odds of QJ2 twotone.
i know this is simple math and ive done it a bunch of times over the years but i just want to double check this is right.
Need to find out how often QJ2 two tone appears as flop in a vacuum. ( disregard our hand or opps hands blocking this) just need to find raw value.
Odds of Q,J or 2 coming on flop = 12/52 (0.23)
Odds of 2nd card being suited to first card and not pair = 2/51 (0.0392)
Odds of 3rd card being not suited to first 2 and not a pair to first 2 = 3/50 (0.06)
0.23*0.0392*0.06 = 0.00054 * 100 = 0.054% or 1 in 185 flops.
This doesnt seem right in my brain. lol is it correct?
6 Replies
i know this is simple math and ive done it a bunch of times over the years but i just want to double check this is right.Need to find out how often QJ2 two tone appears as flop in a vacuum. ( disregard our hand or opps hands blocking this) just need to find raw value.Odds of Q,J or 2 coming on flop = 12/52 (0.23)Odds of 2nd card being suited to first card and not pair = 2/51 (0
I figure it’s 0.027%
There are 22,160 unique flops.
52*51*50
If you don’t care about specific suits and want to call all two tone the same, you just need to count the number of combinations of 2 tones of specific ranks.
For a two tone flop there are 3 combinations of suit order (hi card is different, mid card is different, low card is different) and 4 different suits that the same cards could be and 3 different suits that the different card could be. Therefore there’s 12 combinations for each set of which card is not suited, and 3 orders so 36 flops.
Here’s the list in which the 2 is the different suit.
QcJc2d
QcJc2h
QcJc2s
QdJd2c
QdJd2h
QdJd2s
QhJh2c
QhJh2d
QhJh2s
QsJs2c
QsJs2d
QsJs2h
The list would have the same number for the j being different and the Q being different so 12*3.
So there’s 36 flops that would describe a QJ2 flop being two tone out of a possible 22160 unique flops.
36/22,160=0.02715%
Sent from my iPhone using Tapatalk
I figure it’s 0.027%There are 22,160 unique flops. 52*51*50If you don’t care about specific suits and want to call all two tone the same, you just need to count the number of combinations of 2 tones of specific ranks. For a two tone flop there are 3 combinations of suit order (hi card is different, mid card is different, low card is different) and 4 different suits that the sam
i dont think thats right either. that only accounts for exactly QJ2 flops. what about QJ2,Q2J, JQ2,J2Q,2QJ,2JQ?
then do the suit combos like you did, which would be 6*12*3 = 216 / 22,160 = 0.00974
or no?
i dont think thats right either. that only accounts for exactly QJ2 flops. what about QJ2,Q2J, JQ2,J2Q,2QJ,2JQ?
then do the suit combos like you did, which would be 6*12*3 = 216 / 22,160 = 0.00974
or no?
The order shouldn’t matter since the denominator is 52*51*50.
QsJs2d Is considered the same as Qs2dJs when counting the total number of combinations.
If you care about the order of the flop it would be different but I did the above considering they were the same.
Sent from my iPhone using Tapatalk
Off the top of my head there seem to be some small mistakes here. I am going to do this off the top of my head so I might make mistakes too that someone can fix.
I see the number of combo listed here as 22,160. Typo? Isn't it 22,100? So 36/22100 = about .16% or 1 in about 600 combo way.
(52*51*50)/(3*2*1) = 22,100
Also you have the combo way and the distinct way of doing things and they are not interchangeable in the middle of a conversation but sometimes that is exactly what you see. You still see people say there are 270,725 combos of starting hands. So combos is the most popular use and not saying there are 16,432 distinct ways. Preflop is still popular to use combos and if someone tells you that a specfic situations is done for example 19.50% of the time they are likely giving you the combo % not the distinct % which will be similar but not exactly the same. But now you are often seeing the opposite when talking about flops. Instead of people sying there are 22,100 combo flops, you are instead seeing many saying there are 1,755 distinct flops. Make sure you divide combos by combos and distinct by distinct, apples by apple and oranges by oranges.
So distinct is 3 ways to have QJ2 2-tone and 1755 flops so about .17% or maybe more easily seen as 1 out of 585 flops. Someone check if I flubbed all of this.
Off the top of my head there seem to be some small mistakes here. I am going to do this off the top of my head so I might make mistakes too that someone can fix.I see the number of combo listed here as 22,160. Typo? Isn't it 22,100? So 36/22100 = about .16% or 1 in about 600 combo way.(52*51*50)/(3*2*1) = 22,100Also you have the combo way and the distinct way of doing thing
Only callout in your last statement is that there are not equal weights of strategically identical flops. There’s only 4 instances of 567monotone, but many more of 567 rainbow you can’t weight them the same. You would need to weight them by frequency if using the strategically unique subset or just use the full set. I’ll double check the total number of unique yeah I definitely had a typo and dropped the 3! In “showing the work”. But I don’t think you can use the 1755 without weighting them differently
Sent from my iPhone using Tapatalk
Yeah, the 1755 flops are not all weighted the same, but it damn sure is a fast estimate and similar to combinations. I think there are three different weights possible for the 1755 flops. I think a flop can be weighted 4, 12 or 24 to balance things out based on suits like you said where monotone is different from rainbow and a pair or not would also affect what is possible.
Using the OP's QJ2,
QJ2 monotone is the 4 suits so 4 possibilities
QJ2 two tone creates 12 possibilities
QJ2 rainbow is 24 possibilities
with some logic the same goes for a paired board.
QJJ rainbow would be 12 possibilities
I'll leave the QJJ two tone as a thought experiment for everyone wanting to try.
