US States
NJ Online Casinos
WV Online Casinos
PA Online Casinos
Michigan Online Casinos
Online Casino California
Online Casino Arizona
Online Casino NY
Real Money Casinos
Fastest Payout Casinos
Mobile Casino Apps
New Online Casinos
Casino Payment Methods
Sweepstakes
Sweepstakes apps
No deposit bonus
Daily login bonus
Sweepstakes games
Crown Coins
Funrize
Hello Millions
High 5 Casino
Jackpota Casino
Mcluck
MegaBonanza
PlayFame
Pulsz
RealPrize
Stake.us
Sweepstakes coins
Awards
Search forums
Free Games
Free Blackjack
Free Online Roulette
Free Slots
Bonuses
No Deposit Bonus
Crown Coins Promo Code
Funrize Primo Code
Hello Millions Promo Code
High 5 Casino Promo Code
Jackpota Promo Code
McLuck Promo Code
MegaBonanza Promo Code
Pulsz Promo Code
RealPrize Promo Code
Stake.us Promo Code
Games
Online Slots
Blackjack
Roulette
Poker
CardsChat Freerolls
How to Play Poker
Poker Hands
Poker Strategy
Free Online Poker Game
Poker Bonuses
Poker Tools
Poker Podcast
Poker School
Forum
News
Log in
Join
Search
Search
Search titles only
By:
Log in
Register
Search
Install the app
Install
Forum
Poker Discussion
General Poker
Probability calculations.
JavaScript is disabled. For a better experience, please enable JavaScript in your browser before proceeding.
You are using an out of date browser. It may not display this or other websites correctly.
You should upgrade or use an
alternative browser
.
Reply to thread
Message
[QUOTE="Shoestringx, post: 477915, member: 14693"] Ok so this came up in the Test your Small Stakes Hold'em ....... threads in the Hand analysis section, so I decided to post about it in case anyone was interested. I'll start with somthing very simple, the probability of drawing a specific pocket pair (for this example I'll show with A's). p(A,A) = p(A1) * p(A2) (where p(A1) and p(A2) are probabilities of getting an ace on the first and second card respectively. P(A1) is very simple, there are 52 cards in the deck and 4 aces therefore 4/52. P(A2) is a also very simple, there are 51 remaining cards in the deck and 3 aces, therefore 3/51 p(A,A) = (4/52)*(3/51) = 0.45% That calculation was easy as there is only one possible outcome of cards, and they both need to be Aces. However lets say you wanted to find the probability of finding any A,K then you need to do: p(A,K) = p(A1)*p(K2) + p(K1)*p(A2) = (4/52)*(3/51) + (4/52)*(3/51) = 0.90% Now lets say you are dealt those two A,A to start, what are the odds of flopping a set (or maybe better, but no quads)? p(Aset) = p(A3)*p(X1)*p(X2) + p(X1)*p(A3)*p(X2) + p(X1)*p(X2)*p(A3) =[(2/50)*(48/49)*(47/48)] + [(48/50)*(2/49)*(47/48)] + [(48/50)*(47/49)*(2/48)] = 11.51% I'll take some time later tonight and work out a couple more complicated examples if there is any interest in that (I think Zebranky mentioned an example, I'll probably try that one later). [/QUOTE]
Insert quotes…
Verification
Post reply
Forum
Poker Discussion
General Poker
Probability calculations.
Top