(betting_system) While it will win in the long run, if there are no zer0s, it will win you only one betting unit. E.g. you will lose 1 + 2 + 4 + 8 (= 15) to win 16, or lose 31 to win 32, etc
    $endgroup$
    – Mawg
    Jan 11 '16 at 14:14














2












2








2





$begingroup$


I have a question regarding probability.



Lets assume that roulette has just red and black, no green and an equal amount of both.



Lets also assume we start each game with a bet of 1 dollar.
We also have enough dollars to be able to double our bet up to 40 times.



Now an example game is we bet our 1 dollar on black.
If we lose we bet 2 dollars on black, the total expenditure is now 3 dollars.
If we lose again we bet 4 dollars on black, the total expenditure is now 7 dollars.



If we win we get 8 dollars back. 1 of which is a profit.



Remember we can lose 40 times in a row before we cannot double up any longer.



The odds of getting 2 reds in a row is 25% so getting 40 reds in a row would be very low.



My question is what is the probability of making a profit after one entire game ( a game is where you keep doubling until you can or you profit)



and after 1000 games.






probability







share|cite|improve this question









$endgroup$




I have a question regarding probability.



Lets assume that roulette has just red and black, no green and an equal amount of both.



Lets also assume we start each game with a bet of 1 dollar.
We also have enough dollars to be able to double our bet up to 40 times.



Now an example game is we bet our 1 dollar on black.
If we lose we bet 2 dollars on black, the total expenditure is now 3 dollars.
If we lose again we bet 4 dollars on black, the total expenditure is now 7 dollars.



If we win we get 8 dollars back. 1 of which is a profit.



Remember we can lose 40 times in a row before we cannot double up any longer.



The odds of getting 2 reds in a row is 25% so getting 40 reds in a row would be very low.



My question is what is the probability of making a profit after one entire game ( a game is where you keep doubling until you can or you profit)



and after 1000 games.






probability




probability






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Mar 11 '14 at 13:26









user134679user134679

111




111












  • $begingroup$
    Just remember before setting off to the casino with your doubling scheme that although you will win a little very often, when you loose you loose a LOT. Your expected return is still 0 on your even roulette wheel, and still negative on a wheel with one or two zeroes.
    $endgroup$
    – Tom Collinge
    Mar 11 '14 at 14:33






  • 2




    $begingroup$
    In practice this strategy is foiled by casino bet limits.
    $endgroup$
    – MJD
    Mar 11 '14 at 14:51












  • $begingroup$
    Congratulations, you just discovered the en.wikipedia.org/wiki/Martingale_(betting_system) While it will win in the long run, if there are no zer0s, it will win you only one betting unit. E.g. you will lose 1 + 2 + 4 + 8 (= 15) to win 16, or lose 31 to win 32, etc
    $endgroup$
    – Mawg
    Jan 11 '16 at 14:14


















  • $begingroup$
    Just remember before setting off to the casino with your doubling scheme that although you will win a little very often, when you loose you loose a LOT. Your expected return is still 0 on your even roulette wheel, and still negative on a wheel with one or two zeroes.
    $endgroup$
    – Tom Collinge
    Mar 11 '14 at 14:33






  • 2




    $begingroup$
    In practice this strategy is foiled by casino bet limits.
    $endgroup$
    – MJD
    Mar 11 '14 at 14:51












  • $begingroup$
    Congratulations, you just discovered the en.wikipedia.org/wiki/Martingale_(betting_system) While it will win in the long run, if there are no zer0s, it will win you only one betting unit. E.g. you will lose 1 + 2 + 4 + 8 (= 15) to win 16, or lose 31 to win 32, etc
    $endgroup$
    – Mawg
    Jan 11 '16 at 14:14
















$begingroup$
Just remember before setting off to the casino with your doubling scheme that although you will win a little very often, when you loose you loose a LOT. Your expected return is still 0 on your even roulette wheel, and still negative on a wheel with one or two zeroes.
$endgroup$
– Tom Collinge
Mar 11 '14 at 14:33




$begingroup$
Just remember before setting off to the casino with your doubling scheme that although you will win a little very often, when you loose you loose a LOT. Your expected return is still 0 on your even roulette wheel, and still negative on a wheel with one or two zeroes.
$endgroup$
– Tom Collinge
Mar 11 '14 at 14:33




2




2




$begingroup$
In practice this strategy is foiled by casino bet limits.
$endgroup$
– MJD
Mar 11 '14 at 14:51






$begingroup$
In practice this strategy is foiled by casino bet limits.
$endgroup$
– MJD
Mar 11 '14 at 14:51














$begingroup$
Congratulations, you just discovered the en.wikipedia.org/wiki/Martingale_(betting_system) While it will win in the long run, if there are no zer0s, it will win you only one betting unit. E.g. you will lose 1 + 2 + 4 + 8 (= 15) to win 16, or lose 31 to win 32, etc
$endgroup$
– Mawg
Jan 11 '16 at 14:14




$begingroup$
Congratulations, you just discovered the en.wikipedia.org/wiki/Martingale_(betting_system) While it will win in the long run, if there are no zer0s, it will win you only one betting unit. E.g. you will lose 1 + 2 + 4 + 8 (= 15) to win 16, or lose 31 to win 32, etc
$endgroup$
– Mawg
Jan 11 '16 at 14:14










2 Answers
2






active

oldest

votes


















2












$begingroup$

After one game, you make a dollar unless you lose $40$ times in a row, which happens $2^{-40}$ of the time, about once in $10^{12}$ tries. After $1000$ games, you have to win them all, so you will lose approximately $1000 cdot 2^{-40}$ of the time, or about once in a billion tries. This is not exact-it is actually smaller because some of the probability comes from two losses. The chance of that is greater than zero, but very small.






share|cite|improve this answer









$endgroup$





















    -2












    $begingroup$

    The probability of you losing after an entire game is $frac{1}{2}^{40}$, because you have to get 40 reds in a row to lose, and the probability of getting a red is $frac{1}{2}$. The probability of making a profit is then $1-frac{1}{2}^{40}$.



    Starting with 1$, the (maximum) amount of money you could lose after an entire game is $2^{40}-1$(lost 1$ after round one, 3$ after round two, 7$ after round three...), if you win once, you earn 1$. So playing this game you have a chance of $frac{1}{2}^{40}$ to lose $2^{40}-1$$, and a chance of $1-frac{1}{2}^{40}$ to win 1$, which means that if you repeated the full game forever (restocking to be able to play 40 rounds again), you would neither gain money nor lose money, because $frac{1}{2}^{40}times (2^{40}-1) = (1-frac{1}{2}^{40})times 1$, as worked out below:



    Lose: $frac{1}{2}^{40}times (2^{40}-1) = frac{1}{2}^{40}times 2^{40} - frac{1}{2}^{40} = 1 - frac{1}{2}^{40}$



    Win: $(1-frac{1}{2}^{40})times 1 = 1-frac{1}{2}^{40}$



    Lose = Win: $1 - frac{1}{2}^{40} = 1 - frac{1}{2}^{40}$






    share|cite|improve this answer











    $endgroup$













    • $begingroup$
      A careful reading of the Question will show you that winning (making a profit) does not depend on getting $40$ reds in a row.
      $endgroup$
      – hardmath
      Apr 18 '18 at 16:25










    • $begingroup$
      @hardmath Oops, thanks for pointing that out, I had given the probability of losing instead of that of winning.
      $endgroup$
      – alvitawa
      Apr 19 '18 at 19:43











    Your Answer





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    2 Answers
    2






    active

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    2 Answers
    2






    active

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    active

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    active

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    2












    $begingroup$

    After one game, you make a dollar unless you lose $40$ times in a row, which happens $2^{-40}$ of the time, about once in $10^{12}$ tries. After $1000$ games, you have to win them all, so you will lose approximately $1000 cdot 2^{-40}$ of the time, or about once in a billion tries. This is not exact-it is actually smaller because some of the probability comes from two losses. The chance of that is greater than zero, but very small.






    share|cite|improve this answer









    $endgroup$


















      2












      $begingroup$

      After one game, you make a dollar unless you lose $40$ times in a row, which happens $2^{-40}$ of the time, about once in $10^{12}$ tries. After $1000$ games, you have to win them all, so you will lose approximately $1000 cdot 2^{-40}$ of the time, or about once in a billion tries. This is not exact-it is actually smaller because some of the probability comes from two losses. The chance of that is greater than zero, but very small.






      share|cite|improve this answer









      $endgroup$
















        2












        2








        2





        $begingroup$

        After one game, you make a dollar unless you lose $40$ times in a row, which happens $2^{-40}$ of the time, about once in $10^{12}$ tries. After $1000$ games, you have to win them all, so you will lose approximately $1000 cdot 2^{-40}$ of the time, or about once in a billion tries. This is not exact-it is actually smaller because some of the probability comes from two losses. The chance of that is greater than zero, but very small.






        share|cite|improve this answer









        $endgroup$



        After one game, you make a dollar unless you lose $40$ times in a row, which happens $2^{-40}$ of the time, about once in $10^{12}$ tries. After $1000$ games, you have to win them all, so you will lose approximately $1000 cdot 2^{-40}$ of the time, or about once in a billion tries. This is not exact-it is actually smaller because some of the probability comes from two losses. The chance of that is greater than zero, but very small.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Mar 11 '14 at 13:32









        Ross MillikanRoss Millikan

        297k23198371




        297k23198371























            -2












            $begingroup$

            The probability of you losing after an entire game is $frac{1}{2}^{40}$, because you have to get 40 reds in a row to lose, and the probability of getting a red is $frac{1}{2}$. The probability of making a profit is then $1-frac{1}{2}^{40}$.



            Starting with 1$, the (maximum) amount of money you could lose after an entire game is $2^{40}-1$(lost 1$ after round one, 3$ after round two, 7$ after round three...), if you win once, you earn 1$. So playing this game you have a chance of $frac{1}{2}^{40}$ to lose $2^{40}-1$$, and a chance of $1-frac{1}{2}^{40}$ to win 1$, which means that if you repeated the full game forever (restocking to be able to play 40 rounds again), you would neither gain money nor lose money, because $frac{1}{2}^{40}times (2^{40}-1) = (1-frac{1}{2}^{40})times 1$, as worked out below:



            Lose: $frac{1}{2}^{40}times (2^{40}-1) = frac{1}{2}^{40}times 2^{40} - frac{1}{2}^{40} = 1 - frac{1}{2}^{40}$



            Win: $(1-frac{1}{2}^{40})times 1 = 1-frac{1}{2}^{40}$



            Lose = Win: $1 - frac{1}{2}^{40} = 1 - frac{1}{2}^{40}$






            share|cite|improve this answer











            $endgroup$













            • $begingroup$
              A careful reading of the Question will show you that winning (making a profit) does not depend on getting $40$ reds in a row.
              $endgroup$
              – hardmath
              Apr 18 '18 at 16:25










            • $begingroup$
              @hardmath Oops, thanks for pointing that out, I had given the probability of losing instead of that of winning.
              $endgroup$
              – alvitawa
              Apr 19 '18 at 19:43
















            -2












            $begingroup$

            The probability of you losing after an entire game is $frac{1}{2}^{40}$, because you have to get 40 reds in a row to lose, and the probability of getting a red is $frac{1}{2}$. The probability of making a profit is then $1-frac{1}{2}^{40}$.



            Starting with 1$, the (maximum) amount of money you could lose after an entire game is $2^{40}-1$(lost 1$ after round one, 3$ after round two, 7$ after round three...), if you win once, you earn 1$. So playing this game you have a chance of $frac{1}{2}^{40}$ to lose $2^{40}-1$$, and a chance of $1-frac{1}{2}^{40}$ to win 1$, which means that if you repeated the full game forever (restocking to be able to play 40 rounds again), you would neither gain money nor lose money, because $frac{1}{2}^{40}times (2^{40}-1) = (1-frac{1}{2}^{40})times 1$, as worked out below:



            Lose: $frac{1}{2}^{40}times (2^{40}-1) = frac{1}{2}^{40}times 2^{40} - frac{1}{2}^{40} = 1 - frac{1}{2}^{40}$



            Win: $(1-frac{1}{2}^{40})times 1 = 1-frac{1}{2}^{40}$



            Lose = Win: $1 - frac{1}{2}^{40} = 1 - frac{1}{2}^{40}$






            share|cite|improve this answer











            $endgroup$













            • $begingroup$
              A careful reading of the Question will show you that winning (making a profit) does not depend on getting $40$ reds in a row.
              $endgroup$
              – hardmath
              Apr 18 '18 at 16:25










            • $begingroup$
              @hardmath Oops, thanks for pointing that out, I had given the probability of losing instead of that of winning.
              $endgroup$
              – alvitawa
              Apr 19 '18 at 19:43














            -2












            -2








            -2





            $begingroup$

            The probability of you losing after an entire game is $frac{1}{2}^{40}$, because you have to get 40 reds in a row to lose, and the probability of getting a red is $frac{1}{2}$. The probability of making a profit is then $1-frac{1}{2}^{40}$.



            Starting with 1$, the (maximum) amount of money you could lose after an entire game is $2^{40}-1$(lost 1$ after round one, 3$ after round two, 7$ after round three...), if you win once, you earn 1$. So playing this game you have a chance of $frac{1}{2}^{40}$ to lose $2^{40}-1$$, and a chance of $1-frac{1}{2}^{40}$ to win 1$, which means that if you repeated the full game forever (restocking to be able to play 40 rounds again), you would neither gain money nor lose money, because $frac{1}{2}^{40}times (2^{40}-1) = (1-frac{1}{2}^{40})times 1$, as worked out below:



            Lose: $frac{1}{2}^{40}times (2^{40}-1) = frac{1}{2}^{40}times 2^{40} - frac{1}{2}^{40} = 1 - frac{1}{2}^{40}$



            Win: $(1-frac{1}{2}^{40})times 1 = 1-frac{1}{2}^{40}$



            Lose = Win: $1 - frac{1}{2}^{40} = 1 - frac{1}{2}^{40}$






            share|cite|improve this answer











            $endgroup$



            The probability of you losing after an entire game is $frac{1}{2}^{40}$, because you have to get 40 reds in a row to lose, and the probability of getting a red is $frac{1}{2}$. The probability of making a profit is then $1-frac{1}{2}^{40}$.



            Starting with 1$, the (maximum) amount of money you could lose after an entire game is $2^{40}-1$(lost 1$ after round one, 3$ after round two, 7$ after round three...), if you win once, you earn 1$. So playing this game you have a chance of $frac{1}{2}^{40}$ to lose $2^{40}-1$$, and a chance of $1-frac{1}{2}^{40}$ to win 1$, which means that if you repeated the full game forever (restocking to be able to play 40 rounds again), you would neither gain money nor lose money, because $frac{1}{2}^{40}times (2^{40}-1) = (1-frac{1}{2}^{40})times 1$, as worked out below:



            Lose: $frac{1}{2}^{40}times (2^{40}-1) = frac{1}{2}^{40}times 2^{40} - frac{1}{2}^{40} = 1 - frac{1}{2}^{40}$



            Win: $(1-frac{1}{2}^{40})times 1 = 1-frac{1}{2}^{40}$



            Lose = Win: $1 - frac{1}{2}^{40} = 1 - frac{1}{2}^{40}$







            share|cite|improve this answer














            share|cite|improve this answer



            share|cite|improve this answer








            edited Dec 22 '18 at 22:59

























            answered Apr 18 '18 at 15:51









            alvitawaalvitawa

            13




            13












            • $begingroup$
              A careful reading of the Question will show you that winning (making a profit) does not depend on getting $40$ reds in a row.
              $endgroup$
              – hardmath
              Apr 18 '18 at 16:25










            • $begingroup$
              @hardmath Oops, thanks for pointing that out, I had given the probability of losing instead of that of winning.
              $endgroup$
              – alvitawa
              Apr 19 '18 at 19:43


















            • $begingroup$
              A careful reading of the Question will show you that winning (making a profit) does not depend on getting $40$ reds in a row.
              $endgroup$
              – hardmath
              Apr 18 '18 at 16:25










            • $begingroup$
              @hardmath Oops, thanks for pointing that out, I had given the probability of losing instead of that of winning.
              $endgroup$
              – alvitawa
              Apr 19 '18 at 19:43
















            $begingroup$
            A careful reading of the Question will show you that winning (making a profit) does not depend on getting $40$ reds in a row.
            $endgroup$
            – hardmath
            Apr 18 '18 at 16:25




            $begingroup$
            A careful reading of the Question will show you that winning (making a profit) does not depend on getting $40$ reds in a row.
            $endgroup$
            – hardmath
            Apr 18 '18 at 16:25












            $begingroup$
            @hardmath Oops, thanks for pointing that out, I had given the probability of losing instead of that of winning.
            $endgroup$
            – alvitawa
            Apr 19 '18 at 19:43




            $begingroup$
            @hardmath Oops, thanks for pointing that out, I had given the probability of losing instead of that of winning.
            $endgroup$
            – alvitawa
            Apr 19 '18 at 19:43


















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            Journaliste

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            Pour les articles homonymes, voir Journaliste (homonymie). Journaliste Journalistes Appellation Journaliste Secteur d'activité Média - Communication Professions voisines Journaliste sportif - Reporter-photographe - Éditorialiste Codes CITP 2451, 2642 IDEO (France) 10197 ROME (France) E1106 modifier - modifier le code - modifier Wikidata Un journaliste est une personne qui recueille, écrit ou distribue des informations. L'activité d'un journaliste est le journalisme. Un journaliste peut travailler sur des sujets généraux ou se spécialiser dans certaines questions. La plupart des journalistes ont tendance à se spécialiser et, en collaborant avec d'autres journalistes, rédigent un article ou mettent en forme un reportage qui couvre l'actualité. Par exemple, un journaliste sportif couvre les nouvelles dans le monde du sport, mais ce journaliste peut aussi faire partie d'un journal qui couvre d'autres su...
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