Harry,
Sorry for slow reply.
I am so far sticking with my answer - a four cycle JOB royal drought
occurs ON AVERAGE about once per 2.2 million hands IN THE LONG RUN.
As I attempted to articulate my thinking, I may have found an equation
to answer a somewhat more general question, the expected number of
slumps of some minimum length over a limited interval. If I was all
wet before, I am completely under water now.
You were objecting to my solution at least partially based on lack of
independence of droughts. The fact that a drought in progress
precludes starting another drought does make answering some questions
more difficult. In your marble game analogy, you asked what is the
probability of at least one "drought" in a series of trials. I agree
that this methodology will not answer this question. My methodology
is only to calculate the average number of droughts in an interval. I
believe this was the original question. It does give the probability
of a drought in a relatively short interval, where two droughts are
not possible, but is not adequate over longer intervals. For
instance, I don't have an answer to the probability of one or more 4
cycle droughts over 10 cycles. I, like you, suspect that some number
crunching iteration is likely necessary.
I am defining a drought as the interval between royals, REGARDLESS OF
LENGTH (two royals in a row would be a zero hand drought). I think
this makes droughts independent and simplifies the math. Using this
definition, every drought you experience except your "virgin drought",
the slump from the start of play, begins with a royal.
Every time you play a hand you have a (1/40391) chance of hitting a
royal and starting a new drought.
Each drought, except those that start within 4 cycles of your stop
point, has a 1.83% chance of being 4+ cycles long.
Every hand you play, except for ones within 4 cycles of the last hand
you play, has a (1/40391)*.0183 chance of INITIATING a 4 cycle+
drought during your play. When you are less than 4 cycles from your
stop point, you cannot initiate a 4+ cycle drought., because you will
quit before it can "ripen".
Expected number of 4+ cycle droughts experienced over X hands =
聽聽聽聽聽聽聽聽聽.0183+(X-4*40391)*.0183/40391
You have a .0183 chance of having a 4+ drought right out of the blocks
You have a .0183/40391 chance of initiating a 4+ cycle drought for
every hand played, except those within 4 cycles of your stop point.
You have zero chance of initiating a 4+ cycle drought in the last 4
cycles played, so 40391*4 is subtracted from hands played.
Applying this equation to an 8 cycle interval (your example) I get an
expectation of .0915 4 cycle+ slumps. This is also equal to the
probability of a 4+ cycle drought in an 8 cycle interval, since we
can't squeeze more than one 4+ cycle slump in an 8 cycle interval.
As X=number of hands played becomes very large, the equation approaches
X*.0183/40391 or 1/(2.2 million hands)
btw, In a few places I probably should have used 40390 or 40392 in
place of 40391, and maybe X-1 in place of X, but this is government
work isn't it?
Corrections, objections, comments, ridicule welcome.
AJ
AJ wrote:
> I'll take a shot.
>
> On average, once per cycle you hit a royal and have an "opportunity"
> to start a dry spell. How about cycle/1.83%=40391/.0183=2.2 million
> hands? If you play 2.2 billion hands, you will average about 1000 4
> cycle losing streaks. Perhaps I am cheating by not answering the
> exact question? I think that you are correct that depending on how
> you state the question the math can be much harder.
Well, I don't think there's any question that you've answered my
question which was what is the average occurrance of a 4 cycle RF
drought. If you're correct, then that would be once in 2.2 million
hands.
But I don't quite see the 1.83% probability of a drought within any
given 4 cycle set of hands as being extended in this manner. There
are two problems:
- Can the 1.83% probability within any 4 cycle drought be properly
applied in this manner?
- Is there a problem when it comes to the interdependence of the
possible occurances of the event over any given number of hands?
------------
The math seems to answer the question: Over how many independent
cycles would you expect an event to occur once, on average, if the
probability of that event within any given cycle is 1.83%. For
example, if we were talking about a cycle of 1 (a single hand), this
would suggest that an event with a probability of 1.83% for one hand
would be expected to occur once in every 1/.0183 hands = 54.6 hands.
Thus far, I've stated the obvious and the answer you've arrived at is
once in 54.6 royal cycles. But that treats the problem as if the only
sets of hands we're looking at is each royal cycle as an independent
event -- I hope it's clear that I'm saying that this treats the
problem as if the only opportunity we have for a 4 cycle drought
occurs only once every 4 cycles, or, for example, only twice within
8 royal cycles.
That's not the case. Within 8 royal cycles, (again, for example),
there's aren't just two opportunities for a 4 cycle drought -- there
are (4*40391 + 1) opportunities. I'm counting, of course, each and
every sequence of 4 cycles in hands, where the first opportunity
(using the sympolism of [Hx] represents the xth hand in the sequence
of 8 royal cycles) extends H1-H40391, and the second runs H2-H40392,
etc.
路路路
--- In vpFREE@yahoogroups.com, "Harry Porter" <harry.porter@...> wrote:
The problem is that the probability of a 4 cycle drought within a
sequence of 8 royal cycles of hands isn't a straightforward
calculation because each cycle isn't independent of each other.
------------
By way of example, let's look at a simplified version of this problem:
Let's say that we're dropping a marble that can fall into any one of
10 distinct slots, and the probability of it dropping into each slot
is equal at 10%. The outcome for which we're interested in is when
the marble doesn't fall into any slot other than 10. Clearly the
probability is 10%.
The question I'll pose is what is the probability of this outcome
happening twice in succession at least once over the course of 3
drops. Within those 3 trials, there are two opportunities for the
outcome. (This is analogous to the question of what is the
probability of a 4 cycle drought over the course of 4 cycle + 1 hands,
where again there are two opportunities for the event.)
This probability is equal to the probability of a success in the
1st/2nd drops + the probability of a success in the 2nd/3rd drops -
the probability of a success in the two overlapping sequence of 2
drops (where we want to count that occurrence as only only one success
and not two).
That works out to .01 + .01 - .001 or 1.9%. I'm going to suggest that
as you extend the problem over a greater number of drops and ask about
the probability of this outcome occurring over a greater number of
successive drops, the arithmetic becomes rather convoluted. In fact,
if you shorthand that general question and express the number of drops
as D and the number of successive positive outcomes as S, I don't
think there is a simple equation by which to calculate the desired
probability. It instead requires building an extended calculation
through what is likely some type of iterative method.
While you might program a computer to perform that iteration to arrive
at a probability, you're not going to write a simple equation in terms
of S and D to solve it.
------------
Thus, I think the only approach that will arrive at the answer I'm
looking for (to the same exact question that you took a stab at) is a
brute force one, obviously assisted by a program.
Care to make a call if I've come to the right conclusion or if,
instead, I've managed to lead myself down the proverbial garden path?
