Texas Hold'Em: you have 5, 6 of hearts in your hand....

You are playing Texas hold ’em, and you have two cards in your hand: 5 of hearts and 6 of hearts. On the table are 7 of hearts, K of diamonds and J of clubs. Two more cards will be dealt on the table, and you will then use any five cards to make your best hand.

(a) What is the probability that you will get a straight, i.e., five cards in a row, disregarding suit? (The order of cards is 2, 3, 4, 5, 6, 7, 8, 9, 10, J, Q, K, A, where A can also count as “1”in a straight).

(b) What is the probability that you will get a flush, i.e., five cards of the same suit?

How do i go about taking into account that the cards are in a row for part a ? Not sure how to approach it without that information and how using combinations i could do part b.

Re: Texas Hold 'em Probability

xoticaox said:

You are playing Texas hold ’em, and you have two cards in your hand: 5 of hearts and 6 of hearts. On the table are 7 of hearts, K of diamonds and J of clubs. Two more cards will be dealt on the table, and you will then use any five cards to make your best hand.
(a) What is the probability that you will get a straight, i.e., five cards in a row, disregarding suit? (The order of cards is 2, 3, 4, 5, 6, 7, 8, 9, 10, J, Q, K, A, where A can also count as “1”in a straight).

(b) What is the probability that you will get a flush, i.e., five cards of the same suit?

How do i go about taking into account that the cards are in a row for part a ? Not sure how to approach it without that information and how using combinations i could do part b.

Click to expand...

The same way you were told to at:

viewtopic.php?f=12&t=31920

start counting

For the first card - howmany possible way

for the second card how many possible way

and so on...

Re: Texas Hold 'em Probability

Subhotosh, couple things have me stumped here...
Has 5,6 and 7 is showing.
Possible straights are: 3,4,5,6,7 or 4,5,6,7,8 or 5,6,7,8,9.
r = remaining cards in deck.
If 1st card = 4; has 4,5,6,7 ; has 2 possible chances: 3 or 8 ; 8/(r-1)
If 1st card = 8; has 5,6,7,8 ; has 2 possible chances: 4 or 9 ; 8/(r-1)
If 1st card = 3; has 3,5,6,7 ; has 1 possible chance : 4 ; 4/(r-1)
If 1st card = 9; has 5,6,7,9 ; has 1 possible chance : 8 ; 4/(r-1)

But r = 52 - 3 - 2p (p = number of players) : right?
And to throw in a monkeywrench, other players possibly have 1 or more of the 3-4-8-9's : right?

How in heck do you bring all that into the equation?

Anyhow, IF he's playing by himself(!), meaning 47 cards left, then:
8/47 * 4/46 (1st card 3 or 9) + 8/47 * 8/46 (1st card 4 or 8) = 48/1081 ; agree?

And to you, xoticaox (to see where you're at):
what is the probability of getting 4-same (like 4 3's or 4 Queens) by
randomly drawing 4 cards from a full deck?