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how to calculate these 2 Poker odds?


q4agl

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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 by q4agl
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  • 3 months later...

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.

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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 by koti
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