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I came up with a strategy, which immitates the martingale(doubling) strategy , but it seems that there is a good probability to gain profit, which I want to compute.

So this is how we start:

We do not bet on black/red, but we chose columns, or dozens. Let's say we start with 100 chips.

  • Bet $1$ chip in e.g. a column. The probability to win is $12/37$. If you win, start over again, while if you lose:
  • Bet again $1$ chip to a column. Why again $1$? This is because we will still have profit if we win ( I will have won 3 chips, and lost 2 ). Again, if win go to step $1$, if lose:
  • Bet $2$ chips in a column. Each time we bet the lowest amount of funds, such that we have profit in case we win. In this case, if we win we will have played $4 $ chips , and won $6$, which gives the minimum profit($2 $ chips).

Continuing with this strategy, one has to bet in columns/dozens, the following sequence of chips (each time he loses), in order to have the minimum profit : $$1,1,2,3,4,6,9,14,21,31,47\dots$$ So, here is my question:

Using this strategy, and starting with $100$ chips, what is the probability of gaining $+50\%$, and what is the probability of doubling? You continue until you have doubled, and stop if you have less chips than they are required to put the next bet.

*We have to consider that when we gain some profit, e.g. we have reached $140$ chips, we are able to bet more chips of the above sequence. With $100$ chips we can lose up to 9 times, and then bet $31$ chips. But with $140$ chips we can lose one more time , and then bet $46$.

Due to the complexity of the problem, I would also appreciate a numerical solution, but I do not have the knowledge to run a simulation.

Anyone can figure out a solution using probability theory?

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"The martingale strategy is well-known, however one needs a great amount of initial funds, in order to win, in the long term"... Sorry to disappoint you but the "martingale strategy" (or ANY strategy) makes that you LOSE, in the long term. –  Did May 9 '13 at 10:59
    
What is the stopping rule? Will you continue on until you have doubled? If that is the case, I suspect the answer to both the +50% and doubling probabilities is 1. –  CommonerG May 9 '13 at 10:59
    
@Did Even with the martingale strategy there is a probability to win. (remove the "long term" if you don't like it) –  Dimitrios Ntalampekos May 9 '13 at 11:01
    
@CommonerG You continue until you have doubled, and stop if you have less chips than they are required to put the next bet –  Dimitrios Ntalampekos May 9 '13 at 11:02
    
I see now, I overlooked the bit at the end about number of chips. –  CommonerG May 9 '13 at 11:02
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2 Answers

up vote 2 down vote accepted
+50

We either end as winners with 150 or 151 chips or as losers with between 0 and 50 chips (if we lose early, with between 0 and 33 chips; in general the upper limit is $\frac13$ of the last peak). In a fair game this would mean a winning probability between $\frac12$ and $\frac35$ (just by the relation of up and down movements $+50:-50$ to $+50:-67$). Even this rough estimate (that does not take into account the bank advantage) matches well with Sharkoe's simulation.

A more precise calculation for $p_{k,d}$, the probability to reach $150$ when starting with $k$ and trying to win at least $d$ in the first sequence

  • $p_{k,d}=0$ if $2k<d$
  • $p_{k,1}=1$ if $k\ge 150$
  • $p_{k,d}=\frac{12}{37}p_{k+2b,1}+\frac{25}{37}p_{k-b,d+b}$ with $b=\lceil \frac d2\rceil$

This allows us to compute the probybility exactly as a fraction. However, numerator and denominator have about $785$ digits, so the numerical value from rounding that fraction $$ p_{100,1}\approx0.56097114279613511032732301110367086534$$ should be good enough for us.

Code in PARI/GP (with memoization for values $p_{k,1}$, $100\le k\le 150$), edited to use less memory):

 Start=100;Target=150
 A=vector(Target-Start+1,n,-1);
 getP(k,d)={local(pkd);
    if(2*k<d, 0, if(k>=Target, 1, 
       pkd=if(d==1&&A[k-Start+1]>=0, 
             A[k-Start+1], 
             12/37*getP(k+2*ceil(d/2),1)+25/37*getP(k-ceil(d/2),d+ceil(d/2))
          );
       if(d==1,A[k-Start+1]=pkd);
       pkd))
    }
 getP(Start,1) +.0

The new code after editing does not reflect it, but the "worst" case among the cases occuring is $p_{39,62}\approx0.1842$.

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Assuming I didn't code a billion bugs, running a quick 100,000 samples gave me:

100 -> 150 case: $0.55989 \pm 0.00156976$.

100 -> 200 case: $0.37052 \pm 0.0015272$.

For a smaller case:

10 -> 15 case: $0.56233 \pm 0.00156881$.

10 -> 20 case: $0.39492 \pm 0.00154583$.

For comparison,

$12/37 = 0.324324$

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Thank you for your answer! I would expect that the more chips you have in the beginning, the less probability you have to lose them all. –  Dimitrios Ntalampekos May 12 '13 at 18:46
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