these correct N may I see the forumla please!
 

Pocket pair improving to three of a kind on flop
1 in 8
Im not sure how to do this one.


2 random cards (starting hand)improving to 1 pair on flop
1 in 3.125
Im not sure how to show this BUT I say its6.5-1 and then divided by 2 right? More like 3.250 ?



Gutshot straight draw hitting by river
1 in 6

Wouldnt the actual odds of the gutshot be 1 in 6.5-1?
Theres 13 cards no?

Re: these correct N may I see the forumla please!
 

[ QUOTE ]
Pocket pair improving to three of a kind on flop
1 in 8
Im not sure how to do this one.


2 random cards (starting hand)improving to 1 pair on flop
1 in 3.125
Im not sure how to show this BUT I say its6.5-1 and then divided by 2 right? More like 3.250 ?



Gutshot straight draw hitting by river
1 in 6

Wouldnt the actual odds of the gutshot be 1 in 6.5-1?
Theres 13 cards no?

[/ QUOTE ]

1. 50 unknown cards, 48 don't give you a set. Chance of NOT getting a set on the flop is:
48/50 * 47/49 * 46/48 =~ .882
Chance of getting a set = 1 - .882 = .118
.882:.118 =~ 7.5:1, or 2 in 17

2. 50 unknow cards, 44 are blanks, chance of at least pairing are 1 - (chance of getting 3 blanks). Solved as above.

3. with 2 cards to come, there are 47 unknown cards, 43 do not complete, so you have
1 - (43/47*42/46) =~ .164, which works out to just over 5:1 against.

There are not 13 cards, there are 13 ranks, but not all ranks have 4 representatives still in the deck.

There are a lot of problems (the three you posted included in them) that are easiest to solve by figuring the chance of something not happening, then subtracting that value from one.

Hope that helped.

Re: these correct N may I see the forumla please!
 

First question, depends on whether you're saying 3 of a kind or at LEAST 3 of a kind. Most people want a full house included, so here goes. Say you hold 7-7.

50 C 3 possible flops

48 C 3 possible flops without either of the 7's.

So 48 C 3/50 C 3 is the probability you won't hit a set, so subtract from 1:

1 - 48 C 3/50 C 3 = 1 - 17296/19600 = 11.755102% or 1 in 8.506944444 or about 7.5 to 1.


The second question can be solved the same way as above essentially.


For the third question. Say you hold 7-5 and the flop comes K-8-4, no flush possible.

47 C 2 possible combinations coming on turn/river.

43 C 2 possible combinations without at least one 6.

1 - 43 C 2/47 C 2 = 16.466235% or about 1/6.07 or roughly 5 to 1.