Probability of a higher pocket pair

[im a model]

you say that 22 is good 42.02% of the time. you say brians page is exact. brians page says that 22 is good 32.69% of the time. what am i missing?

[R Gibert]

I realize that

1- [(1225-72)/1225]^9

comes from

1- [(50c2-4c2*(13 - 1))/50c2]^9

but if it were so easily computed, that article would have been about 2 or 3 paragraphs long instead of several pages.

It's amusing that it is as close as it is, but of course your calc is still short of what the article gives, e.g.

0.42025238946 < 0.42062


I've checked my Python program, at appears to be correct to me:
<font class="small">Code:</font><hr /><pre>
from random import *

n = 100000
for r in range(13, 1, -1):
c = 0
d = []
for i in range(50):
d.append(i/4 + 2)
t = r*4 - 8
d[t:t + 4] = [14]*4
d[48:] = [r]*2
print "%2d:" % (r),
for i in range(n):
shuffle(d)
for j in range(0, 17, 2):
if d[j]==d[j + 1] and d[j] &gt; r:
c += 1
break
print "%6.4f" % (float(c)/n)
</pre><hr />

[jay_shark]

If you notice on Brians page there are three columns and to be exact there should be more columns but the result would be negligible so he didn't include them all .

If you add the columns you should get a number close to 42% .
ie , .3269+ 0.08186 + 0.01186 =~ 42% .

Also , remember that my formula is a very good approximation since it assumes that the events are independent . This isn't entirely true but it is damn close to being independent .

In general , you should use this :

If you hold pocket pair X at an n handed table , then the answer is approximately

1- [1225-a)/1225]^n where a is the number of pairs higher than your pocket pair x . . For instance , if you hold pocket 7's , then there are 7*4c2 = 42 combos .
1- [1225-42]/1225]^9 =~26.94%

Now go to Brians page and add the three columns from the row with pocket 7's and you get 0.2380+0.03218+.0023=~ 27.248%

The answers are pretty close which is why I prefer using the "independence assumption" formula to give me the answers to these questions .

[jay_shark]

If you are interested in the exact answer then a better method would be the inclusion/exclusion formula .

Here is a link with some of these calculations .

http://archiveserver.twoplustwo.com/...te_id/1#import

[binions]

[ QUOTE ]
I wrote a Monte Carlo program in Python that produced the following results:

13: 0.0440
12: 0.0860
11: 0.1263
10: 0.1649
9: 0.2011
8: 0.2372
7: 0.2703
6: 0.3029
5: 0.3346
4: 0.3643
3: 0.3916
2: 0.4187

I used 1 million iterations apiece. This is in very close agreement (&lt; 1% difference) to the sum of each row in the last table from the link you gave "and yet not":

13: 0.0439
12: 0.085983
11: 0.1271218
10: 0.1659144
9: 0.2030309
8: 0.238537
7: 0.27248
6: 0.304978
5: 0.33591
4: 0.36546
3: 0.393731
2: 0.42062


[/ QUOTE ]

Looks like another rule of 4, or more precisely, 4% for 1-6 possible overpairs, and then 3% for the 7th-12th possible overpairs.

[binions]

jay

If 6 outs (ie 6 ways to make a pair) is worth about 4%, can we say that with AK, there is a ~4% chance AA or KK is out at a full table?

And that with AQ, there is 12 possible AK, 6, KK, 3 AA and 3 QQ such that there is a ~16% chance of being dominated (24/6 = 4 * 4%) at a full table?

Of course, these numbers are more subject to elimination effect, because if it is folded to you on the button with AQ, AA-QQ + AK is not going to be out as often as if someone raised or even limped UTG.

[jay_shark]

Yes Binions that would be a good approximation .

With AQ , the probability against 9 other opponents that you will be dominated is approximately :

1-[(1225-24)/1225]^9=16.31%

With AK versus A's or K's it is :


9c1*6/50c2 -9c2*3*3*2/50c2/48c2= 4.36%

[magneticskull]

This may be cheating but why doesn't someone with a lot of hands just email PokerStars support "complaining" that their pocket pairs "always lose". In the past, they have replied with amazing statistical details over tens of thousands of hands to the old "AK rigged" complaints etc...

They might even let you just ASK for the stats without the ruse... "dear pokerstars, How often does each pair run into an overpair at full ring/6 max?"

Please keep in mind, I understand this is less fun and perhaps less of a learning experience than mastering the math, but it's REAL DATA [img]/images/graemlins/smile.gif[/img]

[edit] I also fully expect the math you folks have come up with to be correct &amp; likely verified by this sort of experiment.

Best Regards,
M