PDA

View Full Version : Is there an easy way to figure out...

04-10-2005, 06:38 PM
If the flop missed everyone? That is, for X opponents, how likely it is if the flop didn't pair anyone, for each of the three flop cases( 3 ranks, 2 ranks, and one rank)?

Knowing this could be useful in a blindsteal situation. I'm not looking for something that's 100% precise, just something that's easy to get an estimation from, like the "multiply outs by 2 for each street" trick you can use for figuring your odds on the flop.

irchans
04-11-2005, 10:12 AM
The general formula is

p = C[52 - iRanks*4, 2*iOpp]/C[52, 2*iOpp]

where C is the choose function, iRanks is the number of ranks on the flop, iOpp is the number of opponents, and p is the probability that none of those opponents get a pair, trips, or quads.

The table below shows all the probabilities.

<font class="small">Code:</font><hr /><pre>
1 Rank 2 Ranks 3 Ranks

0.959 0.804 0.663 1 opponents miss
0.918 0.641 0.431 2 opponents miss
0.878 0.505 0.274 3 opponents miss
0.837 0.393 0.171 4 opponents miss
0.796 0.302 0.103 5 opponents miss
0.755 0.229 0.061 6 opponents miss
0.714 0.170 0.034 7 opponents miss
0.673 0.124 0.019 8 opponents miss
0.633 0.089 0.010 9 opponents miss
</pre><hr />

For example, if you have 4 opponents and there are 3 ranks on the flop (no pair and no trip), then the probability that none of them has a pair is 17.1%.

04-11-2005, 12:35 PM
Awesome, exactly what I wanted. Thanks.

pzhon
04-11-2005, 05:00 PM
[ QUOTE ]
The general formula is

p = C[52 - iRanks*4, 2*iOpp]/C[52, 2*iOpp]

[/ QUOTE ]
You should be dividing by the number of possibilities given the flop you see, so that should be C[49, 2*iOpp] even if you are ignoring the cards you have.

If you do want to use the information about the cards you have, then if you have n cards in the ranks of the flop, the conditional probability everyone else has none is

C[52-4 iRanks - (2-n), 2 iOpp]
------------------------------
C[47, 2 iOpp]

Keep in mind that this counts having an overpair or flopping a flush or straight as missing the flop. If I have A/images/graemlins/club.gif K/images/graemlins/club.gif and the flop is 4/images/graemlins/club.gif 3/images/graemlins/club.gif 2/images/graemlins/heart.gif, I would think of that as hitting the flop very hard.

04-11-2005, 10:26 PM
Yea, I agree pzhon, but it's nice to just have an estimate in your head, and the easiest way to do that I figure is with pairings. The more coordinated the board is the higher probability someone has a piece of it, but it's nice to have a starting point.

irchans
04-13-2005, 01:47 PM
[ QUOTE ]
[ QUOTE ]
The general formula is

p = C[52 - iRanks*4, 2*iOpp]/C[52, 2*iOpp]

[/ QUOTE ]
You should be dividing by the number of possibilities given the flop you see, so that should be C[49, 2*iOpp] even if you are ignoring the cards you have.

If you do want to use the information about the cards you have, then if you have n cards in the ranks of the flop, the conditional probability everyone else has none is

C[52-4 iRanks - (2-n), 2 iOpp]
------------------------------
C[47, 2 iOpp]

Keep in mind that this counts having an overpair or flopping a flush or straight as missing the flop. If I have A/images/graemlins/club.gif K/images/graemlins/club.gif and the flop is 4/images/graemlins/club.gif 3/images/graemlins/club.gif 2/images/graemlins/heart.gif, I would think of that as hitting the flop very hard.

[/ QUOTE ]

Thanks pzhon,

I incorrectly typed the formula rather than cutting and pasting /images/graemlins/tongue.gif . The table I gave is correct, but

p = C[52 - iRanks*4, 2*iOpp]/C[52 - 3, 2.*iOpp]

as you said.

LINES
05-02-2005, 06:47 PM
The formula that phzon posted is the correct one?