1
$\begingroup$

Suppose I'm gambling using a strategy of doubling my bet whenever I lose to recoup my losses from previous bets. If my initial bet is 1/2048 of my capital, I can bet 10 times before I run out of money. Statistics says that the chances of this happening are ~1/718. As the law of large numbers states, though, this is in no way a guarantee that such an outcome will manifest in any certain number of iterations.

Is it possible, therefore, to calculate the chances that this doesn't happen? For example, is there an equation we can put x (the number of bets) into and determine the likelihood of this 1/718 chance not occurring? My chances of losing within the first ten bets would be extremely small, and my chances after ten thousand would be pretty high, so is there any way to calculate the middle ground?

Improve this question
$\endgroup$

1 Answer 1

Reset to default
3
$\begingroup$

  1. What bets are you making that the chance of losing ten bets out of ten is about 1/718?

  2. I'm pretty sure the law of numbers doesn't actually state what you say it does. (The claim is nevertheless true, though - it's not certain at any finite $n$.)

  3. $P(E)+P(\overline{E})=1$ (where $\overline{E}$ is the complement of $E$, or 'not-$E$'). So if the 1/718 chance of losing it all in ten bets was accurate, then the chance of not losing it all in ten bets is $1-\frac{_1}{^{718}}$. It's a direct consequence of the axioms of probability - see the second item here (the notation there - $\Omega\backslash E$, essentially "everything but $E$" - is the same thing as $\overline E$).

Improve this answer
$\endgroup$
3
  • $\begingroup$ I don't know if I did the math correctly but I thought the formula for finding the probability of an event happening x times in a row was desired/(total)^x In my case the odds of winning are 45% and I think the 718 may have actually been if I were betting 1/1024 of my capital. Does that number sound reasonable? $\endgroup$
    – Matt
    Commented Jul 5, 2014 at 3:37
  • 1
    $\begingroup$ The odds of winning what? Where does the 45% come from? Why wasn't that in your question? (Were we supposed to guess that?) Neither 0.45^10 nor 0.55^10 are 1/718, and probabilities are not odds. Odds are ratios of probabilities (odds against some event $E$, for example, is $(1-P(E))/P(E)$). $\endgroup$
    – Glen_b
    Commented Jul 5, 2014 at 4:25
  • $\begingroup$ 1/718 would be the probability of losing ten times in a row if you chance of losing were about 51.8% $\endgroup$
    – Glen_b
    Commented Jul 5, 2014 at 7:36

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.