perfect perfect

Jeff V · #1 09-26-2003, 11:07 AM

Im not too good at math but was hoping for some help. Suppose I have A5o on a flop of 2J9r. I'm up against a set, what are the odds of catching the 3,4 to make my str8? Thanks.

Nottom · #2 09-26-2003, 11:45 AM

I'm sure someone will correct me if I'm wrong, but just at first glance I'm thinking its 8/47*4/46 (any 3 or 4 on the turn,and then whichever one you didn't hit on the river) or around 66.5-1. You could argue it is 8/45 * 4/44, about 61-1, I suppose since you "know" he has a set of some sort.

BruceZ · #3 09-26-2003, 02:29 PM

Help me... help myself.

I'll take you up on that.

1. How many cards help you on the turn?

2. How many unseen cards are there that could possibly come on the turn? Important note: unseen means you don't know if they help you or not, not that you necessarily actually saw them or know exactly what they are.

3. Using your answers for 1 and 2, what is the probability that you get help on the turn?

4. Repeat steps 1-3 for the river, assuming that you got help on the turn.

5. Multiply your answers for steps 3 and 4. This is your probability.

6. Multiply your probability by 100 to get a percent.

7. Divide 100 BY your percent and subtract 1 to get your odds to 1. For example, if your percent is 25%, divide 100/25 to get 4, and then subtract 1 to get 3-1.

8. Now read the other answer which is correct.

Maybe I will answer more questions like this from now on.

Bozeman · #4 09-26-2003, 03:20 PM

"Important note: unseen means you don't know if they help you or not, not that you necessarily actually saw them or know exactly what they are."

Perhaps you should restate this in a form that is correct or at least makes sense.

Jeff V · #5 09-26-2003, 08:34 PM

turn 1:5.8, river 1:11.5. multiplied = 1:66.7, which would give a probability of .014, a percent of 1.48, and odds 66.5:1. Is this right? Thanks for the help. Jeff

BruceZ · #6 09-27-2003, 01:35 AM

Jeff,

Since your opponent has a set, you know that neither of his 2 cards would help you, so there are really 45 unseen cards instead of 47.

1. How many cards help you on the turn?

8

2. How many unseen cards are there that could possibly come on the turn? Important note: unseen means you don't know if they help you or not.

45

3. Using your answers for 1 and 2, what is the probability that you get help on the turn?

8/45

4. Repeat steps 1-3 for the river, assuming that you got help on the turn.

4, 44, 4/44

5. Multiply your answers for steps 3 and 4. This is your probability.

8/45 * 4/44 = 0.0162

6. Multiply your probability by 100 to get a percent.

1.62%

7. Divide 100 BY your percent and subtract 1 to get your odds to 1. For example, if your percent is 25%, divide 100/25 to get 4, and then subtract 1 to get 3-1.

100/1.62 - 1 = 60.7-1

BruceZ · #7 09-27-2003, 01:59 AM

"Important note: unseen means you don't know if they help you or not, not that you necessarily actually saw them or know exactly what they are."

Perhaps you should restate this in a form that is correct or at least makes sense.

It is correct and makes perfect sense, what are you talking about? [img]/images/graemlins/wink.gif[/img] Our opponent has a set, so we know his two cards wouldn't help us complete our straight. Thus for our purposes, there are 45 total unseen cards that could come on the turn, out of which 8 help us, and 37 don't help us.

not that you necessarily actually saw them or know exactly what they are."

That the problem? That's correct too of course, trivially so, it's just not particularly helpful [img]/images/graemlins/smile.gif[/img]. It would probably be better if it stopped before this phrase, or if it said "unseen cards do not necessarily include all of the ones that you can't actually see". You can't see your opponent's cards, and you don't know exactly what they are, but you do know that they don't help you, and this is enough to remove them from the number of "unseen" cards for the purpose of this calculation. How's that?

PAUL-IN · #8 09-27-2003, 12:44 PM

www.twodimes.net

enjoy! two dimes

Jeff V · #9 09-27-2003, 07:59 PM

Aren't these answers assuming I get my card on the turn? If I or someone were to play this type of hand all the time, assumimg the straight was the only I win would I also factor in the 4 times I miss on the turn? Say we're all in in a tourney or no-limit game. Just trying to get better at this math stuff. Jeff

BruceZ · #10 09-28-2003, 04:51 AM

No, this is the probability of getting your card on the turn AND then getting your card on the river.