One In A Million

[anselm 0]

Just in case someone's not read it yet... Bruce Hayek of the Tilt Boys tells his story of getting aces back-to-back-to-back:http://www.tiltboys.com/html/trip-reports/barge04.htmlIt's under "Thursday Night Poker" and starts with "Then, about 20 minutes later, something happened. Something wonderful..."

[semaj550 0]

The math isn't really all that hard, it just depends on how many hands your sample is. Just for the fun of it, let's assume we're looking at 300 hands and we want to know what the odds are that any given sequence of 3 hands is any combination of AA or KK.So, first the probability of 3 specific hands in a row being AA or KK.let A = hand dealt is AAlet K = hand dealt is KKlet X = event that 3 consecutive hands are either AA or KKP(AorK) = (4/52)(3/51) + (4/52)(3/51) = 0.009050P(x) = P(AorK)^3 = 0.0000007412So the odds of 3 specific hands being either AA or KK is about 1 in 1,349,131. But what are the chances this sequence occurs within a sample of 300 hands? Well, if I am correct, we can just look at 100 sequences each of 3 hands.let Y = event that a given 3 hand sequence is all hands of either AA or KKP(Y) = C(100,1) * 0.0000007412 * 0.9999992588^99P(Y) = 100 * 0.0000007412 * 0.9999266239P(Y) = 0.00007411So the odds of any given 3 hand sequence, in a sample of 300 hands, is about 1 in 13,492.Now, I don't remember my statistics well enough to remember if this assumes you play 3 hands, test to see if it meets the criterion and then deal 3 more hand and repeat meaning that if the last hand of a sequence was AA and the first 2 of the next sequence were AA then KK we would accidentally not count that as having satisfied the criteron. If that's the case the odds would higher than I have given.

[MasterLJ 0]

Probabiltiy of AA or KK: (8*3) / (52*1) = 2/221 = 1 / 110.5not 1/110to be technical.*********AA or KK 3 in a row:0.00000074116203646= 1 / 1,349,232.625Out of 100k hands in a row...hmmm....The trick is in figuring out the number of trials, I don't think it's 99,998.Maybe ~33,333, no ..hmmm Yuo see if you use 99,998 they aren't independent...overlapping sets of hands.But to split them into 33,333 group obviously lowers the chance of 3 in a row, because you are breaking them up.alone in dorm room?

You can use a binomial distribution and find the confidence interval within 95% for the event to occur once.Simply stated this would tell you that in 95% of all series of length N (where N is the number you calculate) 3 consecutive aces or kings occurs.That calculation isn't too bad, but you'd need a computer or SAS to run it.Actuary, do you have access to SAS?

[FatBurger 0]

...although I think the poster you're responding to was mostly kidding.

Yes, I was, though I'm still glad that he took the time to make a detailed post for everyone.

[MasterLJ 0]

Yes, I was, though I'm still glad that he took the time to make a detailed post for everyone.

I got that the second part was a joke, but not the first =P. Anyway, it's a good review for me.

[Actuary 3]

semaj550,you did what I noted, you cant split the sets up, really changes things.*************MasterLJWe don't have independent sets of 3 hands.They interlock, so how do I use the Binomial?**************obviously, with 100k hands, it's still unlikely.I wonder at what number of hands in a series, it becomes >50% to see it at least once.Probably around 750k.

[semaj550 0]

semaj550,you did what I noted, you cant split the sets up, really changes things.

hehe...I was typing my post while your were typing yours, I didn't see it until afterward.Any idea on how to tackle the problem? I've exhausted what I can remember about elementary statistics.

[MasterLJ 0]

semaj550,MasterLJWe don't have independent sets of 3 hands.They interlock, so how do I use the Binomial?

That's a good point.

[Actuary 3]

hehe...I was typing my post while your were typing yours, I didn't see it until afterward.Any idea on how to tackle the problem? I've exhausted what I can remember about elementary statistics.

so I"m walking to my car..to go home..and actually come back here when I got this idea.Let's simplfiy:Lets say: How many times will we flip tails 3 or more times in a row if we flip a coin N times?We have N-2 possible runs of 3:123234....n-1, n-1, n1/2 of our flips are expected to be tails.1/2 the times we flip a tail, we will flip another right after1/2 the times we flip two tails in a row, we flip a 3rdSo 1/4 times we flip a tail, we are on our 3rd tail flip in a row. So its (1/8) * (N-2) times we'd expect.In this case, you would count T T T T T T as 4 successes of 3 in a row.In the AA/KK case, the >3 in a row is too small to consider.I think this makes senseIt's intuitive for a series long enough, that you'd generally expect it too happen the number of times = to Probabiliy X Opportunity

[semaj550 0]

so I"m walking to my car..to go home..and actually come back here when I got this idea.Let's simplfiy:Lets say: How many times will we flip tails 3 or more times in a row if we flip a coin N times?We have N-2 possible runs of 3:123234....n-1, n-1, n1/2 of our flips are expected to be tails.1/2 the times we flip a tail, we will flip another right after1/2 the times we flip two tails in a row, we flip a 3rdSo 1/4 times we flip a tail, we are on our 3rd tail flip in a row. So its (1/8) * (N-2) times we'd expect.In this case, you would count T T T T T T as 4 successes of 3 in a row.In the AA/KK case, the >3 in a row is too small to consider.I think this makes senseIt's intuitive for a series long enough, that you'd generally expect it too happen the number of times = to Probabiliy X Opportunity

Wow, that's really good thinking! I guess that's why you're an actuary. :-)So if we had a sample of 300 hands, and the probability of 3 in a row is still 0.0000007412 then the odds of it happening are (300-2) * 0.0000007412 or 0.0002209 which is about 1 in 4,527. That's sounds very likely since I calculated odds of about 1 in 13,000 assuming each set of 3 hands was an independant trial.Someone give this man a cookie!So now that I'm thinking about that, it's pretty easy to calculate the number of hands before you could expect this to happen.(x-2) * 0.0000007412 = 1x-2 = 1,349,163.5x = 1,349,165.5So we would expect to be dealt some combination of AA or KK in three consecutive hands at some point during each span of about 1.35 million hands.

[dan310 0]

I think this makes senseIt's intuitive for a series long enough, that you'd generally expect it too happen the number of times = to Probabiliy X Opportunity

But doesn't this give us the expected number of AA or KK three time in a row and not the probablitility that it will happen once?

[MasterLJ 0]

But doesn't this give us the expected number of AA or KK three time in a row and not the probablitility that it will happen once?

Once you have the probability that it happens 3 times in a row, you need only divide your series length by 3, and treat them as individual events... I think...If that's the case you could use a binomial distribution.

[SunDrop 0]

Husband?? What was all that one in a million talk?!

I can't believe no one else caught this.

[GrinderMJ 0]

all this hot math action is turning me on

[Verdimme 0]

I once had AA back to back to back.Same guy payed me off the first two times..and folded the third time to a river bet. I showed.

[XX44466XX 0]

No. The odds of being dealt AA is 1 in 220. The odds of being dealt AA or KK is better. I'm not sure exactly what. I bet Al Gore knows.

I'm no Al Gore but the probability of being dealt a pocket pair Jacks or better is 54-1 while merely any pair in the hole is 16-1.

[semaj550 0]

I'm no Al Gore but the probability of being dealt a pocket pair Jacks or better is 54-1 while merely any pair in the hole is 16-1.

To find the probability of being dealt either AA or KK on a single hand you just add their independant probabilities together.P(AA) = (4/52)(3/51) = 0.004525 = 1/221So the chaces of being dealt any specific pocket pair on a given deal are 1 in 221. The chances of being dealt either of any two specific pocket pairs on a given deal are exactly twice that.P(AA or KK) = 0.004525 + 0.004525 = 0.00905 = 1/110.5So the chances of being dealt either are 1 in 110.5 (which was provided earlier).You could also use this to calculate the chances of being dealt any combination of hands. You just find the probability that each event will happen and then add them all together.

[XX44466XX 0]

Will you take my next Math midterm for me?Good stuff man.

[semaj550 0]

Now, if you can tell me how to calculate what the probability is of being dealt either aces OR kings on three consecutive hands at any time during a sample of 100,000 consecutive hands, I'll be impressed.Not that it matters or anything, but I gotta believe there are math geeks out there who live for this kind of thing. On a Friday night. Alone in their dorm room.

Okay, this is probably a little redundant but I thought I'd calculate it anyway. If you read Actuary's great post you'll realize that I'm just computing the answer from his suggestion but here you go:let X = event that a player is dealt AA or KK on each of any 3 consecutive handslet Y = event that event X occurs in a sample of N handsP(X) = 0.0000007412 (calculated in previous posts)P(Y) = (N-2) * 0.0000007412P(Y) = (100,000 - 2) * 0.0000007412P(Y) = 0.07412So the odds against this event occuring in a sample of 100,000 hands are about 25:2.

[Drwnded 0]

Wow, that's really good thinking! I guess that's why you're an actuary. :-)So if we had a sample of 300 hands, and the probability of 3 in a row is still 0.0000007412 then the odds of it happening are (300-2) * 0.0000007412 or 0.0002209 which is about 1 in 4,527. That's sounds very likely since I calculated odds of about 1 in 13,000 assuming each set of 3 hands was an independant trial.Someone give this man a cookie!So now that I'm thinking about that, it's pretty easy to calculate the number of hands before you could expect this to happen.(x-2) * 0.0000007412 = 1x-2 = 1,349,163.5x = 1,349,165.5So we would expect to be dealt some combination of AA or KK in three consecutive hands at some point during each span of about 1.35 million hands.

So, pretty close to one in a million huh? Kind of ended up back where we started Well done guys. I am, as promised, impressed.

[semaj550 0]

So, pretty close to one in a million huh? Kind of ended up back where we started Well done guys. I am, as promised, impressed.

Well there odds of it happening aren't necessarily 1 in a million, it all depends on how many hands we are talking about. For a sample as small as 300 hands it's only about 1 in 4,500. However, you would expect it to happen once in ever 1.35 million hands.And the odds of it happening on 3 specific hands in actually about 1 in 10.6 million. :-P

[JBradburn6 0]

Husband?? What was all that one in a million talk?!

I was gonna go with "So you're saying there's a chance" but this works.

No. The odds of being dealt AA is 1 in 220. The odds of being dealt AA or KK is better. I'm not sure exactly what. I bet Al Gore knows.

Haha, so does Loogie = Al Gore?

[troyomac 0]

I can't believe no one else caught this.

Me neither actually, I thought it was genious.

I was gonna go with "So you're saying there's a chance" but this works.

That one works too

[Actuary 3]

But doesn't this give us the expected number of AA or KK three time in a row and not the probablitility that it will happen once?

yes.semaj550,multiplying the P(x) by N-2 = Expected number of times it will happen...otherwise, you can't have a probability >1..***************************************But, I'm not happy with my reply earlier.There is a difference between taking N-2 independent trials, and taking N-2, from a series.Although, intuitively, since this is a 1/110.5 event individually, the risk of dependence skweing the anwer much isw mitigated.I think...but still want a better answer...I'ma dork************************************

[Kathy Liebert 0]

Baby girl better known as Kathy, you wish id come swing by your way so you and I could "play"......cuz baby Im one in a million ohh ohhh