7

that I can make 10 times of what I initially have?

Here's the formal description. In a fair gamble, I lose or double my wager each with probability 1/2. No matter how much money I have, I always gamble half of my money (the money is infinitely divisible, so that I'm never ruined). If I start with 1 dollar, and I win once I own 10 dollars or more, what's the probability of winning?

Let $X_n$ be the money I have after the nth round. Thus $X_{n+1} = 1.5X_n$ or $0.5X_n$ each with p = 1/2, and the process is a martingale. I tried to use the optional stopping theorem, by setting $\tau$ as the time that my money goes over 10 or goes below $\epsilon$ for the first time, and then let $\epsilon$ approach 0. The problem is, $X_{\tau}$ never actually equals 10. In fact it lies in the interval [10,15). So the theorem would only give a bound on the desired probability:

$E[X_{\tau}] = E[X_{\tau}|X_{\tau} \geq 10] \Pr (X_{\tau} \geq 10) + E[X_{\tau}|X_{\tau} \leq \epsilon] \Pr (X_{\tau} \leq \epsilon) $

Next I take $y$ as the natural log of money and let $f(y)$ be the probability I can win with initial log-money $y$. The following formulas looked promising:

$f(y) = 1, y \geq \ln 10$

$f(y) = 1/2 + f(y - \ln 2), \ln(20/3) \leq y < \ln 10$

$f(y) = f(y - \ln 2)/2 + f(y + \ln (3/2))/2, y < \ln(20/3)$

but still I can't find the right analytic expression of $f(y)$. I'm not even sure if $f$ is continuous at $y=\ln 10$ or $y = \ln (20/3)$.

Any help is appreciated!

wangtiance
  • 71
  • 1
    Here is one thing you could do: you can set your betting strategy in a way such that, it is capped at 10. so you are currently at 8 dollars, then you would bet 2 rather than 4. In this case, the optimal sampling theorem would apply and your chance of winning is exactly $x/10$ where $0\leq x\leq 10$ is your starting capital. otherwise, it'd very difficult to compute the probability explicitly and I do not know how. – Lost1 Jun 29 '14 at 21:23
  • I do not really understand your $f$. how did you get these? get the computer to fit a graph? I am very sceptical by these experessions here and I see no reason whatsoever to see why $f$ should be discontinuous anywhere. – Lost1 Jun 29 '14 at 21:23
  • you might also find http://math.stackexchange.com/questions/612571/can-you-make-money-on-coin-tosses-when-the-odds-is-against-you interesting. – Lost1 Jun 29 '14 at 21:27
  • Thanks. The bet is always set to be 50% of my capital. The $f$ is obtained by the following argument: with 1/2 chance, my capital shrink by 50% or increase by 50%. Therefore, the probability that I win given log-capital y is the average of that given log-capital y+ln(3/2) or y - ln2 – wangtiance Jun 29 '14 at 22:06
  • Oh i see, sorry, i did not look carefully enough, they were in fact formulae. i thought they were explicit expressions. I do not believe it has an analytic expression, tbh. – Lost1 Jun 29 '14 at 22:18
  • $f(y)$ is trivially discontinuous at $\log 10$ and $\log \frac {20} 3$ and any other points that can be reached in finite number of steps from $\log 10$ in the backwards way. This set is a countable dense set, so I don't expect a neat expression for $f$, But you know that your win either happens soon or never happens, so running simulation for 5-10-20 steps and estimating the left-over part will give you an excellent approximation. – A.S. Mar 08 '16 at 06:37

1 Answers1

1

(Using pounds instead of dollars to simplify formatting) So the rule is: You start with £1 In each round, you bet half your money and win and lose with probability 0.5 each, except when you have £10 or more you stop betting.

After n rounds, the expected amount of money in your pocket is £1 since the game is fair. Since you "win" when you have £10 in your pocket, the probability of winning is at most 1/10. Since you actually have from £10 to 315 in your pocket when you win, the probability of winning is actually between 1/10 and 1/15.

The only approach that I would recommend is a computer program where you count the possible outcomes, for example after five rounds you have 0, 1, 2, 3, 4, 5 wins with certain probabilities, after six rounds you have 0 to 5 wins out of 6, or finish with 6 wins and £11.390625 probability 1/64. The sum over (probability of winning x pounds, multiplied by x) over all possible x is 1, so after calculating the odds after each round you find a new upper and lower bound for the probability of winning.

(For example, after six rounds you win with £11.390625 with probability 1/64; the product is 0.177978515625. £1 - £0.177978515625 = £0.822021484375. The probability of winning is therefore between 1/64 + 0.822021484375/15 and 1/64 + 0.822021484375/10 or between 0.0704264322916667 and 0.0978271484375 instead of between 0.0666666666666 and 0.1; these boundaries will converge).

gnasher729
  • 10,113