getting 3 pocket Aces in a row

[Ray Zee]

its much more likely to strike the same spot again, then a non previously struck spot. it isnt a random act.

[FA_man]

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]

Can an act be random? And is lightning an act? But then again, I'm not religious [img]/forums/images/icons/smirk.gif[/img]

[Cyrus]

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

"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]

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]

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]

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]

"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 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]

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. [img]/forums/images/icons/tongue.gif[/img]

[Cyrus]

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]