I'm trying to remember how to calculate probabilities. How do you
calculate the probability of dealing a straight from a single deck?
What about multiple decks (4 for example)?
Thanks,
Bill
XVP Math help, please
I'm busy, busy, busy. So I have to do this fast. Hopefully, I get it
right:
You can be dealt ten different straights from five-high to Ace-high. So
it's 4X4X4X4X4=1024. 1024x10=10,240. But in those straights will be 4
royal flushes and 36 straight flushes. 10,240 minus 40=10,200.
Diivide the cycle, 2,598,960 possible combinations by 10,200 and you
will make a straight every 254.8 games.
···
--- In vpFREE@yahoogroups.com, "billc140" <vphobby2@...> wrote:
I'm trying to remember how to calculate probabilities. How do you
calculate the probability of dealing a straight from a single deck?
What about multiple decks (4 for example)?Thanks,
Bill
billc140 wrote:
> I'm trying to remember how to calculate probabilities. How do you
> calculate the probability of dealing a straight from a single deck?
> What about multiple decks (4 for example)?
mickeycrimm wrote:
I'm busy, busy, busy. So I have to do this fast. Hopefully, I get
it right:You can be dealt ten different straights from five-high to Ace-high.
So it's 4X4X4X4X4=1024. 1024x10=10,240. But in those straights will
be 4 royal flushes and 36 straight flushes. 10,240 minus 40=10,200.
Diivide the cycle, 2,598,960 possible combinations by 10,200 and you
will make a straight every 254.8 games.
With some time on my hands, I'm going to fill that in with what's
likely somewhat of a remedial prelude, but helpful for the uninitiated
in understanding how to approach hand probabilities in general.
All card probability problems take the form of the ratio between the
number of unique hands that fit the bill in question (e.g. dealt
straights) divided by the total universe of unique hands. ("Unique"
means that two hands with the same cards but ordered differently are
counted only once.)
···
------
For a 52 card deck, the calculation of the number of unique hands can
be conceived of in the following manner: The first card dealt can be
one of 52 cards, the second one of 51 ... the fifth one of 48. The
total number of possible deals is 52*51*50*49*48.
However, within that set of deals are hands that are rearrangements of
the same 5 cards. To arrive at the number of distinct 5 card hands
it's necessary to divide the product above by the number of ways 5
cards can be arranged. In placing 5 cards into a hand, you have 5
cards from which to select the first, 4 for the second ... The total
number of rearrangements if 5*4*3*2*1=120.
Divide the first product by 120 and you get Mickey's 2,598,960.
-------
So, now count the number of unique hands in question and divide
Mickey's number and you get the hand cycle.
Divide it by Mickey's number and you express the probability as a
percent -- multiply that percent by the payout on the hand (in bet
units) and you get the percent of overall game return derived from
that particular dealt hand (a handy exercise when you feel like you're
getting nailed in multiplay by a dearth of dealt quads 
------
Most of the dealt hand probability problems are fairly
straightforward. Some can be a little more complex, but still
readily approached if you take time to structure the solution (such as
how often you're dealt two pair, where one pair is Aces)
The calculations can be extended to determining how often you see a
given hand after the draw -- by meticulously examining the frequency
of all the deals with card combinations that might ultimately result
in that hand, backing out the deals that contain other cards that
would be held in lieu of the that particular combination, and then
factoring the probability of completing the hand on the draw from each
particular combination type. (An example would be calculating the
frequency with which royals appears from each of the various number of
card holds.)
- Harry
(and, having pounded that out "stream of consciousness" without taking
time to edit afterwards, I welcome corrections