Conference rusure::math

Assume first that the events the gamblers are betting on meets the usual
test for a random process: stationary and independent.  This assumption 
will not be dropped, but further assumptions should be.  Call the probability
of winning on each event p, and the probability of losing q.

Assume the dollars won is equal to the dollars lost in each bet.

Assume that the gamblers bet the same amount on each event.

Assume also that the gamblers and the house have deep pockets.

Assume that you are dealing with a sufficiently large population of gamblers.

Then consider the set of gamblers who bet exactly n times.  Their f(x) will 
be given by a binomial distribution.  For n sufficiently large, this can be
approximated by a normal distribution, with mean np and variance=npq.  If we
scale x using n, then the mean will become p and the variance pq/n.  All these
f(x) will add up to a total f(x) which will be roughly bell shaped, but not
normal.  It will remain uni-modal.

Drop the assumption that the gamblers bet the same amount on each event and 
allow them to bet based on the outcome of previous events (not bets).  This
will not change their winnings, because the process is independent, so it 
will not change f(x).

Drop the assumption that their bets depend on their past success.  This will
change the result greatly.

Assume that each gambler decided to quit when their stake reaches 0$ or z$.
Then the bell shaped distribution will be modified by two spikes at 0$ and z$.
If the gamblers play long enough, the bell shape will go to zero and the spikes
will total probability 1, so you will have a bimodal distribution.  The relative
size of the two spikes is given by the solution to the gambler's ruin problem.
I was just looking at this last week, but I don't remember the details.

Other strategies have other effects.  Doubling bets on a winning or losing
streak can put a lot of gamblers far from the mean.  Cutting bets in half as
you approach the limits can increase the peak at the mean.

I think a lot of this could be analyzed for small n by a good algebraist.

Re: -.1	You have some good ideas.   I have a couple of comments.

>Then consider the set of gamblers who bet exactly n times.  Their f(x) will 
>be given by a binomial distribution.  For n sufficiently large, this can be
>approximated by a normal distribution, with mean np and variance=npq.  If we
>scale x using n, then the mean will become p and the variance pq/n.  All these
>f(x) will add up to a total f(x) which will be roughly bell shaped, but not
>normal.  It will remain uni-modal.

	I presume that you add the *unscaled* f(x).   The sum should be a
weighted sum depending on the population, p(n), of gamblers who have played n 
times.   p(n) can be presumed to be monotonically decreasing.   The
distribution of the total f(x) will not be symmetrical, but I can believe that
it will be uni-modal.

>Assume that each gambler decided to quit when their stake reaches 0$ or z$.
>Then the bell shaped distribution will be modified by two spikes at 0$ and z$.
>If the gamblers play long enough, the bell shape will go to zero and the spikes
>will total probability 1, so you will have a bimodal distribution.  The 
>relative size of the two spikes is given by the solution to the gambler's 
>ruin problem.   I was just looking at this last week, but I don't remember 
>the details.

	I would be interested in the details of the formalization, and
would be willing to have a crack at the algebra.   I suspect there are also
gamblers whose unconscious strategy is simply to quit when their stake reaches
0$!   This is the limiting case as z scoots off to infinity.

	It occurs to me that I had been assuming that roulette was a cheaper
game than a state lottery, because there's only a 3% rake-off in roulette,
whereas in a lottery it can be 50% or so.   But if your strategy is a 0/z one,
then you can achieve that variance very quickly in a lottery, while in roulette
you may need many bets, and to *repeatedly* bet the same money many times, thus
incurring a much higher final cost.

Regards,
Andrew.

       If we stop each gambler after some random number of wagers (different 
    for each gambler), the distribution is still bell-shaped -- binomial, I 
    think.  This is true when the other conditions are similar to the first 
    part of .1, i.e.

   	. all gamblers have same initial stake
        . each wager is a constant amount
   	. all gamblers have deep pockets
   	. each wager has equal chance of winning or losing
   	  the wagered amount

   The binomial will keep cropping up as long as the gambling experiments
   are Bernoulli (success or failure) trials.  For tyros such as myself, 
   the old science fair marble and pegboard experiment is a good
   intuitive model:

       If we have a board with staggered rows of pegs, laid on its side,
   (see illustration below) and start dropping marbles down a chute, the
   marbles will accumulate at the bottom in a bell-shaped curve.  Assume 
   that every marble hits some peg in every row, and that every hit
   results in an equal chance of bouncing left (losing) or right
   (winning) one column width as it falls down to the next row.

       The bell-shaped distribution seems natural when we realize that
   for a marble to end up all the way on the left or right requires
   that all bounces be in one direction.  Of course it's far more likely
   that at least some bounces will be toward the center.

       A handy point of analogy: the more rows (more trials per gambler)  
   the more columns will  be needed, as the increased number of rows allows 
   some (small) number of marbles to move to the extreme right or left.

       Illustration:
                                         +-- Drop marbles in here
         o = peg                         |
       | | = accumulating column	 |
                                         V
   	        +------------------------ ----------------------+
   		|o o o o o o o o o o o o o o o o o o o o o o o o|
   		| o o o o o o o o o o o o o o o o o o o o o o o |
   		|o o o o o o o o o o o o o o o o o o o o o o o o|
   		| o o o o o o o o o o o o o o o o o o o o o o o |
   		|o o o o o o o o o o o o o o o o o o o o o o o o|
   		| o o o o o o o o o o o o o o o o o o o o o o o |
   		|o o o o o o o o o o o o o o o o o o o o o o o o|
   		| o o o o o o o o o o o o o o o o o o o o o o o |
   	        | | | | | | | | | | | | | | | | | | | | | | | | | 
   		| | | | | | | | | | | | | | | | | | | | | | | | | 
   		| | | | | | | | | | | | | | | | | | | | | | | | | 
   		| | | | | | | | | | | | | | | | | | | | | | | | | 
   		| | | | | | | | | | | | | | | | | | | | | | | | | 
   	        +-----------------------------------------------+

> I think this will seldom be exactly true, and will be approximately true only 
> for specific rules for choosing the random number.  
    
    I should have been more clear.  If the random numbers are uniformally
    distributed, allowing any one gambler to make n wagers, where 1=<n<=k
    (k being the global maximum for all gamblers), then this will indeed
    result in a bell-shaped distribution when both k and the number of
    gamblers are  sufficiently large.  A few quick simulation experiments
    bear this out.
    
    But whether it approaches being binomial in this case, I really don't
    know.

.2>	I would be interested in the details of the formalization, and
>would be willing to have a crack at the algebra.   

As promised, here is a little about the classical ruin problem, from

	_An Introduction to Probability Theory and Its Applications_,
	William Feller,  Wiley, New York, 1957.

	By the way, I don't like this book much.  It is too mathematical
	and too 'frequentist' for me, but it does treat a lot of interesting
	examples.  Readers in this conference would probably find the book 
	easier to understand than I do, but other texts may also cover this 
	problem.


I'm sure I don't understand all of what follows, but here it is.  You can read
the book or fill in the details yourself.


A gambler makes a series of $1 bets, winning with probability p and losing with
probability q=1-p.  He starts with initial stake z and plans to quit when his
stake is 0 or a.  We seek the probability of ruin q(z) or the probability of
success p(z).

After the first step, the stake is z-1 or z+1, so we have the difference
equation

	q(z) = p * q(z+1) + q * q(z-1)

For convenience, we express the boundary conditions as q(0)=1 q(a)=0.

The difference equation has the two general solutions 

	q(z)=1		q(z) = (q/p)^z

The general solution is a linear combination of these, and the particular
solution satisfying the boundary conditions is 

	q(z) = ( (q/p)^a - (q/p)^z ) / ( (q/p)^a - 1 )

For the special case q=p=1/2, we can use l'Hopital's rule to get

	q(z) = 1 - z/a

Feller gives several methods for solving the general and special problems, and
then applies the techniques to a number of related problems.