Search All of the Math Forum:
Views expressed in these public forums are not endorsed by
NCTM or The Math Forum.


Math Forum
»
Discussions
»
sci.math.*
»
sci.math
Notice: We are no longer accepting new posts, but the forums will continue to be readable.
Topic:
Probability game question
Replies:
6
Last Post:
Nov 29, 2000 7:03 AM




Re: Probability game question
Posted:
Nov 29, 2000 7:03 AM


Hugo van der Sanden wrote: > > Wim Benthem wrote: > > > > On 25 Nov 2000 01:54:27 GMT, amyshiningstar@aol.com (Amy Shining Star) > > wrote: > > > > >Suppose you start with $2. > > > > > >The game is as follows: > > > > > >You flip a fair coin. > > >If you get a head, you gain $1. > > >If you get a tail, you have a chance to pick a ball from an urn. > > >With probability 0.6, you will get a green ball. Then you don't lose anything. > > >With probability 0.3, you will get a blue ball. Then you will lose $1. > > >With probability 0.1, you will get a red ball. Then you will lose $2. > > > > > >Repeat the game forever. If you have a negative balance, you must stop playing. > > > > > > > > >What is the probability that the game will never stop? > > >What is the probability that you will end up in debt (game stopped)? > > > > It's possible to find an exact solution. The probability that the game > > will end is only dependent on the amount of money you have left. > > lets define P(n) is the probability that the game will end if you have > > n dollars left. > > Now we can find a differential equation for P(n) > > > > P(n) = 0.5 P(n+1) + 0.3 P(n) + 0.15 P(n1) + 0.05 P(n) > > > > as boundary conditions we have > > P(1) = 1 > > P(2) = 1 > > and lim P(n) = 0 > > n>oo > > > > if you can't solve the differential equation: look up linear > > differential equations somewhere. > > I've made some progress on this one, but I'm now stuck. > > I've defined a sequence d_n, given by: > > d_n = 0 (for n < 2), > d_2 = 1, > d_n = (14 d_{n1}  3 d_{n2}  d_{n3}) / 10 (for n > 2), > > and (if I haven't made any silly errors) this gives: > P(n) = sum_{i=n+2}^\inf{(d_{i+1}  d_i) * d_i / d_{i+1}}
A rec.puzzles post from 'r.e.s.' helped me to realise that I had made silly errors; this should have been: P(n) = sum_{i=n+2}^\inf{ (1  d_i / d_{i+1}) * d_{n+2} / d_i }
.. which simplifies easily.
> A Monte Carlo simulation suggests P(n) is very close to 1  d_{n+2}/2, > at least as tested for some small values of n.
.. and P(n) does in fact simplify to 1  d_{n+2}/2.
Hugo



