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

 Messages: [ Previous | Next ]
 Hugo van der Sanden Posts: 234 Registered: 12/8/04
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:
>

> >
> >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(n-1) + 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_{n-1} - 3 d_{n-2} - d_{n-3}) / 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

Date Subject Author
11/24/00 Amy Shining Star
11/24/00 Steve Lord
11/25/00 mike@nospam.com
11/25/00 mike@nospam.com
11/26/00 Wim Benthem
11/28/00 Hugo van der Sanden
11/29/00 Hugo van der Sanden