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Re: probability question

To: <s-news@wubios.wustl.edu>
Subject: Re: probability question
From: Tony Plate <tplate@blackmesacapital.com>
Date: Tue, 22 Jul 2003 11:52:14 -0600
Cc: "Horace Tso" <Horace_Tso@pgn.com>, "Leeds, Mark" <mleeds@mlp.com>
In-reply-to: <sf1c0e99.013@pgn.com>
Reply-to: tplate@acm.org
I believe that Mark Leeds' problem is different from the Sultan's dowry problem (as described at http://mathworld.wolfram.com/SultansDowryProblem.html) because in the Sultan's dowry problem, the goal is to identify the maximum dowry (the payoff is binary), whereas in the Mark Leeds' problem, the goal is to maximize the expected payoff, where the actual realized payoff is a number between 1 and 100.

These types of problems are known as "optimal stopping problems." Mark Leeds' problem is a nice easy one to solve with dynamic programming. Here's the framework of a solution written as a recursive function. (I'm sure there's also a closed-form solution to this version of the problem.)

> V <- function(k) {
     # V(k) is expected value achieved from k trials
     if (k<=0) {
         return(0)
     } else {
          # v1 is expected value achieved from k-1 trials
          v1 <- V(k-1)
          # actual code removed and replaced by explanation :-)
          # X is a random variable - an integer selected uniformly from 1:100
# The best thing to do is to stop (and accept X) if X is greater than
          # the value we expect if we continue, i.e., stop if X > v1.
          # So, our expected value if we stop is E(X | X > v1) (i.e., the mean
          # of all values between ceiling(v1) and 100).
          # The expected value if we continue is v1.
          # To calculate the expected value of V(k), we multiply these values
          # by their respective probabilities and sum.
          # return Pr(X > v1) . E(X | X > v1) + Pr(X <= v1) . v1
     }
}
> V(0)
[1] 0
> V(1)
[1] 50.5
> V(2)
[1] 63
> V(3)
[1] 70.03
> V(4)
[1] 74.671
> V(20)
[1] 92.48587
>

-- Tony Plate

At Monday 04:02 PM 7/21/2003 -0700, Horace Tso wrote:
Mark, that's also known as Sultan's Dowry Problem. You want to solve for the smallest x such that

H(x) = H(n) - 1

where n is the number of turns, and H(n) is harmonic number.

Check out : http://mathworld.wolfram.com/SultansDowryProblem.html

Cheers.

Horace Tso

>>> "Leeds, Mark" <mleeds@mlp.com> 07/21/03 03:46PM >>>
i've seen the following question before but
i don't remember how to do it so
i was wondering if anyone knows
where i can find the answer. i'm
not asking anyone to do it because
i know it's published somewhere.

you have a game : each turn,
a random number is chosen
between 1 and 100  ( assume uniform distribution )
and given to you. you can either

A) accept the number and stop  the game
and decide that the number you received is
your number.

B) throw out that number ( but you
can't claim it later ) and try for another number

you get say, 20 turns ( numbers
are chosen with replacement ) and you
want to maximize your value where
your value is the number you receive when
you choose to stop.

what is the algorithm for deciding when to stop ?

sorry to bother everyone.


                                                        mark


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Tony Plate   tplate@acm.org


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