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View Full Version : What are the odds? (Not a bad beat post)

09-20-2005, 08:33 PM
I do not know the stats behind this, or the calculations to figure it, so can someone advise?

It is the bubble of a NLHE sng on party. You are dealt Ac2s. What are the odds that another player has a higher ace, or a PP?

Thanks!
Bluetooth

skipperbob
09-20-2005, 08:34 PM
50%/50%...He either has it or he doesn't /images/graemlins/cool.gif

That's it for my "on-topic" post for Sept.

09-20-2005, 08:37 PM
[ QUOTE ]
50%/50%...He either has it or he doesn't /images/graemlins/cool.gif

That's it for my "on-topic" post for Sept.

[/ QUOTE ]

Heh. It can't be that simple. There must be a mathematical way to find the exact odds.

valenzuela
09-20-2005, 08:51 PM
Why dont ppl figure it out by themselves...is not that hard /images/graemlins/mad.gif.
44/50 * 43/49 * 42/48 and so on.

bigt439
09-20-2005, 08:54 PM
[ QUOTE ]
[ QUOTE ]
50%/50%...He either has it or he doesn't /images/graemlins/cool.gif

That's it for my "on-topic" post for Sept.

[/ QUOTE ]

Heh. It can't be that simple. There must be a mathematical way to find the exact odds.

[/ QUOTE ]

We've been trying to find it forever. 50/50's all we got. (Buy sng power tools, you can figure out stuff like that). You could also just use simple probability math.

valenzuela
09-20-2005, 08:59 PM
bah, ill do it.
1-(47/50* 46/49 * 45/48 * 44/47 * 43/46 * 42/45) * 46/49 = 6141/ 17150

RikaKazak
09-20-2005, 09:03 PM
I usually do this, in my head just for quick guesses, pp is 16:1 and ace is 9:1 when you have one, those aren't 100% right, but yeah, close enough for just "i'm 4 handed what do I do" situations. so like 5:1 or something close to that. that A2 is good. If you want exact odds, do the math, it's pretty easy, factor in you have an ace and a 2 so less ace combos etc. etc.

valenzuela
09-20-2005, 09:07 PM
wtf u want PP and aces combined /images/graemlins/mad.gif..grrrr...well I gave u aces( including AA).
anyway I will do both...i will take long.

valenzuela
09-20-2005, 09:14 PM
Im not smart enough to exclude A2, so Ill include A2
1-(47/50 * 43/49 * 45/48 * 41/47 * 43/46 * 39/45) = 0,453462178.
edit: Im now working on a exacter number, I realize my fraction multiplication has some flaws.

skipperbob
09-20-2005, 09:16 PM
[ QUOTE ]
[ QUOTE ]
50%/50%...He either has it or he doesn't /images/graemlins/cool.gif

That's it for my "on-topic" post for Sept.

[/ QUOTE ]

Heh. It can't be that simple. There must be a mathematical way to find the exact odds.

[/ QUOTE ]

O.K. Blue: But no more whining about my "off-topic" nonsense /images/graemlins/smile.gif"O.K."? /images/graemlins/smile.gif

There are 1,098,230 ways to deal a pair and 377,463 ways to deal an "Ace-high" (given that you already have an Ace. Therefore there are 1,213,642 ways to deal "either" a pair or an A-h (h = kicker higher than duece)...given the lack-of-exclusivity between the two events...Since there are 2,644,000 possible 5-card hands; the odds of him having one of the given hands is 1,213,642/2,644,000 or pretty damn close to 50/50 /images/graemlins/blush.gif

RikaKazak
09-20-2005, 09:18 PM
LOL, finally made someone else do it /images/graemlins/smile.gif /images/graemlins/smile.gif just post wrong math, and in 30 seconds people will jump all over it;)