Texas Holdem (consisting 6 players)

i got AA as starting hand, what are chances of there being another A on one of the 5 cards on table? what are chances of 2 other A on the 5?

 

1. So my question is, when u have pair as start. What are chances of you getting "3 of a kind" and "four of a kind"?

 

2. Unlike commonsense, I believe starting hand of 10 J is slightly better than K A, since u can make straights on both sides. Am i right? who can calculate it? please

Edited May 24, 20169 yr by q4agl

Probability of not getting 3 of a kind is(48/50)(47/49)(46/48)(45/47)(44/46).

 

http://www.pokernews.com/poker-tools/poker-odds-calculator.htm

 

Above may help. Google Texas holdem - odds

You have two aces in your hand so there are two still "out". Assuming there are still 52- 2= 50 cards you don't know, 2 aces and 48 non-aces, then the probability than an "A" is the first of the 5 cards left is (2/50), the probability that the next is not an ace is (48/49), then (47/48), then (46/47), then (45/46), then (44/45). The probability that you get an ace first then four non-ace is (2/50)(48/49)(47/48)(46/47)(45/46)(44/45)= (2/50)(44/49)= 88/2450= 44/1225. The probability that 1 card will be and ace and the other four non-aces, in any order, is 5 times that: 5(44/1225)= 44/245.

 

To look at two aces, do much the same. The probability the first two cards are aces is (2/50)(1/49) and then the last three must be aces so (2/50)(1/49) = 1/1225. There are then $\begin{pmatrix}5 & \end{pmatrix}= \frac{5!}{2!3!}= 10$ so the probability of 2 aces and 3 non-aces in any order is 10/1225= 2/245.

I believe starting hand of 10 J is slightly better than K A, since u can make straights on both sides. Am i right? who can calculate it? please

This is incorect because straights are less probable than two pairs, one pair, A high or K high.

Somone here Im sure will be able to do the calculation for you (mind you, the suits matter too)

or just google the starting hands in holdem.

Edited September 20, 20169 yr by koti