Odds of making pair with AK ?

R

RickAversion

Visionary
Silver Level
Joined
Mar 2, 2013
Total posts
597
Chips
0
You have 6 outs. So, can you use "Rule of 2 and 4", but for 5 cards?
Rule of 10? Basically, 60% odds to getting an A or K?

P(another A or K)
= 6/50 + 6/49 + 6/48 + 6/47 + 6/46
= .62 ?
 
nabmom

nabmom

Community Guide
Community Guide
Joined
Dec 24, 2009
Total posts
6,371
Awards
13
Chips
325
I found this online, since there are many smarter people then me on poker odds. But this seems logical

In every case, there are 6 cards (the three remaining Aces, and the three remaining Kings) that will pair one of your hole cards.
Not pairing with the first flop card = (50 - 6) / 50
Not pairing with the second flop card = (49 - 6) / 49
Not pairing with the third flop card = (48 - 6) / 48
Not pairing with the turn = (47 - 6) / 47
Not pairing with the river = (46 - 6) / 46
The probability of not pairing either hole card = 44/50 x 43/49 x 42/48 x 41/47 x 40/46 = 0.512568
The probability of pairing at least one of the hole cards = 1 - 0.512568 = 0.487432
 
R

RoTs

Enthusiast
Silver Level
Joined
Jun 3, 2008
Total posts
88
Chips
0
I found this online, since there are many smarter people then me on poker odds. But this seems logical

In every case, there are 6 cards (the three remaining Aces, and the three remaining Kings) that will pair one of your hole cards.
Not pairing with the first flop card = (50 - 6) / 50
Not pairing with the second flop card = (49 - 6) / 49
Not pairing with the third flop card = (48 - 6) / 48
Not pairing with the turn = (47 - 6) / 47
Not pairing with the river = (46 - 6) / 46
The probability of not pairing either hole card = 44/50 x 43/49 x 42/48 x 41/47 x 40/46 = 0.512568
The probability of pairing at least one of the hole cards = 1 - 0.512568 = 0.487432
Just missing one adjustment subtracting 2 cards for every player dealt into the hand
 
R

RickAversion

Visionary
Silver Level
Joined
Mar 2, 2013
Total posts
597
Chips
0
Just missing one adjustment subtracting 2 cards for every player dealt into the hand

You sure about that? How does having more players in the round increase or decrease your odds of hitting a pair? It doesn't matter if the cards are in the deck or in other player's hands. Bottom line is that there are 6 outs among 50 unknown cards. Having 5 other players in the round does not increase your odds to 6 out of 40 cards. This is also how pot equity is calculated. It ignores number of players in the round.

Looks like it's 50/50 to hit a pair if you go to the river.
 
R

RoTs

Enthusiast
Silver Level
Joined
Jun 3, 2008
Total posts
88
Chips
0
Not positive but since you watch a lot of wsop videos ever noticed how they show you your winning percentage (they even adjust based on someone folding a previous out).
Unless your all in, each street has to take into account how many cards the dealer has left in their hand to get a more accurate estimate.
Never really looked into the pairing of a whole card by river all I know its about 32% chance of pairing on the flop
 
Last edited:
DaReKa

DaReKa

Rock Star
Silver Level
Joined
Nov 1, 2012
Total posts
264
Chips
0
Not positive but since you watch a lot of WSOP videos ever noticed how they show you your winning percentage (they even adjust based on someone folding a previous out).
Unless your all in, each street has to take into account how many cards the dealer has left in their hand to get a more accurate estimate.
Never really looked into the pairing of a whole card by river all I know its about 32% chance of pairing on the flop


The other players cards are ignored as you don't know what they are. The odds are adjusted when you do know, but you can't subtract the other players cards since you don't know whether or not they contain an A or K.
 
C

cander128

Rock Star
Silver Level
Joined
Mar 13, 2013
Total posts
158
Chips
0
You have 6 outs. So, can you use "Rule of 2 and 4", but for 5 cards?
Rule of 10? Basically, 60% odds to getting an A or K?

P(another A or K)
= 6/50 + 6/49 + 6/48 + 6/47 + 6/46
= .62 ?

This is the correct math, actually closer to 62.5%, other players cards are unknowns so can't be taken in to consideration.
 
D

DunningKruger

Legend
Silver Level
Joined
Feb 7, 2013
Total posts
1,030
Chips
0
I found this online, since there are many smarter people then me on poker odds. But this seems logical

In every case, there are 6 cards (the three remaining Aces, and the three remaining Kings) that will pair one of your hole cards.
Not pairing with the first flop card = (50 - 6) / 50
Not pairing with the second flop card = (49 - 6) / 49
Not pairing with the third flop card = (48 - 6) / 48
Not pairing with the turn = (47 - 6) / 47
Not pairing with the river = (46 - 6) / 46
The probability of not pairing either hole card = 44/50 x 43/49 x 42/48 x 41/47 x 40/46 = 0.512568
The probability of pairing at least one of the hole cards = 1 - 0.512568 = 0.487432

Since there seems to be some confusion itt, the answer is right here. An alternative method of calculation is 1 - (44 5)/(50 5). ~48.743%
 
runnerx289

runnerx289

Rock Star
Silver Level
Joined
Jan 10, 2013
Total posts
166
Chips
0
what about hitting the flop?
 
runnerx289

runnerx289

Rock Star
Silver Level
Joined
Jan 10, 2013
Total posts
166
Chips
0
Not positive but since you watch a lot of WSOP videos ever noticed how they show you your winning percentage (they even adjust based on someone folding a previous out).
Unless your all in, each street has to take into account how many cards the dealer has left in their hand to get a more accurate estimate.
Never really looked into the pairing of a whole card by river all I know its about 32% chance of pairing on the flop

WSOP do that for the viewers benefit.
 
Poker Odds - Pot & Implied Odds - Odds Calculator
Top