# Double or Halve Riddle [closed]

You start off with a \$100 pool. Every second, your pool doubles in value with probability $$p$$ or gets cut in half with probability $$1-p.$$ If you drop below 98.76543210¢ at any point, you lose and the host of the game show laughs at you. Furthermore, you stubbornly refuse to stop playing. What's the probability that you eventually lose? Note: Assume that you and the host are both immortal, and that thermodynamic restrictions do not prevent gameplay in the longue duree. ## 1 Answer Meta: This is in effect a standard question in the theory of random walks. Does that make it off-topic as a textbook problem? I don't think so. The relevant meta-question is this one whose excellent and highest-voted answer gives this closely related question as an example of something that's a "math puzzle" rather than a "math problem". OK then. Assuming that there's no weird business with rounding of fractional cents, this question is equivalent to the following: You start with a number $$x=7$$. You repeatedly add 1 (with probability $$p$$) or subtract 1 (with probability $$1-p$$). What's the probability that you eventually reach $$x=0$$? (Why 7? Because 6 halvings of \$100 take you to \$1.5625 (above the threshold in the question) and 7 halvings take you to \$0.78125 (below the threshold).)

And the answer, which you can find in any suitable textbook, is

The probability is 1 if $$p\leq\frac12$$; otherwise it's $$\bigl(\frac{1-p}{p}\bigr)^7$$.

If you don't want to go out and find a suitable textbook, you can also deduce it from the accepted answer to that question I linked above: see that formula for $$c_1,c_2$$ and the way in which the answer to the original question is derived from it, and then put $$a=-7$$ instead of $$a=-1$$.

Meta again: Does that mean that this should have been closed as a duplicate? I think it's just different enough not to be: you can figure out the answer to this one from the answer to that one, but you can't just see it there.