What are the odds of 7 pocket aces in 350 hands

[SaintAces]

assuming no cheating

[uuDevil]

You can calculate the probability of getting AA precisely 7 times in 350 hands in Excel by using the statement
"=Binomdist(7,350,1/221,false)" which gives a result of .000985, which is equivalent to odds of 1014:1.

If you wanted to know the probability of getting AA 7 times or fewer in 350 hands, you would use "=binomdist(7,350,1/121,true)" which results in 0.99977, or odds of 1:4326.

If you wanted to know the probability of getting AA more than 7 times in 350 hands, that would be 1-0.99977=0.00023, or odds of 4326:1.

Edit: And in case anybody cares, the most likely number of AA hands in 350 deals is 1.

[Wake up CALL]

I do not see how this can possibly be correct.

[Izverg04]

[ QUOTE ]
I do not see how this can possibly be correct.


[/ QUOTE ]
Why is that? Calculation is exactly correct.

It's easy to quickly get an order of magnitude estimate. Probability of seeing N(2 Aces) here is Poisson-distributed with mean=350/221=1.6 and sigma=sqrt(mean)=1.3, so value of 7 is (7-1.6)/1.3=4 sigma away. 4 sigma confidence interval corresponds to O(10^-4) probability.

[uuDevil]

I'm not a mathematician or statistician, but here is what I did:

The binomial distribution describes the probability of outcomes of an experiment that can result in either success or failure where the probability of a success is p, x the number of successes, n the number of trials.

b(x:n,p)= COMBIN(n,x)*p^x*(1-p)^(n-x)

or in Excel, binomdist(x,n,p,false)

For this problem, n=350, p=1/221, giving the following results for x=1 to 7 (values truncated, not rounded):


k P(x=k) Odds P(x<=k) Odds P(x>k) Odds
0 0.20447 3.89 0.2044 3.8905 0.7955 0.2570
1 0.32530 2.07 0.5297 0.8876 0.4702 1.1267
2 0.25802 2.88 0.7878 0.2694 0.2122 3.7126
3 0.13604 6.35 0.9238 0.0824 0.0761 12.132
4 0.05364 17.6 0.9774 0.0230 0.0225 43.442
5 0.01687 58.2 0.9943 0.0057 0.0056 176.72
6 0.00441 225. 0.9987 0.0012 0.0012 821.17
7 0.00098 1014. 0.9997 0.0002 0.0002 4325.6

(Ughh! Is there a way to make a decent table?)

So, for instance, the odds against getting AA more than 6 times in 350 tries are 821:1.

[uuDevil]

[Bozeman]

For our purposes, the meaningful stat is probably 7 or more, which happens about .13% of the time for a random set of 350 hands. If instead you chose these 350 hands because 349 only gave you 6, the prob. (6+ out of 349) is more like .6% of the runs of 350 hands that start with AA will have 7+.

Craig

[Izverg04]

Nobody corrected my typo, so I'll have to do that myself. 4 sigma corresponds to O(10^-3) probability. Sorry.

[mostsmooth]

the chances of getting AA 7 times or less in 350 hands has got to be much better than 1 in 4326.

[BruceZ]

It's not 1 in 4326, it's odds of 1-to-4326 against, which means it happens 4326 in 4327 times.

[mostsmooth]

[ QUOTE ]
It's not 1 in 4326, it's odds of 1-to-4326 against, which means it happens 4326 in 4327 times.

[/ QUOTE ]
so i was right? 8-)

[uuDevil]

[ QUOTE ]
the chances of getting AA 7 times or less in 350 hands has got to be much better than 1 in 4326.

[/ QUOTE ]
This statement is true.

[ QUOTE ]
If you wanted to know the probability of getting AA 7 times or fewer in 350 hands, you would use "=binomdist(7,350,1/121,true)" which results in 0.99977, or odds of 1:4326.

[/ QUOTE ]
This statement is also true.

The odds AGAINST getting AA 7 times or less in 350 hands are 1:4326. I.E. this is FAVORED to occur-- 1 time you will get AA more than 7 times and 4326 times you will get AA less than 7 times. Writing it this way may be confusing, but I think this is the common convention.

From now on, I think I'll leave these questions to BruceZ et. al. /images/graemlins/grin.gif

[pzhon]

[ QUOTE ]

It's easy to quickly get an order of magnitude estimate. Probability of seeing N(2 Aces) here is Poisson-distributed with mean=350/221=1.6 and sigma=sqrt(mean)=1.3, so value of 7 is (7-1.6)/1.3=4 sigma away. 4 sigma confidence interval corresponds to O(10^-4) probability.


[/ QUOTE ]

/images/graemlins/spade.gif It is not good to use a normal approximation to a Poisson distribution with a mean that low. If you don't have to compute it in your head, just use the actual Poisson distribution: mean^n / (n! e^mean), where here the mean is 350/221.

The normal approximation to the Poisson distribution is more appropriate for a larger mean and events closer to the mean. The normal approximation's tail is much smaller than the Poisson distribution's tail for a large number of standard deviations above the mean.

/images/graemlins/heart.gif For an integer-valued variable to be 7+, you should use something more like the range 6.5+.

/images/graemlins/diamond.gif 3+ standard deviations is about 0.135% (one tail), or 740:1. 4+ standard deviations is about 43 times as unlikely, 0.00317%, or about 32000:1.

Side note: 3 sigmas is roughly 1/1000, so the 6-sigma quality control movement in manufacturing is supposed to be an effort to reduce defects to 1/1,000,000, despite the fact that 6 sigmas is about 1/10^9 rather than 1/10^6. It's funny to start a quality control program with an error.

/images/graemlins/club.gif If you hear someone say, "I just had aces 7 times in 350 hands!" you don't know how surprising that should be. How many times would they have aces 5 times, but make a mistake or exaggerate? Most extremely unlikely streaks you hear about or think you have personally observed, without objective records, are surprisingly vulnerable to this. If you would miscount 2% of the time, but less exciting streaks are 50 times as common, then half of the time you think you see a longer streak you are wrong.

Were the 350 hands chosen to start and end with aces, or was it a natural set of 350 hands? How much does that player play? How many people will that one player tell? How likely is it that it really happened to a "friend of a friend" but the teller is pretending it happened to him? It is much less surprising for an event to happen at least once in history than for it to happen tomorrow.

[mackthefork]

Edit: And in case anybody cares, the most likely number of AA hands in 350 deals is 1.

[/ QUOTE ]

I don't like to disagree with you but I think the most likely number of AA hands in 350 is 2, though I may be missing something, it's certainly marginal.

[dansalmo]

I once got pocket kings 7 times in a 2 hour session, and all 7 held up. (15-30 game) The odds of getting them and habing them hold up is must be more than 500,000 to 1

I also once made 3 quads, 3 full houses, and 2 sets in one 30 minute dealer rotation. The old guy who gave me the seat (it was his option to take it, but then he changed his mind) to this day thinks I am the luckiest guy he has ever seen play. (15-30)

[maurile]

Interesting thread because I got pocket aces 8 times last Saturday in 10 hours of play (B&M). They held up 5 times.

[pzhon]

[ QUOTE ]
Edit: And in case anybody cares, the most likely number of AA hands in 350 deals is 1.

[/ QUOTE ]

I don't like to disagree with you but I think the most likely number of AA hands in 350 is 2, though I may be missing something, it's certainly marginal.

[/ QUOTE ]

The probabilities of getting each number of pocket aces in 350 hands:

0: 20.45%
1: 32.53%
2: 25.80%
3: 13.60%
4: 5.36%
5: 1.69%
6: 0.441%
7: 0.0985%
8: 0.0192%
9: 0.0033%
10: 0.0005%

The median result changes from 0 to 1 at 153 hands, and from 1 to 2 at 371 hands. A Poisson approximation would say 154 and 371, 221(log_e 2) rounded up and 221 (1.67835) rounded up. The 1.67835 is the value of a nonelementary function that I don't believe simplifies.

I haven't proved it, but my guess is that between (k - 1/3)221 and (k + 2/3)221 hands, the median number of aces is k, for k sufficiently large. Since 350 is comfortably between 2/3 and 1 2/3 of 221, the median number of times you get pocket aces is 1.

[mackthefork]

Mine was an estimate (no calculator), I did not realise the probability of 0 was so high nor that over 2 was even more probable than zero. Thank you for your detailed reply, it is much appreciated.

Sincerely

ML

[pzhon]

[ QUOTE ]
I haven't proved it, but my guess is that between (k - 1/3)221 and (k + 2/3)221 hands, the median number of aces is k, for k sufficiently large. Since 350 is comfortably between 2/3 and 1 2/3 of 221, the median number of times you get pocket aces is 1.

[/ QUOTE ]

Apparently, I wasn't the first to conjecture this (J. Chen and H. Rubin did in 1986, proving some bounds), and it was proven by 2002: www.math.uu.se/~sea/pogamma.pdf (http://www.math.uu.se/~sea/pogamma.pdf)