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

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Hugo van der Sanden

Posts: 234
Registered: 12/8/04
Re: Probability game question
Posted: Nov 29, 2000 7:03 AM
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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(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 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







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