Re: Fourier Transfer and a game?!?!
- To: mathgroup at smc.vnet.net
- Subject: [mg54173] Re: Fourier Transfer and a game?!?!
- From: Maxim <ab_def at prontomail.com>
- Date: Sat, 12 Feb 2005 01:57:51 -0500 (EST)
- References: <cuf5kg$god$1@smc.vnet.net>
- Sender: owner-wri-mathgroup at wolfram.com
On Thu, 10 Feb 2005 08:22:40 +0000 (UTC), elparedblanco
<cire1611 at gmail.com> wrote:
> let's say we have a probabability of picking an amount of money from a
> vat. I can go into the vat either 0,1,2,3 times. The probability with
> which I pick from the vat is described by the vector {.4, .35,.15,.1}.
> I call this frequency.
>
> I can pull only four amounts of money from the vat. The amounts are
> {$2500,$5000,$7500,$10000}. The probability of picking each amount is
> described by the vector {.35,.35,.15,.15}. I call this severity1.
>
> Process is this. First I choose how many times I can go into the vat.
> Then I go in that many times. I always replace what I pick out. so it
> is possible to win $30000.
>
> The Question is what's the probabililty of winning certain amounts of
> money, such as 15,000 or 7,500 or any number, given the fact that I can
> pick multiple times?
>
>
> ANSWER = {.4000, .1225, .1409, .0935, .0995, .0499, .040, .0257, .016,
> .0074 .0034, .0010, .0003}
>
Suppose we have a model with three vats which open independently with
probabilities p1, p2, p3 respectively. Then the generating function (GF)
for the amount of money taken from the first vat is ((1 - p1)
+ p1*(.35*z^2500 + .35*z^5000 +.15*z^7500 + .15*z^10000)), and similarly
for the other vats. Of course then the solution would be simple, because
to find the GF for the sum taken from the three vats we just have to
multiply the three GFs. So if we could find p1, p2, p3 such that the
probability of all three vats opening, which is p1*p2*p3, equals .1 and
the equations for other combinations hold as well, we would be done.
In[1]:=
sol = Solve[
{(1 - p1)*(1 - p2)*(1 - p3) == .4,
p1*(1 - p2)*(1 - p3) + (1 - p1)*p2*(1 - p3) +
(1 - p1)*(1 - p2)*p3 == .35,
p1*p2*(1 - p3) + p1*(1 - p2)*p3 + (1 - p1)*p2*p3 == .15},
{p1, p2, p3}][[1]]
Out[1]=
{p1 -> 0.253124 - 0.401596 I, p2 -> 0.253124 + 0.401596 I,
p3 -> 0.443751}
Other solutions are just the permutations of the first one. The solution
is complex-valued, but what if we substitute it in the GF?
In[2]:=
Module[
{gf = .35*z^2500 + .35*z^5000 +.15*z^7500 + .15*z^10000},
((1 - p1) + p1*gf)*((1 - p2) + p2*gf)*((1 - p3) + p3*gf) /.
sol // Expand // Chop //
Replace[List @@ #,
{k_.*z^p_. -> {p, k}, k_ -> {0, k}},
{1}]&
]
Out[2]=
{{0, 0.4}, {2500, 0.1225}, {5000, 0.140875}, {7500, 0.0935375}, {10000,
0.0994875}, {12500, 0.049875}, {15000, 0.03995}, {17500, 0.02565}, {20000,
0.015975}, {22500, 0.007425}, {25000, 0.003375}, {27500, 0.0010125},
{30000, 0.0003375}}
It would be interesting to give a rigorous justification for this trick.
Maxim Rytin
m.r at inbox.ru