Re: 15 out of 16 times (math, not laundry)

[ph@netcom.com (Peter Hendrickson)]

>>Pretend the casino is run out of a church.  "Parishioners" arrive and
>>enter a confessional to place their bets.  The "priest" cannot see who
>>is placing each bet.  Each "parishioner" plays until he or she is
>>broke.  "Parishioners" arrive at a steady rate and will do so
>>indefinitely.

> Let me just make sure I understand what you mean.  I believe you are saying:

>  Conjecture A:

>    A.1   As parishoners play and leave, the division of wealth approaches the
>	   `odds' of the game.  Thus if the odds are .51 house (of God), .49
>	   parishoner, then eventually the house will end up with 51 cents
>	   out of every dollar `played'.  Just as it would if the church were
>	   playing against one very wealthy parishoner (i.e., the `world').

>    A.2   Since there are a large number of parishoners, enough games can
>	   always be played to make the distribution match the odds.

>   If this is _not_ what you mean to say then I apologize for missing your
>   point; read no further---just send me explanations to clear up my
>   mis-understanding.  If Conjecture A is accurate statement of your belief,
>   then please step across this line.

I agree with both conjectures.

>   ----------

> Let me walk through your model, one parishoner at a time.  Please read this
> with an open mind; it could be true.

>> Each "parishioner" plays until he or she is broke.

> Lets say the odds of the game are .51 to .49.  Each parishoner has $100. 
> Each parishoner plays until broke.

> At some point in play, the distribution of wealth with respect to _that
> player_ may be arbitrarily close to c=$51, p=$49.  What, though, is the
> distribution at the _end_ of that game?  Since each game only ends when the
> p=$0, the distribution is c=$100, p=$0.  On to the next parishoner.

> After the 9th, but before the 10th parishoner, the distribution must be
> c=$900, p[10]=$100.  It can't be worse than that for the church, or we
> wouldn't have moved on to the 10th parishoner.  It can't be better for the
> player because each has only $100 to wager.  After the n'th, c=$100n,
> p[n+1]=$100.

> Conjecture A predicts that as n, the number of players, goes to infinity,
> c, the fraction of money won by the church, approaches C, the probability
> the church will win a single trial.  But in fact, the model shows that as n
> approaches infinity, c goes to 1.

There is a slight difference between what Conjecture A predicts and
this statement.  Conjecture A predicts that as b, the number of bets,
goes to infinity the fraction of bets won will approach C, the
probability that the church will win a single trial.

> Where could one disagree with this interpretation of the model?

You should think about what you mean by "fraction of money".  I think
there is a seductive error here.  In one sense, we mean the amount of
money placed on bets, but we also mean the actual bank notes in play.
These concepts address two different things.  Whether or not banknotes
are recycled by the parishioners will not affect the church's
winnings.

> [...Deleted parts which I think are answered above...]

>>The chance of the "church" to win or lose is the same on every
>>bet, regardless of who places it.

> That is true.  But the only way the player can realize his mathematical
> expectations is if he is allowed to continue playing even after he is out
> of money (i.e., so he can climb back out of the hole).  

Each parishioner has a high probability of losing their savings and a
low probability of winning everything owned by the church.  It is
possible for any single parishioner to win everything, but it is
unlikely.

> Ok, the first player goes out, but the infinity of players after him
> can make up for that, right?  Wrong, because on his way to winning
> back the first players money, if the second player goes broke, _his_
> game is over.  Now its up the third guy, ad infinitum
> (literally)..... just because the series is infinite doesn't mean
> the sum is.

> No set of players, all of whom go broke, break the church.  Therefore, for
> the series to end it must be instigated by a set of players that includes
> at least one who doesn't go broke (i.e., the church goes broke instead). 
> In fact, a single player who doesn't go broke ends the series without any
> help from other players.

> Thus, to stem the tide of pious donations (i.e., the church's
> winnings), a single player with enough money to `outlast' the church
> is required.

The player needs to be lucky.

Let's say the church's assets are H dollars.  In order for it to lose
everything, it has to have a series of bets whose sum is a negative
value less than -H.  This series has a beginning - the point at which
the church's assets dropped below H and moved down to 0.  If
parishioners play until they win or are broke, the player who took the
church below H will be the same player who wins everything.  (I am
assuming fixed size bets, but the conclusions can be generalized.)

This player wins because he or she was fortunate enough to place the
first bet in the series.  Having more capital means that more bets can
be placed.  That increases the probability of placing the first bet in
the winning series, but does not affect the odds of the church losing
everything.

> Hope you found this interesting but not insulting,

I found it interesting.  Your message was written clearly.

I've seen this question and similar ones come up again and again in
discussions of gambling, trading, and insurance.  It would be nice if
having a large body of capital would allow one to "make money off the
noise", but it isn't so.

It has been observed that small traders in the futures markets tend to
lose money to large traders.  One way this has been explained is that
the large traders outlast the small traders with their larger capital
and that is how they make money.  I think a more likely explanation is
that the large traders tend to make good trades.

Peter