Ok, I did some google searches, and found the following entry from what appears to be a work on probability and statistics:
Towards the end of the page, there is a section on the “Gambler’s Ruin” problem with an infinitely rich adversary. This seems to me to be very close to the situation described in the OP.
Anyway, the book defines “p” to be the gambler’s probability of winning and “q” to be his probability of losing. He starts off with stake “z,” and the question is asked what is the probability of eventual ruin?
When q >= p, eventual ruin is basically certain. However, when p > q, the probability is given as (q/p)^z. So if the gambler starts out with a decent-sized stake and a decent edge over the house, there is a good chance he will never get to zero.
This of course assumes that the text I found is legit.
Agree, which is why I said it was “very close to the situation described in the OP.”
But it seems analogous – in either case, there’s a good chance that you’ll drift away forever.
Actually, it occurs to me that one can choose a gambler’s ruin problem that is isomorphic to the problem posed in the OP. Just choose p and q such that if you had a mini-gambler’s ruin problem where you start with 1 chip and the two absorbing states are 0 and 3, the odds of ruin are 50/50.
You can do this by taking the second formula from the bottom of page 488 in the text I linked to and setting q[sub]z[/sub] equal to one half; Z equal to 1; and M equal to 3. If you to this and solve for (q/p), it seems you get a cubic equation -
(q/p)^3 - 2 * (q/p) + 1 = 0.
The only root that makes sense is if q/p = (sqrt(5)-1) /2.
Substituting this figure into the formula under “infinitely rich adversaries” on the next page gives us the probability of eventual ruin as
[(sqrt(5) - 1)/2] ^1000
My calculator gives this figure as 1.02 x 10^(-209).
So by my calculation, the odds of the number in the OP ever hitting zero are approximately one in 10 ^ 208.
I discount it because it doesn’t make any sense. If q/p =1, then p=q = 1/2. If you start with 1 chip, and you’re asking whether you’ll get to 3 or to 0 first, and p = q = 1/2, then the odds of “ruin” are gonna be greater than 50/50.
Besides, if you substitute 1 for q/p into the equation I used, you end up dividing by zero.
**
What’s an order of magnitude between friends?
(Actually, I hit the “1/x” key on my calculator and read off the answer.)
Imagine that you start with the number 1000. You then flip a coin, heads you add 2 to the number, tails you subtract 1. My question is this. If you flip the coin an infinite number of times will the number ever be reduced to 0?
The best odds, are getting the 1,000 tails initially, since by giving heads +2, for every tail -1. At a certain point, I fail to see why it would be necessary to carry the experiment any further. To calculate the odds of one coin toss of getting tails is easy enough: 1/2. For two tails in a row it’s 1/4, for three, 1/8 and so on. Ten tails in a row is 1/1024, or 1/2 to the tenth power. Right? It just keeps doubling. I calculated what the odds would be for 26 tails in a row, simply doubling my last entry each time before my 8 decimal calculator ran out of space. 67,108,864/1 odds of just getting 26 tails in a row. Rounding out, that’s one in 67 million. This could be roughly the same odds of winning a lottery, perhaps. And this is with millions and millions of people playing this game, for maybe a couple of winners. And again, this is only the odds for 26 tails coming up in a row. Even if billions of people played the game every day, I doubt any lottery would even come close to ever making a pay out on a lottery that needed something to the equivalent of getting 1,000 tails in a row to win. The number would be 1/2 to the 1,000th power. Not impossible, because one could give odds on it, but a number this size is so astronomical, as to pretty much make it so, I would think.
I do believe that Phi[sup]-1000[/sup] is the answer. On the other hand, though, p = q = 1/2 does make a certain amount of sense, because those are actually the odds of success and failure; the difference is how much you get as a result. So, I’m thinking, I’m not totally convinced that the analysis done in that book isn’t actually applicable to this situation. I’ll think about it some more, though.
The thing is, I was constructing a gambler’s ruin problem with the same chance of ruin as the number in the OP eventually hitting zero. Think about it – in a gambler’s ruin problem where p actually equals q, the chance of eventual ruin is 1.
:smack: I was using a calculator to find the roots instead of turning my brain on. Never mind.
I considered going to the extreme case where P(+2)=1 and seeing that one is still a solution, where it can be discounted, but I don’t think this helps.
Well thanks for the feedback. It’s probably pretty obvious that I am a biologist not a mathematician and I am thus way out of my depth in this kind of discussion, I understand these things are difficult to address with logic (but I can’t do the math ). My reasoning was along the lines of if it’s possible to do something and you keep doing it for long enough it’s bound to happen eventually. But I can certainly see where the opposing viewpoint is coming from especially as concisely expressed in the last post. Could anyone who has been calculating the probability give a fairly concise explanation of their workings suitable for us non-genius types? Thanks
). My reasoning was along the lines of if it’s possible to do something and you keep doing it for long enough it’s bound to happen eventually. But I can certainly see where the opposing viewpoint is coming from especially as concisely expressed in the last post. Could anyone who has been calculating the probability give a fairly concise explanation of their workings suitable for us non-genius types? Thanks