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Topic:
Probability game question
Replies:
6
Last Post:
Nov 29, 2000 7:03 AM




Re: Probability game question
Posted:
Nov 28, 2000 6:52 PM


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 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.
The roots of the characteristic equation of d_n are a = (2 + sqrt(14)) / 10, b = (2  sqrt(14)) / 10 and c = 1, which gives me:
d_n = ((a  b)  a^n(1  b) + b^n(1  a))/((a  b)(1  a)(1  b))
.. which further simplifies to:
d_n = 2  (10 / sqrt(14)) (a^n (1  b)  b^n (1  a))
I also have:
d_{n+1}  d_n = (a^n  b^n) / (a  b)
.. but for the other factor, the best I have is:
d_n / d_{n+1} = 1  5 (a^n  b^n) / (2 sqrt(14)  a^n(3 + sqrt(14)) + b^n(3  sqrt(14)))
.. and I don't yet see how to simplify the product of these two enough to evaluate or telescope the sum. I'd appreciate a hint.
Hugo



