From: collins@newton.apple.com (Scott Collins)

To: ph@netcom.com (Peter Hendrickson)

Message Hash: d7200b1cce0b58813a7f2a71bf8a864ab550486a9a419edc10b4450dc16f6e83

Message ID: <9404192201.AA13670@newton.apple.com>

Reply To: *N/A*

UTC Datetime: 1994-04-19 23:58:18 UTC

Raw Date: Tue, 19 Apr 94 16:58:18 PDT

```
From: collins@newton.apple.com (Scott Collins)
Date: Tue, 19 Apr 94 16:58:18 PDT
To: ph@netcom.com (Peter Hendrickson)
Subject: Re: 15 out of 16 times (math, not laundry)
Message-ID: <9404192201.AA13670@newton.apple.com>
MIME-Version: 1.0
Content-Type: text/plain
>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.
----------
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.
Where could one disagree with this interpretation of the model?
a. Maybe the church has 10 confessionals, or 1000, or 10,000.
Serializing the players might be a `paper' advantage to the church
that doesn't occur in reality.
b. Players can have any amount of money, not just $100 dollars.
c. What if the church goes broke?
(a) Imagine that the church has at most k confessionals, and thus can play
no more than k simultaneous games. Fill all k. All other players are
waiting in line for an open spot. The next parishoner can't play until an
existing player goes broke. The distribution of wealth during play by the
(k-1+10)th player is exactly as before, except now it is +/-$100(k-1).
(b) has no impact. As above, at the end of each game the fraction of money
won by the church with respect to that player is 1 (assuming it's the
player and not the church that `went out').
(c) If the church goes broke, all bets are off, literally but not
figuratively. The distribution of wealth is c=0, P=1 (P for all players as
opposed to p for a single player). This also does not match the
expectation of .51.
>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). 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.
Hope you found this interesting but not insulting,
Scott Collins | "That's not fair!" -- Sarah
| "You say that so often. I wonder what your basis
408.862.0540 | for comparison is." -- Goblin King
................|....................................................
BUSINESS. fax:974.6094 R254(IL5-2N) collins@newton.apple.com
Apple Computer, Inc. 5 Infinite Loop, MS 305-2D Cupertino, CA 95014
.....................................................................
PERSONAL. 408.257.1746 1024:669687 catalyst@netcom.com
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- 1994-04-19 (Tue, 19 Apr 94 16:58:18 PDT) - Re: 15 out of 16 times (math, not laundry) -
*collins@newton.apple.com (Scott Collins)*- 1994-04-20 (Wed, 20 Apr 94 10:50:42 PDT) - Re: 15 out of 16 times (math, not laundry) -
*ph@netcom.com (Peter Hendrickson)*

- 1994-04-20 (Wed, 20 Apr 94 10:50:42 PDT) - Re: 15 out of 16 times (math, not laundry) -