Let me outline the computation for you. Then, if you don't think
it is
a Bayesian-like approach, please supply another name. [BTW, I don't
think Bayesian statistics requires independent events; if the
events
are independent however, the statistics are easier to compute,
since
you don't need to know the joint distributions, but rather only the
independent distributions of the 2 events]
My understanding of Bayesian analysis is that it requires events to
be dependent, if they are independent there is no conditional
probability. If you have two events A and B which are dependent then
given that event A has occurred you can modify your estimation that
event B has occurred by applying Bayes theorem.
I don't see how your recursive formula below implies statistical
dependence. If I'm playing a 99% ev game with a goal of winning $100
and a stop loss at $1,000 if after 1000 hands I am down (500) what
does that tell me about the probability or ev of the 1001 hand,
nothing more than what I knew when I started playing, the ev is still
99%, the pdf hasn't changed.
1) Compute the PDF for the fist hand of VP: PDF(1) = Starting
State X
PDF (game), where X stands for convolution, and the starting state
has
a probiliaty of 1 at 0 (no win or loss)-- that is a delta function.
2) Adjust the PDF(1) to take into account the loss limit and the
win
limit. Call this PDF(1)'. To do this, set the PDF(1) <= Loss limit
to
0 and PDF (1)>= Win Limit to zero. Also store the Total Probability
<=
Loss Limit {I will call this the RoR for the 1st hand} and store
the
total prob >= win limit (this is the Prob. of Success, I call it
PoS
for 1 hand)
3) Now compute the PDF for the second hand, PDF(2)= PDF(1)' X PDF
(game).
4) Next adjust PDF(2), computing PDF(2)' as above in step(2). This
yeilds the RoR & PoS for 2 hands.
5) Continue this procedure until: you reach some large number of
hands,
the total Prob of PDF(n)' becomes 0, or you get sick of it.
So, in the end, one gets RoR(n), PoS(n), PDF(n)' for n=1 to a big
number. Now note that for each RoR(j), PoS(j) or PDF(j)' , with
=2,
depends on PDF(j-1)' , etc. Hence, each PDF(j)' is dependant on
all
prior PDFs. [BTW,If you know another way to compute the PDFs, or a
closed form expression for the convolution, please, please, let me
know!]
So, the quantities we are interested in here, are, I think,
conditional
probabilities of a sort: That is, when we ask, what is the
probability
of not losing all our money by hand 10000, we are really asking,
what
is the prob. of not losing all our money by hand 10000 , given a
certain distribution for hand 9999 (which depends conditionally on
hand
9998, and so on).
>
> I'm not sure I see how you would apply Bayesian conditional
> probabilities since all the events are independent, at least if
the
> rng's are working properly. The only thing that is conditional,
is
> any strategy change or denomenation change that a player chooses
to
···
--- In vpFREE@yahoogroups.com, "cdfsrule" <groups.yahoo@v...> wrote:
--- In vpFREE@yahoogroups.com, "thymos_one" <thymos_one@y...> wrote:
> make based on their wins or loses.
>