getting 3 pocket Aces in a row

[BruceZ]

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]

"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. [img]/forums/images/icons/grin.gif[/img]

"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]

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]

"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]

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]

"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]

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]

This seems entirely plausible, but there are no batteries in my calculator just now.

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

[Robk]

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]

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.