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valueplayer
04-21-2003, 08:34 PM
what's the prob. for that?
1/220 * 1/220 * 1/220 = 1/10,648,000 ?????
Thank you.

BruceZ
04-21-2003, 09:36 PM

Ray Zee
04-21-2003, 11:34 PM
its really only getting them twice in a row. see why? and actually its really only getting them the next hand. 220 to one. see that. if not think awhile as this concept will help you think about probability in the correct way.

Cyrus
04-22-2003, 12:05 AM
(P/up)+(sLUG)+ [f(shuffle)]^m + Ins(mark) + |cut| + {Restore[Ch(pass)]} + D2(conf)

DanS
04-22-2003, 08:03 PM
Oh mighty Z,
Yes, I see why this is true. It was obvious to me instantly, so I won't spoil it for those to who(m) it comes across as a trick question.

I have two questions for you: what is the name for this phenomenon (if there is one, and I get the feeling like I learned it in Psych 101), and how do you take advantage (poker wise, not Montana/Deliverance wise /forums/images/icons/tongue.gif ) of people who think this way without even realizing it?

Dan

Cyrus
04-23-2003, 09:00 AM
"What is the name for this phenomenon ?"

All those mistaken beliefs are variations of the Gambler's Fallacy (aka Monte Carlo Fallacy, etc). The gambler sees 5 Blacks in a row in roulette and either believes a Red is "overdue" or that the Black "streak" will continue. (In fact, each result has again the same probability of occuring.)

To the mathematicians the "Red overdue" is a mistaken belief in negative serial autocorrelation of a non-correlated process. The "Black streak" is a mistaken belief in positive serial autocorrelation. (Ask BruceZ for a translation to English.)

In the case of back-to-back, or three times in a row, of a "rare event" such as getting pocket Aces, the gambler confuses the probability before the fact of unrelated events happening in a sequence, with the probability of each event happening separately in the sequence.

"How do you take advantage (poker wise) of people who think this way without even realizing it?"

Here's one : If the other players are allowed to make or see my hand and then, in a round soon after, I have approximately or exactly the same hand in the hole, I can reasonably assume that the average opponent will estimate the probability of me having again something like I had previously, to be smaller than it really is.

A distant variation of that theme is a play immediately after a multiway pot in which a lot of blood was shed. The next round is often an opportunity for someone to steal quietly (but steal early) with relative impunity.

David Ottosen
04-24-2003, 05:46 AM
Yes but according to Zeno's paradox, you will never get pocket aces. In fact, you will never even get one ace, as the dealer will never be able to deal it to you.

Bama Boy
04-24-2003, 09:37 AM
I was trying to search some archives to see if I could figure it out, but I think I was doing it wrong.

Also, what would the odds be of not getting a pocket pair in 100 hands?

Thanks, you math guys are terrific!

joeg
04-24-2003, 09:53 AM
I calculated the odds of not getting a pocket pair in 100 hands as 0.2% as follows

you have a 6% chance of getting a pocket pair so have a 94% chance of not getting a pocket pair, not getting a pocket pair two hands in a row is 94% * 94% or 88%, for n hands the answer is .94^n or for 100 hands 0.94^100 which roughly 0.002 or 0.2%.

this seems a little low to me, are my calculations correct?

Mangatang
04-24-2003, 05:20 PM
It's like the old saying, "Lightning never stikes in the same place twice." Once lightning has stuck a spot, it is just as likely that it will stike that spot again as is any other previously non-stuck spot.

Ray Zee
04-25-2003, 12:30 AM
its much more likely to strike the same spot again, then a non previously struck spot. it isnt a random act.

FA_man
04-27-2003, 04:45 AM
I understand that these events are unrelated, and the gamblers fallacy, but I dont agree.

The question isnt what is the chance of being dealt AA again. The question is to be dealt AA 3 times in a row.

Are you telling me you are willing to bet me with 1:1 odds that I will flip a coin to tails 3 times in a row? This is substantially different than waiting for tails to come up twice and then betting on the next flip alone - at which point the bet would be even.

Am I totally bonkers on this?

rigoletto
04-28-2003, 08:12 AM
Can an act be random? And is lightning an act? But then again, I'm not religious /forums/images/icons/smirk.gif

Cyrus
04-29-2003, 05:21 AM
"I understand that these events are unrelated, and the gamblers fallacy, but I dont agree. The question isn't what is the chance of being dealt AA again. The question is [what is the probability of being] dealt AA 3 times in a row."

Yes, this was the original question and the mathematical answer was provided by BruceZ, in this thread.

But Gambler's Fallacy was mentioned in the context that at every round the player has the same chances of getting AA. When a player gets AA two times in a row, the fallacy lies in the belief that getting AA for a third time straight is a different (and smaller) probability than the probability of being dealt AA in any round.

"Are you telling me you are willing to bet me with 1:1 odds that I will flip a coin to tails 3 times in a row?"

Well, depends who gets to take those odds. /forums/images/icons/smirk.gif

"This is substantially different than waiting for tails to come up twice and then betting on the next flip alone - at which point the bet would be even [i.e. p=0.5]."

Correct about that call. And Mr Z said so, too, in so many words. Needs a little deciphering, is all.

Andy B
05-04-2003, 12:25 PM
The chance of being dealt a pocket pair on any given hand is 1/17. The chance of being dealt a pocket pair three hands in a row is (1/17)^3, which is .0002. The chance of being dealt no pair for 100 hands in a row is (16/17)^100 which the othe guy says is .002. This seems entirely plausible, but there are no batteries in my calculator just now.

Hootie McBoob
05-07-2003, 01:05 PM
i agree with you, fa. the question was pretty clear to me -- three times in a row ex ante (not a third time in a row).

zeno's paradox -- is that meaning you can only get halfway to a pocket pair, then halfway from that point, then halfway from that second point, then halfway from that third point etc. etc. never reaching the pair of bullets?

Mano
05-07-2003, 03:04 PM
What are the odds of getting dealt Aces three times in a row in a given 8 hr session (lets say in the next 250 hands or so)?

Cyrus
05-08-2003, 04:48 AM
"What are the odds of getting dealt Aces three times in a row in a given 8 hr session (lets say in the next 250 hands or so)?"

The probability of being dealt pocket Aces 3 times, in the next 3 hands, at any time, has already been given (http://www.twoplustwo.com/forums/showthreaded.php?Cat=&amp;Board=probability&amp;Number=244 834&amp;page=2&amp;view=expanded&amp;sb=6&amp;o=14&amp;fpart=) by BruceZ in this thread.

The probability of having a specific number of successes within a specific number of trials could be obtained from the well-known Binomial Probability Formula* that gives the number of exacty X successes in exactly n trials.

"Binomial" means having a Yes or No situation, ie with only 2 possible outcomes, call them Success and Failure. So, in order to convert our problem to a binomial probability situation, we will have to treat "get 3 Aces" as a Success and "not get 3 Aces" as a Failure. However, this would be WRONG, because in the number of Failures we would have included the times when we get more than 3 times Aces, i.e. 4 times in a row, 5 times in a row, etc. Clearly, if we want to be practical, getting pocket Aces 4 times in a row should be considered as something we want, rather than a Failure!

So, the question becomes "What is the probability of getting pocket Aces at least 3 times in a row in the next n hands?"

When the number of n trials (hands) is small enough, and in order to get the answer, we can map out all the possible outcomes and calculate each probability. If we had only n=5 hands to examine, for example, we would calculate the probability of getting pocket Aces

P(3)= 3 times in a row
P(4)= 4 times in a row
P(5)= 5 times in a row

and then we would add up those probabilities. The sum would be the Probability of getting pocket Aces at least 3 times in a row in the next 5 hands.

But now n=250 hands! We have to calculate everything from P(3) upto P(250) and, although we would be adding admittedly very small probabilities, this would be the only way, short of a closed formula, of getting the exact answer. (ETFan has provided such a formula, for a large number of trials, in a BJ website, if I remember correctly. I'll try to locate it.)

--Cyrus

__________________________________________________ _________

* Binomial Probability Formula :

n = number of trials

X = number of successes

p = probability of success

q = 1-p = probability of failure

P(X) = probability of getting exactly X successes in n trials

P(X) = [C(n,X)] * [p^X] * [q^(n-X)]

whereby C(n,X) denotes the number of getting X successes from n trials without regard to order, i.e. the number of all possible Combinations when choosing X out of n.

Cyrus
05-08-2003, 06:26 AM
I have located the website and it provides an Excel program which, as it turns out, is based on a BASIC program, rather than a closed formula as I thought. I plugged into the Excel n=250 , X=3 , P(1)= 1/221 = 0.0045248868, and the result was 0.00002287224378960670 or 0.002%.

That seems to be the probability of seeing at least 1 streak of at least 3 Aces in a row in the next 250 hands --- courtesy of ETFan, BJ expert and creator of the program.

I hope I haven't made any errors, especially after the decimal point. /forums/images/icons/tongue.gif

Cyrus
05-08-2003, 06:44 AM
Just to be very clear on this, and although I consider everything in this post as plenty obvious, please read

<ul type="square">...Clearly, if we want to be practical, getting pocket Aces 4 times in a row should be considered as something we want, rather than a Failure! Same thing when getting more than 1 streak of 3 Aces in a row, once more NOT a Failure!..

So, the question becomes "What is the probability of getting at least 1 streak of at least 3 times of pocket Aces in a row in the next n hands?"
...
[/list]

Also

<ul type="square">...If we had only n=5 hands to examine, for example, we would calculate through the above-mentioned formula the probability of getting pocket Aces

P(3)= exactly 3 times in a row in the next 5 hands
P(4)= exactly 4 times in a row in the next 5 hands
P(5)= exactly 5 times in a row in the next 5 hands

and then we would add up those probabilities.
...[/list]

BruceZ
05-08-2003, 06:50 AM
We have to calculate everything from P(3) upto P(250) and, although we would be adding admittedly very small probabilities, this would be the only way, short of a closed formula, of getting the exact answer.

The probability of getting 3 AA in a row is (1/221)^3, so the probability of not getting it starting on a given hand is 1-(220/221)^3. The probability of getting 3 AA in a row in the next 250 hands is 1 minus the probability of not getting it in the next 250 hands which is exactly 1 - [1-(220/221)^3 ]^248 = 1 in 43,254 or .0023%. The reason this is correct is because the probability of a sequence of 3 AA in a row starting on any hand, or not starting on any hand, is independent of it starting on any other hand. There are 248 hands on which a sequence can start, since it cannot start on the last two.

The binomial distribution is not used in this problem, since that applies to getting various numbers of successes in n trials. Here we just want 1 success in 250 trials, where a success is defined as getting 3 AA in a row. Whether or not we get more than 3 in a row is irrelevant.

Also, even if we did want P(3) + P(4) + P(5) + ...+P(250), a rational person would most likely just compute
1 - [ P(1) + P(2) ].

Cyrus
05-08-2003, 07:04 AM
"The probability of getting 3 AA in a row is (1/221)^3, so the probability of not getting it starting on a given hand is 1-(220/221)^3."

I agree up to that part. /forums/images/icons/grin.gif

"The probability of getting 3 AA in a row in the next 250 hands ..."

How many times "3 AA in a row"? In 250 hands, we could have for example, 4 times that this occurs (streaks), in hands #4-5-6, #150-151-153, #201-202-203 and #247-248-249.

How about "more than 3 times in a row"? We could have for instance AA in hands #181-182-183-184. Where are those cases included?

I would appreciate your response.

The problem is far from being as easy or straightforward as you are taking it to be.

(And I hope you are dead certain about the correctness of your analysis of at-least- m-occurences-in-n-trials, seeing as you rushed in with that headline daming mine.)

ACPlayer
05-08-2003, 09:07 AM
This is the same as binomdist with number of successes = 1, trials = 248 probability of 1/221 power 3. Logically too this is correct. The answer is the same, ie 2.3e-5.

This is the correct answer for getting exactly one run of 3 AA in a row in 250 hands.

Cyrus
05-08-2003, 09:49 AM
"This is the correct answer for getting exactly one run of 3 AA in a row in 250 hands."

I wrote as much in my post. But there's more to the problem than BruceZ realizes! Don't let Bruce's inexplicable pique prevent you from seeing this. All the exact answers are indeed trivial to obtain but the problem is not after the trivial answers Bruce assumes.

The original question needed some elaboration, which I tried to provide in my "Odds" post : We are essentially after the number of at least m occurences in n trials, when n is sufficiently 'large'. If someone has the closed-form formula for that, I would be honestly delighted to see it. Otherwise, I have nothing better or more to offer on this.

--Cyrus

ACPlayer
05-08-2003, 12:25 PM
I reviewed your odds post and dont get it.

If the question is what is the odds of getting 3 AA in a row in the next 5 hands - per your example, why add P(4) and P(5). Those are irrelevant to the question.

If the question is what is the of getting at least 3 AA in a row in the next 5 hands then that is exactly P(3)

If the question is what is the probability of getting at most 3 AA in a row. Then the event you are looking for is AAAx (where x is not A). Again you can calculate a closed form equation based on BINOMDIST by calculating the chances of AAAx and using an appropriate number of events.

I do think that Bruce erred when he said you could not use binomial dist because the successes was just one. My post was to point out that it is possible and correct to use Binom dist function for this example.

Cyrus
05-08-2003, 02:43 PM
"If the question is what is the of getting at least 3 AA in a row in the next 5 hands then that is exactly P(3)."

Yes.

The original question was this : "What are the odds of getting dealt Aces three times in a row in a given 8 hr session (lets say in the next 250 hands or so)?"

In my post titled "Odds", I explained that, if indeed we want to get the probability of being dealt exactly 3 Aces in a row and exactly once during the next 250 hands, then this could be obtained straightforwardly. We could use the BinomDist formula, as you correctly surmised, and despite Bruce's objection.

However, as I took pains to point out, and "if we want to be practical", we should go after the probability of at least 1 streak of at least 3 Aces in a row. I explained that this is what a poker player would be more interested to know IMHO. I just find that to be more interesting than the purely theoretical information of getting "exactly 1 streak of exactly 3 Aces in a row" during the next 250 hands. (As I wrote, that last probability does not include the times we get 4 Aces in a row, or twice 3 times in a row Aces. The process assumes those occurences as "Failures"...)

What is your formula for obtaining the probability of "at least once at least m occurences in n trials"? I would be extremely interested to know. (Bruce's notion is unfortunately extremely inadequate.)

ACPlayer
05-08-2003, 03:11 PM
In my post titled "Odds", I explained that, if indeed we want to get the probability of being dealt exactly 3 Aces in a row and exactly once during the next 250 hands, then this could be obtained straightforwardly. We could use the BinomDist formula, as you correctly surmised, and despite Bruce's objection.

This is not correct. You are actually calculating the odds of getting atleast 3 Aces in 250 hands and not exactly 3 Aces. If you wanted to calculate the probabilty of getting exactly 3 Aces you would first calc the probability that in the next 4 hands you would be dealt 3 Aces thrice followed by not AA once. Then apply Binomdist to find the probability of that event happening across the number of trials.

You are correct that we are calculating the probability that the event will happen only once (or however many we specify in the function). I would have to think about this question.

Ulysses
05-08-2003, 09:34 PM
This seems entirely plausible, but there are no batteries in my calculator just now.

Computer broken too, Andy? /forums/images/icons/grin.gif

Robk
05-08-2003, 10:09 PM
Sorry Bruce, but I think you are way off on this one. The problem is indeed quite complicated. Your method is wrong because the outcome of your "trials" are clearly not independent. For example, if the the first trial is not a success, then there was a hand in the first three that is not AA. Clearly this affects the probability that there will be a streak of AA starting with the second hand. To solve this you need generating functions, and a decent amount of work, if I remember correctly.

BruceZ
05-08-2003, 10:24 PM
Cyrus's method of adding P(exactly 3 in a row) + P(exactly 4 in a row) + P(exactly 5 in a row)... does not work as an exact solution because it counts the same cases multiple times. For example, we could get exactly 3 in a row AND exactly 4 in a row. We also have the problem of how to compute these various probabilities. I already said that the exact solution does not use the binomial distribution. The binomial distribution is used for computing the probability of k successes out of n independent trials with probability of success p. In this problem we do not have independent trials with a fixed probability because the trials overlap. If we get exactly 3 in a row on a given hand, it becomes impossible to get exactly 4 in a row on that hand. Once we have one success say on hands 1-3, the probability of a second success on hands 2-4 becomes much more likely, just 1/221. If you only consider the probability of getting multiple non-overlapping sequences of 3 or more, then there are no longer 248 trials, because we have eliminated 3 of them with a single success. In fact we can eliminate up to 5 of them, for example if the first success starts on hand 3. So the number of trials depends on the successes. That's not a binomial distribution.

My solution is NOT equivalent to using the binomial distribution to compute the probability of getting exactly 1 success as AC_Player has suggested. That would be a different number, though it happens to be virtually the same for all practical purposes. My solution computes the probability of getting exactly ZERO successes which is 248 failures, and subtracts this from 1, which gives the probability of getting one success OR MORE, since when we get zero we certainly didn't get more than one. A success has probability (1/221)^3, and this is the probability of getting 3 OR MORE AA in a row on a given hand. This is actually an exponential distribution. If you want, you can consider it a trivial case of the binomial distribution with the number of successes equal to 0, and you can also use the function binomdist to compute it if you subtract what it gives you from 1, and it gives exactly what I computed.

Here are some corrections to the wording in my original post:

The probability of getting 3 AA in a row in the next 250 hands is 1 minus the probability of not getting it in the next 250 hands which is exactly 1 - [1-(220/221)^3 ]^248 = 1 in 43,254 or .0023%. The reason this is correct is because the probability of a sequence of 3 AA in a row starting on any hand, or not starting on any hand, is independent of it starting on any other hand.

The use of this formula by definition implies that the probabilities of a sequence starting on various hands are independent; however, the text should say that we may treat them as independent, and the number is exact to the reported accuracy; however, the formula itself has an implicit approximation because these sequences are not strictly independent. The reason, as I've said above, is because if we get 3 in a row, that changes the probability of getting another 3 in a row. The exact expression for the probability of no successes would be:

P(1)P(2|1)P(3|1,2)P(4|1,2,3)*...*P(248|P(1,2,3,... 247)

Where P(1) means the probability that a success did NOT start on hand 1, and P(a|b,c...) means the probability that a success did NOT start on a, given that it did not start on b nor c nor... These conditional probabilities ARE always independent, which is why they can be multiplied. Now the P(4|1,2,3) = P(4|2,3) = P(5|3,4) and so on, since each term only depends on the probability of a sequence starting on the previous 2 hands, and these are all equal to P(3|1,2), so we can simplify this to:

P(1)P(2|1)P(3|1,2)^246

This could be computed exactly with effort. Note that my method sets each of these terms equal to P(1) = 1-(220/221)^3. This makes the probability of zero successes a little larger, so it makes the probability of 1 or more success a little smaller, so this is a lower bound, and it comes out to 0.0022975%. To get an upper bound, just take 248*(1/221)^3. This simply adds together the 248 probabilities of getting a success starting on any of 248 hands. Look familiar? This is just like the approximation we used before to get the probability of someone having aces out of multiple opponents. It is an upper bound since we are counting multiple times all the times a sequence starts on more than one hand. This upper bound is 0.0022976%. Note that this upper bound differs from the lower bound by only 1 in the 7th decimal place, and the 5th significant digit. The actual answer is between these tight bounds. This was the basis for my initial suggestion that the events were independent, but that was only true to within the accuracy of my spreadsheet at the time. If you say this is "extremely inadequate", then I'd say you are extremely odd.

I don't know what your computer program is computing, but it seems to be close. You get .002287%, and I get .002298%. It may have a small round off error, or perhaps it is only a simulated result, or perhaps it is computing something different. Why don't you give a link, and I'll try to determine what it is computing.

The binomial distribution is not used in this problem, since that applies to getting various numbers of successes in n trials. Here we just want 1 success in 250 trials

Make that 1 or more success, which is equivalent to 0 successes.

Also, even if we did want P(3) + P(4) + P(5) + ...+P(250), a rational person would most likely just compute
1 - [ P(1) + P(2) ].

I originally thought you were summing the binomial probabilities of 3 or more successes, and in that case this should be 1 - [ P(0) + P(1) + P(2) ]. You are actually adding something different, so this comment isn't relevant, and it isn't germane to the solution at hand.

BruceZ
05-08-2003, 10:41 PM
...and let me know if you don't think I've shown that treating these as independent is very strong and completely adequate. It turns out that it is accurate to assume either independence or mutually exclusive events in most practical poker problems of this type. I have done both above in obtaining the upper and lower bounds.

BruceZ
05-08-2003, 11:46 PM
Well, it looks like I don't have a lower and upper bound, just two upper bounds. The probability of one hand failing to start a sequence increases the probability of future hands failing to start a sequence, so the probability of successes is smaller than the independence assumption computes, not larger as I said. I will continue to work on a lower bound if I have time. I'm still confident that this method yields sufficiently accurate results. The 248*(1/221)^3 is off only due to multiple successes. Since the probability of multiple successes is largest for getting 4 in a row, and since this happens only 1/221 of the time we get 3 in a row, I suspect our answer is accurate to within the originally stated accuracy of 2 significant figures (0.0023%).

Now the P(4|1,2,3) = P(4|2,3) = P(5|3,4) and so on, since each term only depends on the probability of a sequence starting on the previous 2 hands, and these are all equal to P(3|1,2), so we can simplify this to:

P(1)P(2|1)P(3|1,2)^246

I believe I went a little too far with this particular simplification, and the sequences are actually dependent no matter how far apart they are, though weakly. The reason is that a failure eliminates the possibility for sequences longer than 3 as well.

Cyrus
05-09-2003, 01:11 AM
Dear Bruce,

For reasons of personal nature, which you shared with the world in that long post about yours truly a month ago, you have obviously been overtaken by a case of pique. The upshot is that, despite your knowledge, and in your haste to prove that I write "nonsense" or that I do "terrible analysis", you now find yourself putting up long posts that attempt to withdraw, correct or annul your previous statements ("not germane" LOL).

When you come to realize that this is not the Other Topics page, that this not a debate about politics and that you, whether you like it or not, are way off the mark on this one, and when your temperature falls enough, then perhaps you will find it in you to apologize.

Because an apology is most definitely in order.

Regards,

--Cyrus

PS : To rei-terate. Do you have the formula for "at least 1 streak of at least m occurences in n trials"? I don't. Show us yours.

Cyrus
05-09-2003, 01:16 AM
Robk is correct, of course.

And I'm surprised that this website, a poker website for christ's sakes, has not dealt with this before! What next, the null hypothesis?

"Sorry Bruce, but I think you are way off on this one. The problem is indeed quite complicated. Your method is wrong because the outcome of your "trials" are clearly not independent. For example, if the the first trial is not a success, then there was a hand in the first three that is not AA. Clearly this affects the probability that there will be a streak of AA starting with the second hand. To solve this you need generating functions, and a decent amount of work, if I remember correctly."

"Decent amount of work" is correct, Robk.

BruceZ
05-09-2003, 03:01 AM
To rei-terate. Do you have the formula for "at least 1 streak of at least m occurences in n trials"? I don't. Show us yours.

Yes, for m=3 and n=250 which the original problem asked. First I gave one which is accurate for all practical purposes. Then I gave an exact closed form formula for which the terms need to be evaluated. Now here I will derive yet another one, from which you may compute the answer as accurately as you please, and from which I will show that the original formula was indeed very accurate, and gives an answer in agreement with the computer program you found (I'd still like this link).

Using inclusion-exclusion to do the dirty work for us: 248*(1/221)^3 counts all sequences of 3 or more, but it counts all sequences of 4 twice, sequences of 5 3 times etc. In addition, it counts all times we get 2 or more distinct sequences of 3 or more at least twice. We must subtract these cases. If we say the probability of sequences of 4 or more is 247*(1/221)^4, this over counts these sequences, but that's OK because we are going to subtract it, and we are now finding a lower bound. Later we will get an upper bound, and then show how these can be made as tight as you please. For 2 or more non-overlapping sequences, if one sequence is 3 long, then the other non-overlapping sequence has a maximum of 244 places it can start, so the maximum number of places the two sequences can start is C(245,2), and the probability is less than C(245,2)*(1/221)^3^2. Note that this allows the sequences to be longer than 3. So putting it together we get a lower bound of:

248*(1/221)^3 - 247*(1/221)^4 - C(245,2)*(1/221)^3^2 = 0.0028722218%.

This is even closer to the computer result. One more iteration produces a very tight upper bound:

248*(1/221)^3 - 247*(1/221)^4 - C(245,2)*(1/221)^3^2 + 246*(1/221)^5 + C(253,2)*(1/221)^4^2 = .0028726885%

So the answer lies between .0028722218 and .0028726885. Further iterations will narrow this range as tightly as you like, or until you run out of significant digits in Excel, which I discovered is less than it prints out, and seems to be about 13 places.

Now, as for the personal issue, I have not withdrawn, corrected, or annulled anything in support of the statements that your analysis was terrible, and that you wrote nonsense. Your use of the binomial distribution and the summation of probabilities you outlined to obtain an exact formula in your original post were terrible for the reasons I have explained, and for reasons which other posters have alluded to, and with whom you even agreed, apparently not really understanding their implications. Your insistence that the binomial distribution should be used for this end despite my "objections" was nonsense. The thing that wasn't germane was in regard to another comment, not concerning why your method was terrible, it's still terrible. The other things I corrected, withdrew, or annulled in the course of perfecting my own computation and the assessment of its accuracy. That's a whole different issue, nothing to do with your attempt at an exact formula. It was not my purpose originally to provide a conceptually exact formula, only one which provided accurate answers. The original formula I gave is one I use routinely to obtain answers to these types of problems with a high degree of confidence. It should be used in most poker problems of this type instead of the exact methods I described later because these are simply not worth the effort, based on my experience working many problems both exactly and approximately. The idea that there is something about this elementary formula I do not understand is also nonsense, and you should have considered that.

So in short I don't see that I have anything to apologize for other than doing a hell of a lot of work here, and the statements I made regarding your methods are accurate. Your analysis in this thread was not the only one that has problems, just the only one I have gotten around to addressing. There will be more. The statements you have made about my assessment of your methods were not accurate, so I think that if anyone should apologize it should be you, but I really don't care.

You are correct this is not the other topics forum. This forum has always been most congenial since its inception up until now. I expect it to remain that way; however, I will not allow incorrect mathematics to go unchallenged, and I will not allow threads to become sidetracked by your diversionary tactics. You cannot bluff me in here. I notice you have not seen fit to address any mathematics I have written or even my criticisms of your method, only to restate your own misconceptions which I have already corrected once, and to make personal references.

Robk
05-09-2003, 02:02 PM
The mean of recurrence time is given by u = (1 - p^r)/qp^r. With p = 1/221, q = 1-p, r = 3, we have that we expect to be dealt aces 3 times in a row once every 10,843,923 hands. The probability of no streak of r in n trials is given (asymptotic approxiamtion, good for n large) by
q(n) = (1-px)/(r + 1 - rx)*x^n+1 where x is the unique positive solution of 1-qs(1 + ps + ... + (ps)^r-1) = 0. If you want to know the answer just plug away.

Robk
05-09-2003, 02:17 PM
What used to be in here is a lie. I hope no one read it LMAO! My approximation gives .00248%.

Cyrus
05-09-2003, 02:22 PM
"This forum has always been most congenial since its inception up until now."

I do not recall being anything but congenial, at least in this forum. But if your idea of congeniality is the tone and the attitude you have adopted throughout this thread, then I'm sorry but you have no idea what the word means.

"I will derive yet another one, from which you may compute the answer as accurately as you please."

Another "exact approximation". I see. Well, I asked for something very specific, but I see that I cannot get through. The formula for "at least one streak of m occurences in n trials, when n is sufficiently large".

Your answer refers to boundaries. But this is not what I'm after.

"...gives an answer in agreement with the computer program you found (I'd still like this link)."

Why do you think that I'm asking for a formula when the best I have is a computer program? (As to the link, I would have provided it if it wasn't a restricted Math site. You could contact the website's owner is you're interested in joining up. Send me a private message if you do.)

"Your insistence that the binomial distribution should be used for this end despite my "objections" was nonsense. The thing that wasn't germane was in regard to another comment, not concerning why your method was terrible, it's still terrible."

(patiently) I made very clear in my post that we could obtain an answer using the BinomDist formula if we want the probability of getting exactly one streak of exactly 3 Aces in a row in the next n hands. You are supposed to prove that this is "nonsense" or "terrible analysis" using mathematics. And not your kind of "congenial" diversions.

"The idea that there is something about this elementary formula I do not understand is also nonsense, and you should have considered that."

The formula for the problem I have posited is not an "elementary" one. But from now on I will respectfully and pre-emptively accept that you understand everything and that you have the answers for everything.

I hope this is congenial enough for you and alleviates your angst.

--Cyrus

Robk
05-09-2003, 02:41 PM
q(n) = (1-px)/(r + 1 - rx)*x^n+1*q

is the correct formula. Note the extra q in the denominator. This is why math shouldn't be done on message boards.

BruceZ
05-09-2003, 04:20 PM
I lost a digit. Those should be .00228...

BruceZ
05-09-2003, 07:02 PM
Another "exact approximation". I see. Well, I asked for something very specific, but I see that I cannot get through. The formula for "at least one streak of m occurences in n trials, when n is sufficiently large".

First of all, there are no "exact approximations". There are formulas which produce exact answers, there are formulas which can only produce approximate answers to within fixed upper and lower bounds, there are formulas which can produce answers to within any desired accuracy but never exact (infinite series), and there are formulas which can provide approximations to any degree of accuracy up to and including exactness. The formulas based on inclusion-exclusion I gave for opponents holding various hands such as AA when you hold KK are of this latter type. That is where this "exact approximation" stuff first came up. Those formulas could always be used to obtain an exact answer for any situation. We obtained an exact answer for AA vs. KK with 9 opponents, and we obtained exact answers for other hands with 4 players. For some hands with 9 players, we used the formula to produce approximate answers only, but these same formulas could have been used to produce the exact answers as well, it was simply not desirable to do so. What you seem to not understand is that the formulas derived by the inclusion-exclusion principle are exact when carried out to all the terms, and they represent closed form formulas. They are only approximate if you choose to make them approximate. The present case is another example, albeit one for which it would certainly be tedious to write out all the terms, and even more tedious to generalize it. I don't know why anyone would want to, but that doesn't mean it can't be done in theory.

In the end, everything is an approximation anyway because whatever you use to compute it on has finite precision, and the results are generally only used and only useful to very limited precision. So I think this obsession with exact closed form formulas is largely unjustified.

Your answer refers to boundaries. But this is not what I'm after.

Well excuse me! First of all, this thread began with a very specific question, not a general one. Failing a general solution, when someone does a lot of work to produce several very accurate approximations, the normal response is normally "thank you" , and not "your method is extremely inadequate" because it is only accurate to the zillionth decimal place. But that's fine, I don't spend my time posting solutions just for your approval.

Second of all, what you want in all likelihood does not exist anywhere. These renewal process problems are usually done with the Laplace transform methods like robk has presented, and these are also only carried out to approximation. Few texts on this subject would provide a solution to this problem based on simple combinatorics which I have provided. On this forum I have found a special case of a renewal process with p=1/2 that yielded a closed form solution in terms of the Fibonacci sequence. I have not found this solution described in the texts which deal with such processes, only the Laplace approximations. I am considering looking into whether my result is generally known, and I am also working on generalizing it.

Third of all, I gave you an exact closed form formula in terms of conditional probabilities. If you're too lazy to compute all the terms that isn't my problem.

I made very clear in my post that we could obtain an answer using the BinomDist formula if we want the probability of getting exactly one streak of exactly 3 Aces in a row in the next n hands. You are supposed to prove that this is "nonsense" or "terrible analysis" using mathematics. And not your kind of "congenial" diversions.

But you have not shown how this probability can be computed exactly with the binomial distribution, nor can you, because this probability is not binomially distributed. I wrote a whole paragraph, of which you only made fun of the length (did you bother to read it?) where I detailed the reasons why this is the case. What BinomDist gives you is the probability of exactly 1 success of probability p in n independent bernoulli trials, or at most 1 success, or at least 1 success if you subract the result from 1. We don't have independent bernoulli trials, so the whole thing is irrelevant to your analysis except as an approximation, and I know you don't want an approximation, so why you continue to push the binomial distribution is completely beyond me. I don't know what you intend to do with it. If you are going to make a claim that you can use this distribution to compute this probability, you had better damn well show how it can be done. If you say you can square the circle, the onus is not on me to prove that you can't.

Both AC_Player and myself have corrected you that if you do use the binomial distribution to approximate this, you can get the probability of at least 3 in a row also. Furthermore, I also indicated why the sum of these exact probabilities you indicated do not produce the correct probability. In the words of the late Peter Griffin "I gave you an explanation. I don't owe you an understanding".

The formula for the problem I have posited is not an "elementary" one. But from now on I will respectfully and pre-emptively accept that you understand everything and that you have the answers for everything.

But the specific problem we were solving (at least 1 run of length 3 or more in 250 hands) lends itself to extremely simplistic methods which are extremely accurate. I completely understand the problem, the limitations of the simplistic methods when applied to the problem, and the errors in the analysis which you continue to advocate. Do you?

I do not recall being anything but congenial, at least in this forum. But if your idea of congeniality is the tone and the attitude you have adopted throughout this thread, then I'm sorry but you have no idea what the word means.

No one spends more time and energy than I do helping people on this forum. I get pms all the time from people who really appreciate it. Some people don't feel they know enough to ask questions publicly, so they learn by reading or asking questions privately. I'm more than generous with my time, and have almost infinite patience with people of all levels of competence. You've asked some questions, and I've been particularly generous trying to help you. However, my patience quickly runs out when my efforts are unappreciated, misunderstood, and even ridiculed. I will always challenge incorrect ideas masquerading as truth. Normally they are quickly reasoned out to a conclusion, and not irrationally defended to the death. You need to be able to separate criticism of ideas from personal criticism. My characterization of what you wrote was accurate, and it becomes more accurate as you continue to defend something which is blatantly wrong. Your characterization of what I wrote was not accurate.

Cyrus
05-10-2003, 01:51 AM
"First of all, there are no exact approximations."

You're telling me. I used your term (http://www.twoplustwo.com/forums/showthreaded.php?Cat=&amp;Board=probability&amp;Number=242 600&amp;Forum=All_Forums&amp;Words=BruceZ&amp;Match=Username&amp;S earchpage=2&amp;Limit=25&amp;Old=allposts&amp;Main=241800&amp;Sear ch=true#Post242600), only to show how preposterous the concept is. You see, I had asked in a previous thread for an exact answer and you insisted that you gave one but that it differed from the exact one by "a small amount". And this I'm supposed to "appreciate" and stop "pestering you".

Indeed, this is an infrequent but bothersome problem with your posts, or at least the ones I have perused the short time I have : You are very cavalier with terminology!

And that comes across despite your sincere willingness and the effort you put in trying to help others, in cases like the wild Goedel misconcenptions (http://www.twoplustwo.com/forums/showthreaded.php?Cat=&amp;Board=genpok&amp;Number=225198&amp;p age=&amp;view=&amp;sb=&amp;o=&amp;vc=1) (comingling the statement "P is neither true nor false" with the statement "P cannot be proven to be true nor false"!) or your recent misuse (http://www.twoplustwo.com/forums/showthreaded.php?Cat=&amp;Board=genpok&amp;Number=254115&amp;p age=2&amp;view=expanded&amp;sb=6&amp;o=14&amp;fpart=) of the term Expected Value ("Suppose you play three 4 hour sessions. In the first session you win \$200. In the second session you win \$400. In the third session you lose \$300. Your average win or EV is = \$100/session or \$25/hour.". No, Bruce, this is not "EV". This is just Average Win. Your indeed very thoughtful and helpful post could create such a confusion in the entry level player's mind about what EV is, that the worth of your post diminishes sharply.)

Do not, please, mistake my unwavering insistence for clarity and exactitude for anything but that. Calling this attitude names ("argumentative troll") is characteristic of neither a mathematician nor an aspirant teacher.

"No one spends more time and energy than I do helping people on this forum."

I have not disputed this. As a matter of fact I have given you a very high contributor's rating for precisely those efforts of yours. I for one appreciate them.

So, a parting word, which is not original and has already (http://www.twoplustwo.com/forums/showthreaded.php?Cat=&amp;Board=genpok&amp;Number=233317&amp;p age=&amp;view=&amp;sb=&amp;o=&amp;vc=1) been suggested to you, in paraphrasing : Being precise does not have to lead to significantly long or less comprehensible posts.

Take care.

--Cyrus

BruceZ
05-10-2003, 10:03 AM
or your recent misuse of the term Expected Value ("Suppose you play three 4 hour sessions. In the first session you win \$200. In the second session you win \$400. In the third session you lose \$300. Your average win or EV is = \$100/session or \$25/hour.". No, Bruce, this is not "EV". This is just Average Win. Your indeed very thoughtful and helpful post could create such a confusion in the entry level player's mind about what EV is, that the worth of your post diminishes sharply.)

No Cyrus, you are wrong, but I'm glad you brought it up, and I will clarify this on the other forum as well. I don't know if I created any confusion, but your attempt to be pedantic here may very well have done so, but that's OK because I will clear it up now to everyone's benefit. I'm not going to debate semantics, though my semantics are technically correct, because what is important are not semantics but the understanding one has of concepts, and how to correctly apply them in practice. Inability to do this effectively is a common shortcoming of many bookish types and mathematicians who only understand theory, as opposed to those with a more practical bent, such as those with a background in engineering, physics, or statistics. At the same time, some of these practical folks may have lapses in fundamental theory. Different fields even use the same terms to mean slightly different things, the notion of independence being one example. My own background is in all these fields, so I will attempt here to bridge the gap.

EV stands for expected value. An elementary statistics book will tell you that the terms "mean", "average", "expectation" and "expected value" are all synonymous. One may compute a mean, average, expectation, or expected value over a finite number of samples, as I have done in my example. Now it is true that the term expected value is normally used by mathematicians to describe the mean of a random variable or of a probability distribution. As such, it can only be obtained from an infinite number of samples of the random variable, or in the limit as the number of samples approaches infinity by the law of large numbers.

When a gambler says "my EV is \$50/hr", this can mean one of two very different and equally valid things. He can mean that he feels that the game he is in worth \$50/hr to him, and if he plays until infinity under identical conditions, this would be his average win per hour. How can he know this EV? After all, nobody can really play to infinity. If the game were a relatively simple one like blackjack, he would simply need to perform a mathematical calculation or simulation taking into account the rules, number of decks, deck penetration, bet size, and strategy. If the game were more complex like poker, perhaps he made some estimates based on how he perceives his skill relative to that of his opponents, and how much that should be worth in dollars. Perhaps, on the other hand, he did not perform such a calculation, but instead simply averaged his actual winnings over some number of hours of play, as I did in my example. If he played for a very long time, the law of large numbers tells us that his result should be very close to the EV he would obtain if he had played until infinity. If he didn't play for such a long time, his result could differ significantly from this value, but it would still be an average of his results nonetheless, and he would use this value to estimate the EV he would obtain if he were to play forever. My simple example used only 3 sessions for simplicity, but the method was meant to be illustrative of how one would perform this computation over many more sessions, in which case one would arrive at a number very close to the EV if one played forever.

Hence there are two different EVs at play here, one which is a theoretical result that he would obtain if he could play forever, and this is an inherent characteristic of the player in the game. The other is a practical result that he computes after a finite number of hours. The latter of these is the more practical notion of EV for complex games such as poker, since the other one can never be known precisely. The term "sample average" is often used to describe the average of a finite number of samples, and the sample average is an estimator of the long term EV.

Just as an average can be computed from a finite number of samples, a standard deviation can be computed from a finite number of samples too. This is what I did in my example. This is a "sample standard deviation", and it is an estimate of the theoretical standard deviation one would obtain if one played forever. Perhaps you have a calculator (or Excel) which gives you the mean and standard deviation of a series of numbers. You don't seem to have a problem calling this a standard deviation, it even says standard deviation next to the key. You should not have a problem calling the mean an EV either, though I admit this is somewhat atypical and potentially confusing. If you study statistics you learn that variance, which is the square of the standard deviation, is the second moment of a generating function when the mean is taken to be zero. The first moment is expected value. If you don't understand this that's OK, my intent is simply to link the notions of EV and standard deviation.

Actually the question of whether EV and standard deviation must be regarded as fixed numbers inherent to a situation, or if they can be regarded as probability distributions whose true value can never be known but only estimated from finite information is a debate which gives rise to different branches of statistics and estimation theory. It is tied to the question of the very definition of probabillity, and whether probability is an inherent quantity, or one which can only be obtained through experiment. Your argument that an average win is not an EV is tantamount to arguing that probabilities can never be known, and are simply arbitrarily assigned quantities, rather than experimentally measured ones. You would not be alone in making this argument, but that does not make the practical argument incorrect or invalid. This is also related to fundamental questions regarding the apparent probabilistic nature of the universe as described by quantum mechanics. I'm digressing to indicate that this issue actually runs much deeper than our discussion.

"First of all, there are no exact approximations."

You're telling me. I used your term, only to show how preposterous the concept is. You see, I had asked in a previous thread for an exact answer and you insisted that you gave one but that it differed from the exact one by "a small amount". And this I'm supposed to "appreciate" and stop "pestering you".

I used this term as a title to a post in a humorous sort of way in response to your question, which was something like, "how can something be both exact and approximate". I feel that you still do not appreciate this concept, and that this is a serious shortcoming which underlies your inability to extract value from this thread as well. The fact is that I GAVE AN EXACT ANSWER TO THE PROBLEM IN QUESTION. Not just an approximation, not an exact approximation, but a genuine exact answer. I did it in an earlier thread, and I did it very concisely at the bottom of the thread you referenced in the post titled "Much simpler way!" The whole purpose of that discussion was to show that one may begin with a very simple first-order approximation (do you know what that is?) which is often very accurate in itself, and then continue to refine this by adding higher order terms until one eventually arrives at the exact answer, and this can always be done for problems with a finite number of outcomes. THIS IS AN EXTREMELY POWERFUL AND IMPORTANT CONCEPT. You will not find it described in this kind of detail in very many places. In the thread that you referenced I provided more explanations to you than most anyone would require, and still you failed to grasp the importance of this. Sorry if this sounds harsh, but this makes it obvious that you possess no qualification to evaluate my explanations of anything of a mathematical nature, NONE WHATSOEVER!

in cases like the wild Goedel misconceptions (comingling the statement "P is neither true nor false" with the statement "P cannot be proven to be true nor false"!)

I never claimed these statements are equivalent. What I said was "It may be a surprise to a lot of people to learn that not all things are either true or false". That is a true statement! Then I said "There are some things that are neither true nor false, but in a third state of 'undecideable', and you can prove that". That is also true in the field of quantum mechanics, and I had this in mind when I wrote that post. This has nothing to do with Gödel. The purpose of these statements was to spark controversy and debate, and to set up the refined explanation which followed.

I have freely admitted that I was very sloppy in the terminology I used later in that post when discussing Gödels theorems, and I subsequently posted a multi-page post which corrected this and explained the matter clearly.

Calling this attitude names ("argumentative troll") is characteristic of neither a mathematician nor an aspirant teacher.

Now let's not commingle comments made in jest in a forum intended to let one's hair down and blow off steam, with statements made on a math forum. My comments here are deadly serious. If I see something that is garbage, I will call it garbage. If you can't handle that, then don't post on a math forum.

Being precise does not have to lead to significantly long or less comprehensible posts.

True, though there is compromise between quantity of information conveyed, time spent, and the probability that one may occasionally be imprecise. This is one reason why twoplustwo should be commended for maintaining both high quality and high quantity of information it publishes. That isn't an easy thing to do. I think I negotiate this compromise pretty well, and I do try to always be precise, and I was precise in two of the three examples you cited here. It is my hope and belief that other's agree. If they didn't, then I would not post nearly as much.

Responses from others are welcome, and from you as well even though you said it was your final post.