Assumed I have infinite money and bet 10 on red or black and every time I lose I will double my bet until I win and then start again with 10 would I make profit?
I did a bit code for that:
http://jsfiddle.net/ua84gkaL/5/
[Math] roulette with infinite money
gamblingprobability
Related Solutions
Let's first look at the first question:
If you bet $1 on black for 100 consecutive spins, how much money will you end up with in expectation?
So you want to know what your final return will be, at the end of 100 spins. Call this $R$. That is just giving it a name, but what is your final return? You can see that it is the sum of the returns from each bet. So let the return on the $i$th bet be $R_i$, then note that $R = R_1 + R_2 + \dots + R_{100}$. So the expected return is $E[R] = E[R_1] + E[R_2] + \dots + E[R_{100}]$ by linearity of expectation.
So to calculate $E[R]$, we'll be done if we calculate each $E[R_i]$. Let's try to calculate a particular $E[R_i]$. You bet $1$ dollar, and you get back $2$ if the ball lands on black, and $0$ if it doesn't. In other words, you gain $1$ dollar if it lands on black, and lose $1$ dollar if it doesn't. The probability of the former is $18/38$, and that of the latter is $20/38$. In other words, $R_i$ is $1$ with probability $18/38$, and $-1$ with probability $20/38$, so the expected value of $R_i$ is $E[R_i] = \frac{18}{38}(1) + \frac{20}{38}(-1) = \frac{-2}{38}$. Now, as this is the same for each $R_i$, we have $E[R] = E[R_1] + E[R_2] + \dots + E[R_{100}] = \left(\frac{-2}{38}\right)100 \approx -5.26$.
For the second question, let the amount you walk away with be $W$. Let $p = 18/38$, the probability that your bet on black succeeds. There are $5$ possible outcomes:
- you win your first bet: probability $p$
- you lose your first bet, and win your second: probability $(1-p)p$
- you lose your first two bets, and win the third: probability $(1-p)^2p$
- you lose your first three bets, and win the fourth: probability $(1-p)^3p$
- you lose all four bets: probability $(1-p)^4$
In the first four outcomes, you walk away with $16$ dollars, so the probability of that happening (let's call it $q$) is $q = p + (1-p)p + (1-p)^2p + (1-p)^3p = 1 - (1-p)^4 = 1 - (20/38)^4 \approx 0.92$.
[More simply, you could think of it as just two outcomes: (a) that you win some bet, which has probability $q = 1 - (1-p)^4$, and (b) that you win no bet (lose all bets), which has probability $(1-p)^4$.]
In other words, $W$ is $16$ with probability $q$, and $0$ with probability $1-q$. So the expected amount of money you walk away with is therefore $E[W] = q(16) + (1-q)0 = (1-(1-p)^4)16 \approx 14.77$.
[Aside: Note that this is less than the $15$ you came in with. This shows that you can't win in expectation even with your clever betting strategy; a consequence of the optional stopping theorem.]
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 $p_{39,62}\approx0.1842$.
Best Answer
Yes you would make a profit. Assume you always bet red and to make things minimally simpler let us say you always start with bet $1$ (not $10$) and double when you lost. Let us consider the run until the first red you hit. So there are $n$ black and then red.
First you loose $1 + 2 + 2^2 + \dots + 2^{n-1}$ then you win $2^n$. The former sum is $2^n- 1$. So not matter what happens in the end you win $1$, every time that red comes up.
(This answer assumes a $1/2$ red, $1/2$ black no zeros as rules there are not uniform but it does not change the ultimate outcome for the variants I know of.)