I apologize for the confusion: the weird symbol ¡°¡ú¡±
started out as a ¡°right pointing arrow¡± but the unicode was lost
when I posted the message. Below, I¡¯ll use the word ¡°to¡± in place
of ¡°¡ú¡±.
¡°P(P¡¯em quad) is 1 in 2361¡± is from the WinPoker analysis of the
game.
Also, the return for P¡¯em is 99.9531% according to WinPoker
analysis, but is 99.9536 according to Skip Hughes on his vphomepage.
I¡¯ll use the WinPoker figure, but hope someone can enlighten me
here.
I wondered, among the 4-of-a-kinds that I hit while playing P¡¯em, on
average, how many of them started out from a pair (2 to quad) vs.
from 3-of-a-kind (3 to quad).
The frequency of (3 to quad) is easy to get because the optimal
strategy is to always hold all 3-of -a -kinds WHENEVER IT¡¯S
PHYSICALLY POSSIBLE TO DO SO.
From there, instead of trying to figure out the frequency of the (2
to quad), I cheated and simply made the observation that the
reciprocals of these two numbers must add up to 1/2361.
Getting a dealt quad in P¡¯em is my second least favorite hand in
this game, I think the worst hand is when you have a choice between
two pairs and you ended up with a full house!----because THAT
involves more than a sense of helplessness.
Also, since I bypassed a high pair to hit my first P¡¯em royal, I
have made an adjustment to the optimal strategy: Royal (INCLUDING
Ace high royals) > High pair. The cost of this adjustment is 0.0011%
(game now returns 99.9520% instead of 99.9531%) and I am willing to
accept this trade-off (my thinking here is that from the choices:
Royal Ace high/ High pair, the chance of ending up with a quad is
the same as the chance of ending up with a royal ¨C they are both 1
in 1128, and so I¡¯ll go for the royal even though it will cost me an
average of 19 cents each time, once every 10 (?) hours or so,
playing for quarters) but then may be what I REALLY NEED to do is to
pay less attention when I play this game!
I use the number of certain rare hands (4-Deuces in Deuces Wild type
games, 4-Aces in Double Bonus, 4-of a kind in P¡¯em) as a yardstick
to independently check the approximate number of total hands I
played in each game. The trustworthiness of this type of yardsticks
no doubt increases with the number of hands played, but for the
small number of hands that I¡¯ve played, my ¡°quad yardstick for P¡¯em¡±
is not very trustworthy, to wit:
Total no. of P¡¯em hands played = 267,900; Quads = 125; S.F.s = 3;
Royals = 2.
So I am over in quads, WAY under in S.F.s, and WAY over in Royals
(royal cycle for this game is an amazing 351,818), and of course,
still nowhere near ¡°long term¡±; but thanks to the two royals, I¡¯ll
probably be ahead in this game for a VERY long time.
L.Wluiki
···
*******************************************************
P(dealt quads in P'em, therefore useless) is 1 in 20,825 or 1 in
every 8.8 quads; at 800 hph, 1 in 26 hours.
P(P'em quad) is 1 in 2361, among them, there are slightly more (by
20%) ¡°2 to quads¡±:
the ratio of (2 to quads) to (3 to quads) is 29 to 24, or 1.21 to
1.
The frequency of (2 to quad) is 1 in 4318, the frequency of (3 to
quad) is 1 in 5208.
L.Wluiki
*********************************************************************
--- In vpFREE@yahoogroups.com, Bill Velek <billvelek@a...> wrote:
I'm still working on this. I know that you're sharper than I am
at this
math, so I'm assuming that you are correct, but I'm trying to see
if I
can reason my way through this.
I think that my original math that concluded that there are 1 in
4,165
combinations was accurate for what I had in mind, but my logical
approach itself was wrong because I see now that that figure
represents
the chances only when stacks are randomly selected. In other
words, if
I were to play the game by always selecting the left-sided stack
and
never the right-sided stack, then I believe the results would be
precisely 1 in 4,165. I used that approach because I initially
was
thinking of those instances when a person doesn't know which stack
to
pick because part of the quad does not appear as the top card in
either
stack. But as I continued to ponder this, I realized that that
was not
the correct approach because it is contrary to what happens when
strategy is applied. Back to the drawing board :-/
Okay, so we know that in order to be even possible to end with
quads,
one of two possible states must exist:
1.) we either have a pair on the far left-side, meaning that the
other
pair to make quads is in one of the two stacks that we can select,
and
might or might not appear as a top card; or ...
2.) we must have trips in one of the stacks that we can select,
which
necessarily means that the top card will _always_ match with one
of our
other two cards, HOWEVER, sometimes proper strategy requires that
we
reject that pair.
Realizing that strategic discards of a pair will reduce the number
of
quads completed, I now realize that this problem is much more
challenging than I have time to play around with. But since I
like to
broaden my mind in regards to how we compute probabilities, I'd
sure
like to hear from you as to how this is done.
Thanks.
Bill Velek