"the Streak"; for Steve J and Harry P

[mkl54321_mkl54321]

Thank you for the calculation work on the likelihood of the current
RF draw drought.

Steve, your two posts kind of contradict each other. The question
was, what is the likelihod of this event happening? Clearly one royal
in 283 attempts has the same likelihood of happening REGARDLESS of
when the successful draw occurs. You saw fit to "chide" me that I
should "know better" about some misinterpretation YOU made of the
situation, i.e., that the second streak should be treated as having
ended at 155 (when it clearly hasn't; in fact, add three more misses
from tonight's play to the current total). You seem to imply that the
expressed royal frequency should be, not 1 in 282, but 1/(127+155)/2.
This would be correct had the second streak ended today. The really
relevant calculation of royal frequency can only occur with present
data. If we wait until I do hit a royal, then the result is skewed
(upward) by ENDING with that royal (given that we could take another
sample some twenty or thirty draws later and expect the same number
of "hits" for the sample). Similarly, if I had started keeping track
immediately after connecting on a one-card draw, that would bias the
result DOWNWARD. But I didn't do that--I in fact, as I already
stated, I started keeping track AFTER I (we) had already missed an
inordinate amount of draws in a row; the first streak was probably
more like 160 or 170, in actuality.

Harry: Steve's numbers for the two streaks being accurate (an
assumption I am making for the moment), it seems that the odds of the
two streaks BOTH happening are (.06)(.03) (rounding his figures to
the nearest percent). This gives 0.0018 or just a little under 2/10
of one percent. This is so close to your figure--but off by a decimal
point--is your calc perhaps wrong? My "gut" is that the less than one
percent figure is closer to the truth.

(BTW, do you agree that it doesn't matter WHEN the single succesful
draw occured, that what we are doing is calculating the chances of a
1/283 success rate for a 1/47 proposition?)

[Steve_Jacobs]

Thank you for the calculation work on the likelihood of the current
RF draw drought.

Steve, your two posts kind of contradict each other. The question
was, what is the likelihod of this event happening? Clearly one royal
in 283 attempts has the same likelihood of happening REGARDLESS of
when the successful draw occurs. You saw fit to "chide" me that I
should "know better" about some misinterpretation YOU made of the
situation, i.e., that the second streak should be treated as having
ended at 155 (when it clearly hasn't; in fact, add three more misses
from tonight's play to the current total). You seem to imply that the
expressed royal frequency should be, not 1 in 282, but 1/(127+155)/2.

Whatever the "correct" frequency is, it certainly is NOT 1 in 282. You
are being selective with your data.

This would be correct had the second streak ended today. The really
relevant calculation of royal frequency can only occur with present
data. If we wait until I do hit a royal, then the result is skewed
(upward) by ENDING with that royal (given that we could take another
sample some twenty or thirty draws later and expect the same number
of "hits" for the sample). Similarly, if I had started keeping track
immediately after connecting on a one-card draw, that would bias the
result DOWNWARD. But I didn't do that--I in fact, as I already
stated, I started keeping track AFTER I (we) had already missed an
inordinate amount of draws in a row; the first streak was probably
more like 160 or 170, in actuality.

Makes no difference. The first streak ended when you hit the royal.
the misses that occured before the royal belong to one and only one
streak. You are trying to also attach those misses to the streak AFTER
the royal, and that is simply bogus math.

Harry: Steve's numbers for the two streaks being accurate (an
assumption I am making for the moment), it seems that the odds of the
two streaks BOTH happening are (.06)(.03) (rounding his figures to
the nearest percent). This gives 0.0018 or just a little under 2/10
of one percent. This is so close to your figure--but off by a decimal
point--is your calc perhaps wrong? My "gut" is that the less than one
percent figure is closer to the truth.

This too is somewhat misleading (see below). But even if this was the
right number, 0.18 percent is about a 3 sigma event. Still not
spectacularly unusual.

(BTW, do you agree that it doesn't matter WHEN the single succesful
draw occured, that what we are doing is calculating the chances of a
1/283 success rate for a 1/47 proposition?)

You can't simply multiply the probabilities of failure together. The
probability of experiencing exactly one royal out of 283 attempts
(one-card draws) is computed as follows:

P = (1/47)*[(46/47)^282]*comb(283,1)
= 0.0137

So, your estimate is too low by a factor of 283/47 = 6.02.

This event happens 1.37 percent of the time, or once in 73 trials.
This is about a 2.2 sigma event.

To put this in perspective, consider the opposite 2.2 sigma event
-- hitting two royals very close together. There have been several
posts talking about hitting two royals on the same day or same week.
If the probability of a royal is 1 in 40,000, then a 2.2 sigma "lucky"
draw of two royals corresponds to a spacing of 550 hands of play.
For every player who suffers a dry streak like yours, there should
be a player who hits two royals within an hour of each other
(assuming all players continue after hitting a royal, which seems
quite unlikely).

I think this illustrates an important point. If you want to claim that
a VP machine is dishonest, looking at only the frequency of royals
is not very meaningful. If you take only 50 players and have them
play long enough to each face 283 one-card draws to a royal, then
there is about an 11% chance that one of the players will not
get a royal, and another 39% chance that a player will only hit
one royal, so that overall there is only a 50/50 chance all players
in the group will hit 2 or more royals in 283 attempts.

···

On Wednesday 10 September 2003 05:30 am, mkl54321 wrote:

[Harry_Porter]

In the interest of saving time, I'm going to comment on Kevin's post
via the extracts in Steve's, and touch briefly on Steve's ...

mkl54321 wrote:

> Clearly one royal
> in 283 attempts has the same likelihood of happening REGARDLESS of
> when the successful draw occurs.

I willing to buy on to this. I've satisfied myself that you've use a
"reasonably" random starting and ending point to your streak for the
sake of discussion. I think both Steve and I agree on the calculation
under this assumption -- a calculation that he states in a modestly
different fashion, but which is identical.

Whatever the "correct" frequency is, it certainly is NOT 1 in 282.

Again, I buy on to your 283 total observations and that's been the the
basis of my calculation (a success, 282 failures). At the end of
Steve's post, he uses the same assumption.

> Harry: Steve's numbers for the two streaks being accurate (an
> assumption I am making for the moment), it seems that the odds of
> the two streaks BOTH happening are (.06)(.03) (rounding his
> figures to the nearest percent). This gives 0.0018 or just a
> little under 2/10 of one percent. This is so close to your
> figure--but off by a decimal point--is your calc perhaps wrong? My
> "gut" is that the less than one percent figure is closer to the
> truth.

You need to multiply the result by 1/47 (the probability of having the
streaks separated by one and only one RF). Then you must multiply
the results by the number of ways in which you can have any two
streaks separated by one RF since surely you're not arguing that your
result is any more anomalous than the guy next to you who happens to
have the same streak, except their royal was on the 100th observation.

The anomaly here is that only one RF was observed in total and all
possibilities under which this might equally likely occur must be
taken into account. Otherwise, I can argue that my last streak of
1000 hands was a one in a google-gillion shot (and it was, as is
anybody else's 1000 hand streak).

You can't simply multiply the probabilities of failure together.
The probability of experiencing exactly one royal out of 283
attempts (one-card draws) is computed as follows:

P = (1/47)*[(46/47)^282]*comb(283,1)
= 0.0137

Steve, I haven't looked at your exact numbers but he difference
between your .0137 and my .0140 are obviously a matter of rounding at
some point and nothing more.

Kevin, if you look at Steve's write-up and mine, hopefully between the
two modestly different approaches you'll come to some type of
satisfaction that what we're telling you is on target.

You're playing in the company of others feeling similar pain somewhat
frequently through the year, if not a given month. And that doesn't
take into account all the other severe droughts others have
experienced in the casinos.

- Harry

[rgmustain]

I recently drew one card to a RF ending a streak of between 400-500
attempts without a hit in over a year and 9 months (Interestingly, it
was exactly the same draw, the 10 of clubs). As bad as this seems I
was only 3-4 RFs behind "normal" since I hit several 3 card draws
during this period.

Currently, my wife is in a very bad streak of 1 royal in over 400
gambling hours (well over 200,000 hands). She is now 8-9 royals
behind over the last couple of years.

However ... My wife once had 3 royals in one day and drew A-K of
hearts in all three cases on 3 different machines over 8 hours. I
once had 3 royals in 24 hours (over two days). I once had deuces 8
times in one day, unfortunately, I wasn't playing DW machines. I
don't consider any of this unusual.

I'm sure many you can give examples even more bazaar then these.
Let's hear some of them to help lighten up the discussion.

Dick

--- In vpFREE@yahoogroups.com, "Harry Porter" <harry.porter@v...>
wrote:

In the interest of saving time, I'm going to comment on Kevin's post
via the extracts in Steve's, and touch briefly on Steve's ...

mkl54321 wrote:
> > Clearly one royal
> > in 283 attempts has the same likelihood of happening REGARDLESS

of

> > when the successful draw occurs.

I willing to buy on to this. I've satisfied myself that you've use

a

"reasonably" random starting and ending point to your streak for the
sake of discussion. I think both Steve and I agree on the

calculation

under this assumption -- a calculation that he states in a modestly
different fashion, but which is identical.

> Whatever the "correct" frequency is, it certainly is NOT 1 in

282.

Again, I buy on to your 283 total observations and that's been the

the

basis of my calculation (a success, 282 failures). At the end of
Steve's post, he uses the same assumption.

> > Harry: Steve's numbers for the two streaks being accurate (an
> > assumption I am making for the moment), it seems that the odds

of

> > the two streaks BOTH happening are (.06)(.03) (rounding his
> > figures to the nearest percent). This gives 0.0018 or just a
> > little under 2/10 of one percent. This is so close to your
> > figure--but off by a decimal point--is your calc perhaps wrong?

My

> > "gut" is that the less than one percent figure is closer to the
> > truth.

You need to multiply the result by 1/47 (the probability of having

the

streaks separated by one and only one RF). Then you must multiply
the results by the number of ways in which you can have any two
streaks separated by one RF since surely you're not arguing that

your

result is any more anomalous than the guy next to you who happens to
have the same streak, except their royal was on the 100th

observation.

The anomaly here is that only one RF was observed in total and all
possibilities under which this might equally likely occur must be
taken into account. Otherwise, I can argue that my last streak of
1000 hands was a one in a google-gillion shot (and it was, as is
anybody else's 1000 hand streak).

> You can't simply multiply the probabilities of failure together.
> The probability of experiencing exactly one royal out of 283
> attempts (one-card draws) is computed as follows:
>
> P = (1/47)*[(46/47)^282]*comb(283,1)
> = 0.0137

Steve, I haven't looked at your exact numbers but he difference
between your .0137 and my .0140 are obviously a matter of rounding

at

some point and nothing more.

Kevin, if you look at Steve's write-up and mine, hopefully between

the

two modestly different approaches you'll come to some type of
satisfaction that what we're telling you is on target.

You're playing in the company of others feeling similar pain

somewhat

···

frequently through the year, if not a given month. And that doesn't
take into account all the other severe droughts others have
experienced in the casinos.

- Harry

[deuceswild1000]

Probably not that bizarre, but it was fun having four deuces delt to
me twice in about an hour. Both times the deuces occupied the
number 1 and 2 and the numbvere 4 and 5 position.

DWK

···

--- In vpFREE@yahoogroups.com, "rgmustain" <rgmustain@a...> wrote:

I'm sure many you can give examples even more bazaar then these.
Let's hear some of them to help lighten up the discussion.

Dick