I am a German student who will finish his A-levels in around half a year, and for quite a long time I have this question on mind that I'm unable to find an answer to.

You have a value y=100 Now, you get a random number between 1 and 100. If the number is between 1 and 45, subtract 1 from y. If the number is between 46 and 100, add 1 to y. The game is stopped when y = 0. The random number part is now repeated infinitely, only stopped when y reaches 0.

Now, what is the probability of y ever reaching zero? Is it 1 or is it a certain percentage, and if it is, is there a way to calculate it? (Those numbers are just examples by the way, the key is just that the probability of it falling is lower than the one of it rising)

I'm really looking forward to your answers, I'm keen on finding out how this works!

  • $\begingroup$ Obviously the probability is not $\infty$, you meant to ask whether it is $1$, or more formally, whether $y$ will reach zero almost certainly. $\endgroup$ – Mark Fischler Dec 20 '16 at 23:10
  • $\begingroup$ Look into gamblers ruin problem - This is a well studied problem. One thing I have to ask is why do you need $1$ to $45$ ? $\endgroup$ – Chinny84 Dec 20 '16 at 23:10
  • $\begingroup$ Thank you Mark, I obviously meant 1 and not Infinity. I edited it now. $\endgroup$ – Pocket Quads Dec 20 '16 at 23:16
  • $\begingroup$ Also, those numbers are just examples as well, could be anything as long as it's lower than 50. Also, I have looked into gamblers ruin quite a lot, in fact this is where I got this question from the first place. The problem is that I can't seem to find any solution for infinite iterations. $\endgroup$ – Pocket Quads Dec 20 '16 at 23:19
  • $\begingroup$ One approach: Calculate the expected "value" on iteration $t$ (i.e., $100 + 55/44 t$) and the variance, which goes as $1/\sqrt{t}$. $\endgroup$ – David G. Stork Dec 20 '16 at 23:23

The probability of ever reaching $y=0$ will be $$ \left( \frac{45}{55}\right)^{100} $$ which is about 2 chances in a billion.

  • $\begingroup$ Thanks a lot. Anyone who wants to know why this is true as well should have a look at the PDF that was recommended in the comments of the question. $\endgroup$ – Pocket Quads Dec 21 '16 at 5:40

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