I thought I knew a little about statistics, but this simple question has me stumped. Given that royal flushes occur every 40,390hands on average, how likely is a person to receive one or more RF's during a single run of 40,390 hands? If you could show me how to compute this, I'd appreciate it!
Don
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I thought I knew a little about statistics, but this simple question has me stumped. Given that royal flushes occur every 40,390hands on average, how likely is a person to receive one or more RF's during a single run of 40,390 hands? If you could show me how to compute this, I'd appreciate it!
Don
The simplest way of doing it is to figure out what the chance of not
hitting one is. 40,389/40,390 is the chance of not hitting it in any
given hand, and just raise that to the power of 40,390. There's about
a 63% chance of hitting something in any given cycle. Except for
extremely short cycles, how long the cycle is doesn't affect this
figure much.
Approximately 64% of a 1-in-n event hitting at least once in n trials
- approximately 1-1/e when n is large.
JBQ
On 2/1/06, Donald Ross <hedonist144@yahoo.com> wrote:
I thought I knew a little about statistics, but this simple question has me stumped. Given that royal flushes occur every 40,390hands on average, how likely is a person to receive one or more RF's during a single run of 40,390 hands? If you could show me how to compute this, I'd appreciate it!
Don
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Tom Robertson and Quero have given good answers, but I thought I'd
spell out the steps for you.
Chance of NOT getting an RF on one play is 40389/40390 = 0.999975241
Chance of NOT getting an RF during 40390 plays is 0.999975241^40390
= 36.79%.
(the "^" means "raise to the power of")
Therefore, the chance of getting one or more RF's in 40390 plays is
1-36.79% = 63.21%.
--Dunbar
I thought I knew a little about statistics, but this simple
question has me stumped. Given that royal flushes occur every
40,390hands on average, how likely is a person to receive one or
more RF's during a single run of 40,390 hands? If you could show me
how to compute this, I'd appreciate it!
--- In vpFREE@yahoogroups.com, Donald Ross <hedonist144@...> wrote:
Don---------------------------------
What are the most popular cars? Find out at Yahoo! Autos
[Non-text portions of this message have been removed]
Thank you Tom, Quero, and Dunbar for your help. So would the chance of getting one or more RF's in 20,000 hands be approximately 31.6%? I'm not sure how to calculate this either.
Tom Robertson and Quero have given good answers, but I thought I'd
spell out the steps for you.
Chance of NOT getting an RF on one play is 40389/40390 = 0.999975241
Chance of NOT getting an RF during 40390 plays is 0.999975241^40390
= 36.79%.
(the "^" means "raise to the power of")
Therefore, the chance of getting one or more RF's in 40390 plays is
1-36.79% = 63.21%.
--Dunbar
I thought I knew a little about statistics, but this simple
question has me stumped. Given that royal flushes occur every
40,390hands on average, how likely is a person to receive one or
more RF's during a single run of 40,390 hands? If you could show me
how to compute this, I'd appreciate it!
Don---------------------------------
What are the most popular cars? Find out at Yahoo! Autos
[Non-text portions of this message have been removed]
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dunbar_dra <h_dunbar@hotmail.com> wrote:
--- In vpFREE@yahoogroups.com, Donald Ross <hedonist144@...> wrote:
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[Non-text portions of this message have been removed]
If the chance of not getting a royal flush in one hand is
40389/40390, then the chance of NOT getting a royal flush in 20000
hands is (40389/40390)^20000=60.9%
So the chance of getting one or more is 39.1%.
--Dunbar
Thank you Tom, Quero, and Dunbar for your help. So would the
chance of getting one or more RF's in 20,000 hands be approximately
31.6%? I'm not sure how to calculate this either.
dunbar_dra <h_dunbar@...> wrote:
Tom Robertson and Quero have given good answers, but I thought
I'd
spell out the steps for you.
Chance of NOT getting an RF on one play is 40389/40390 =
0.999975241
Chance of NOT getting an RF during 40390 plays is
0.999975241^40390
= 36.79%.
(the "^" means "raise to the power of")
Therefore, the chance of getting one or more RF's in 40390 plays
is
1-36.79% = 63.21%.
--Dunbar
>
> I thought I knew a little about statistics, but this simple
question has me stumped. Given that royal flushes occur every
40,390hands on average, how likely is a person to receive one or
more RF's during a single run of 40,390 hands? If you could show
me
how to compute this, I'd appreciate it!
>
> Don
>
>
> ---------------------------------
>
> What are the most popular cars? Find out at Yahoo! Autos
>
> [Non-text portions of this message have been removed]
>vpFREE Links: http://members.cox.net/vpfree/Links.htm
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software Gambling
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--- In vpFREE@yahoogroups.com, Donald Ross <hedonist144@...> wrote:
--- In vpFREE@yahoogroups.com, Donald Ross <hedonist144@> wrote:
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It sounds like you were trying to get the chance of hitting a royal in
20,000 hands by dividing the chance of hitting one in 40,000 hands by
2. That works for the average number of royals you'd hit, but not for
the chance of hitting one or more. To do it that way correctly, you'd
have to take the square root (dividing the exponent, not the chance of
hitting the royal, by 2) of the chance of not hitting one in 40,000
hands and then subtract that from 1.
Thank you Tom, Quero, and Dunbar for your help. So would the chance of getting one or more RF's in 20,000 hands be approximately 31.6%? I'm not sure how to calculate this either.
dunbar_dra <h_dunbar@hotmail.com> wrote:
Tom Robertson and Quero have given good answers, but I thought I'd
spell out the steps for you.Chance of NOT getting an RF on one play is 40389/40390 = 0.999975241
Chance of NOT getting an RF during 40390 plays is 0.999975241^40390
= 36.79%.(the "^" means "raise to the power of")
Therefore, the chance of getting one or more RF's in 40390 plays is
1-36.79% = 63.21%.--Dunbar
--- In vpFREE@yahoogroups.com, Donald Ross <hedonist144@...> wrote:
I thought I knew a little about statistics, but this simple
question has me stumped. Given that royal flushes occur every
40,390hands on average, how likely is a person to receive one or
more RF's during a single run of 40,390 hands? If you could show me
how to compute this, I'd appreciate it!
Don---------------------------------
What are the most popular cars? Find out at Yahoo! Autos
[Non-text portions of this message have been removed]
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Donald Ross wrote:
I thought I knew a little about statistics, but this simple question
has me stumped. Given that royal flushes occur every 40,390hands on
average, how likely is a person to receive one or more RF's during a
single run of 40,390 hands? If you could show me how to compute
this, I'd appreciate it!
=========================================================
Don,
There have been a couple of replies to this post. In case you want
to change your question from "one or more" to "two or more"
or "between 2 and 4," etc., I thought you might want a
generalization.
Let p = 1/40,390 be the probability of a royal flush and let
q = 1 - p = 40,389/40,390.
Let X denote the number of royal flushes and P(X) the probability of
getting that number of royal flushes. If you play 40,390 hands, you
could get anywhere from 0 to 40,390 royal flushes. The sum of all
the probabilities is 1, so we must have:
(*) P(0) + P(1) + P(2) + P(3) + ... + P(40,390) = 1
To calculate P(X) for any X, use
P(X) = C(40,390, X) times p^X times q^(40,390 - X), where C(N, X) is
a combination of N things taken X at a time [can be done by
hand...easier to use your calculator if it can handle these].
When you say "one or more," note that in (*) you would have to add
up P(1) + P(2) + P(3) + ... + P(40,390), but it is easier to just
calculate P(0) and subtract it from 1 on the right side. So
P(X >= 1) = 1 - P(0) = 1 - C(40,390, 0) * p^0 * q ^ 40,390
= 1 - 1 * 1 * q^40,390, which is about .6321 or 63.21% (as reported
by other replies). Now if you changed your question to "three or
more," your work might look something like this:
P(X >= 3) = 1 - P(0) - P(1) - P(2)
~= 1 - .37 - .37 - .18 = .08, or 8%.
I see that you later asked the same question for 20,000 hands. Then
we get
P(X >= 1) = 1 - P(0) = 1 - C(20,000, 0) * p^0 * q ^ 20,000
= 1 - 1 * 1 * q^20,000, which is about .3905 or 39.05%.
Jeff
P(X) = C(40,390, X) times p^X times q^(40,390 - X), where C(N, X) is
a combination of N things taken X at a time [can be done by
hand...easier to use your calculator if it can handle these].
You can also do it with a spreadsheet:
P(X)=(1/40391)^X times (40390/40391)^(40391-X) times Combin(40391:X)
X_ P(X)
0_ 36.7874887% (probability of 0 royals)
1_ 36.7883995% (probability of 1 royal)
2_ 18.3941998% (probability of 2 royals)
3__ 6.1312481% (probability of 3 royals)
4__ 1.5327361% (probability of 4 royals)
5__ 0.3065245% (probability of 5 royals)
6__ 0.0510824% (probability of 6 royals)
7__ 0.0072966% (probability of 7 royals)
8__ 0.0009119% (probability of 8 royals)
9__ 0.0001013% (probability of 9 royals)
10_ 0.0000101% (probability of 10 royals)
--- In vpFREE@yahoogroups.com, "jeffcole2003oct" <jeff-cole@...> wrote:
10 0.0000010% (probability of more than 10 royals)
Nightoftheiguana published figures for the probability of
getting various numbers of royals in one cycle, given that the royal
cycle is exactly 40,390 hands.
I've always used the BINOMDIST function on EXCEL to figure
this out, and the numbers I got by doing this are close to the ones
shown, but not exactly the same. For practical purposes, it doesn't
matter to me whether I have a 1.58% chance of hitting 4 royals in 1
cycle or 1.53%. I am not using these numbers in mathematical
calculations so precision is unnecessary.
Nevertheless I am curious whether the Binomial Distribution
is better or worse than the other technique, and why. The first column
represents the figures I got from BINOMDIST and the second shows what
NIGHT published, using whatever tool he used.
If they are approximately comparable (as they appear to be),
BINOMDIST took me about 30 seconds to do and I don't know what tool
NIGHT was using or how long it took.
36.78748870%
36.7874887% (probability of 0 royals)
36.78666222%
36.7883995% (probability of 1 royal)
18.57267057%
18.3941998% (probability of 2 royals)
6.251098102%
6.1312481% (probability of 3 royals)
1.577933817%
1.5327361% (probability of 4 royals)
0.318640142%
0.3065245% (probability of 5 royals)
0.053619183%
0.0510824% (probability of 6 royals)
0.007733611%
0.0072966% (probability of 7 royals)
0.000975982%
0.0009119% (probability of 8 royals)
0.000109481
0.0001013% (probability of 9 royals)
Bob Dancer
For the best in video poker information, visit www.bobdancer.com
or call 1-800-244-2224 M-F 9-5 Pacific Time.
[Non-text portions of this message have been removed]
Even though my spreadsheets (using Lotus for Windows) come up with
some strange results when they're using extremely large numbers, I did
a spreadsheet to double check nightoftheiguana's work and, with 1 tiny
exception probably due to rounding, I got the exact same answers. I
assume they're right. Yours probably approximates more. That they
diverge more and more in percentage terms as the number of royals
increases tells me that the approximations have a cumulative effect,
also.
I have no idea how spreadsheets are programmed. Just in doing
relatively simple calcuations in which I know the answer is a whole
number, I'll often get the right whole number, but an extremely small
decimal value in addition. The right answer might be 3671 and the
spreadsheet will say it's 3671.000000000524, for example. Or the
answer will be 0 (exactly), but it will say that it's 7 x 10^-12.
Nightoftheiguana published figures for the probability of
getting various numbers of royals in one cycle, given that the royal
cycle is exactly 40,390 hands.I've always used the BINOMDIST function on EXCEL to figure
this out, and the numbers I got by doing this are close to the ones
shown, but not exactly the same. For practical purposes, it doesn't
matter to me whether I have a 1.58% chance of hitting 4 royals in 1
cycle or 1.53%. I am not using these numbers in mathematical
calculations so precision is unnecessary.Nevertheless I am curious whether the Binomial Distribution
is better or worse than the other technique, and why. The first column
represents the figures I got from BINOMDIST and the second shows what
NIGHT published, using whatever tool he used.If they are approximately comparable (as they appear to be),
BINOMDIST took me about 30 seconds to do and I don't know what tool
NIGHT was using or how long it took.36.78748870%
36.7874887% (probability of 0 royals)
36.78666222%
36.7883995% (probability of 1 royal)
18.57267057%
18.3941998% (probability of 2 royals)
6.251098102%
6.1312481% (probability of 3 royals)
1.577933817%
1.5327361% (probability of 4 royals)
0.318640142%
0.3065245% (probability of 5 royals)
0.053619183%
0.0510824% (probability of 6 royals)
0.007733611%
0.0072966% (probability of 7 royals)
0.000975982%
0.0009119% (probability of 8 royals)
0.000109481
0.0001013% (probability of 9 royals)
Bob Dancer
For the best in video poker information, visit www.bobdancer.com
or call 1-800-244-2224 M-F 9-5 Pacific Time.
Nightoftheiguana published figures for the probability of
getting various numbers of royals in one cycle, given that the royal
cycle is exactly 40,390 hands.
I used 40391. You could get even more precise with wizard's numbers:
http://wizardofodds.com/videopoker/tables/jacksorbetter.html
19933230517200/493512264= 40390.54745193
no royal:
(40389.54745193/40390.54745193)^40390.54745193= 0.3678748870393
one royal:
(40389.54745193/40390.54745193)^40389.54745193 times
(1/40390.54745193) times 40390.54745193= 0.3678839952098
two royals:
(40389.54745193/40390.54745193)^40388.54745193 times
(1/40390.54745193)^2 times 40390.54745193 times 40389.54745193 /2=
0.1839419976049
Nevertheless I am curious whether the Binomial Distribution
is better or worse than the other technique, and why. The first column
represents the figures I got from BINOMDIST and the second shows what
NIGHT published, using whatever tool he used.
Looks like BINOMDIST is slightly off, I'm not sure why.
--- In vpFREE@yahoogroups.com, "Bob Dancer" <bob.dancer@...> wrote:
Looks like BINOMDIST is slightly off, I'm not sure why.
On second thought, I bet you used 40390
Looks like BINOMDIST is slightly off, I'm not sure why.
On second thought, I bet you used 40390
No, I used 40,390, and, with 1 tiny exception, got exactly the same
numbers you did using 40,391.
found this:
http://support.microsoft.com/default.aspx?kbid=827459&product=xl2003
says for n<1030, formula is:
COMBIN(n,x)*(p^x)*((1-p)^(n-x))
but for n>1030, an "alternative algorithm" is used
that's probably the source of some roundoff error
try using COMBIN(n,x)*(p^x)*((1-p)^(n-x)) instead of BINOMDIST(x, n,
p, false)
--- In vpFREE@yahoogroups.com, Tom Robertson <thomasrrobertson@...> wrote:
>> Looks like BINOMDIST is slightly off, I'm not sure why.
>
>On second thought, I bet you used 40390No, I used 40,390, and, with 1 tiny exception, got exactly the same
numbers you did using 40,391.
If this is really true Bob, I would have thought that you of all people would have used the
Poisson distirbution, and saved a few more seconds (and keysrokes), LoL
P( N RF's in 1 cycle) = 1/n! * 1/e.
Simple, quicker and much more accurate than your method (! = factorial, 0! = 1, 1! =1)
For N = 4
1.532736128266 % double precision floating point
1.5327361 % Noti's Combi result (excell)
1.53283100488101 % Poisson, double precision FP
1.577933817 % Bob's Approx:
--- In vpFREE@yahoogroups.com, "Bob Dancer" <bob.dancer@...> wrote:
I've always used the BINOMDIST function on EXCEL to figure
this out, and the numbers I got by doing this are close to the ones
shown, but not exactly the same. For practical purposes, it doesn't
matter to me whether I have a 1.58% chance of hitting 4 royals in 1
cycle or 1.53%. I am not using these numbers in mathematical
calculations so precision is unnecessary.
cdfsrule was concerned about the number of keystrokes for me: If this is
really true Bob, I would have thought that you of all people would have
used the Poisson distirbution, and saved a few more seconds (and
keysrokes), LoL
(and I'll pretend he was challenging the message and not the messenger
<g>)
For both the BINOMDIST and POISSON distribution in EXCEL, there are
canned formulas where you plug in values. I have a BINOMDIST spreadsheet
set up and saved, so it was just a matter of entering the 40390 and
doing some cut-and-paste, and then I had the answer. I don't have such a
spreadsheet set up for POISSON, although it wouldn't be difficult to do
so.
More to the point, though, is which one is more relevant. My math is
"high intermediate level" relative to others on this forum, I suspect,
but there are certainly a number more knowledgeable than me in
mathstuff. The reason I was using the Binomial Distribution rather than
the Poisson Distribution is because that was the distribution I thought
was more applicable. Are the math guys here positive that the Poisson is
better?
Bob Dancer
For the best in video poker information, visit www.bobdancer.com
or call 1-800-244-2224 M-F 9-5 Pacific Time.
[Non-text portions of this message have been removed]
For this problem, the binomial distribution is exact and the Poisson
distribution is a (very close) approximation.
On Feb 3, 2006, at 10:51 AM, Bob Dancer wrote:
cdfsrule was concerned about the number of keystrokes for me: If this is
really true Bob, I would have thought that you of all people would have
used the Poisson distirbution, and saved a few more seconds (and
keysrokes), LoL(and I'll pretend he was challenging the message and not the messenger
<g>)For both the BINOMDIST and POISSON distribution in EXCEL, there are
canned formulas where you plug in values. I have a BINOMDIST spreadsheet
set up and saved, so it was just a matter of entering the 40390 and
doing some cut-and-paste, and then I had the answer. I don't have such a
spreadsheet set up for POISSON, although it wouldn't be difficult to do
so.More to the point, though, is which one is more relevant. My math is
"high intermediate level" relative to others on this forum, I suspect,
but there are certainly a number more knowledgeable than me in
mathstuff. The reason I was using the Binomial Distribution rather than
the Poisson Distribution is because that was the distribution I thought
was more applicable. Are the math guys here positive that the Poisson is
better?
By "binomial distribution" Steve is refering to:
COMBIN(n,x)*(p^x)*((1-p)^(n-x))
Which may or may not be the same as EXCEL BINOMDIST(x, n, p, false)
n=sample size, x=number of hits, p=probability of getting a hit
--- In vpFREE@yahoogroups.com, Steven Jacobs <jacobs@...> wrote:
For this problem, the binomial distribution is exact and the Poisson
distribution is a (very close) approximation.
Night said: By "binomial distribution" Steve is refering to:
COMBIN(n,x)*(p^x)*((1-p)^(n-x))
Which may or may not be the same as EXCEL BINOMDIST(x, n, p, false)
n=sample size, x=number of hits, p=probability of getting a hit
Fair enough. I'm willing to trust Excel on this. Since we know Poisson
is a close approximation, and we saw that the numbers NIGHT came out
with (using Poisson) were indeed close to BINOMDIST. Further, Excel
would have every reason to have their BINOMDIST be appropriate, I'm
content to continue to use it.
Whether or not CDSFRULE is concerned with the number of keystrokes I
"waste" by doing so.
Bob Dancer
For the best in video poker information, visit www.bobdancer.com
or call 1-800-244-2224 M-F 9-5 Pacific Time.
[Non-text portions of this message have been removed]