Two math/chance to happen questions

[hagelito]

Two math/chance to happen questions, game Texas hold'em.

Hand 1:
Flop QJ9 rainbow, I have T8, how big is the chance that one of 3 opponents have flopped the bigger straight?


Hand 2:
At river (X = don't remember but not a heart): 9h 7h 8h X Th
I got 66 and made the low straightflush, how likely is it that someone has the higher straighflush? I think I had 5 opponents at the flop.


Another thing:
I think I win about 72-80% of my sessions, but I always feel unlucky when I play. I usually don't win any pot the first 45-90 minutes (don't matter what cards I get), but many of the new players that enter the board win their first hand (happens very often but never happens to me).

Btw, this unlucky feeling only apply to the first hour(s), then pretty much always when I have about 30-50% of my stack left, all wins kick in within a short period of time - and I'm on plus or on a big plus (within a few hands compare to how many I've played).

Does anyone of you also (almost) always feel unlucky when you play online poker? And what is the usual average time for you to win your first pot? (online poker small stakes)

Thanks in advance,
Hagel.

[cardcounter0]

1) ~2.77%

2) 22.22%

[hagelito]

[ QUOTE ]
1) ~2.77%

2) 22.22%

[/ QUOTE ]

The first one I knew it wasn't very big chance that I was beaten, but that the second one was as high as 22%....that I couldn't imagine.

I didn't see any straightflush in maybe my first 6-8 months of play, I didn't get my first straightflush until after +2 years of play, then when I get my second I lose to a bigger one, that hurts /images/graemlins/wink.gif I'm glad I'm playing limit...

Ok, thanks for your answer.
/Hagel.

[pif]

*1) there r 12 hands with KT (4K*3T) from 1082 possible hands in the flop give u 1.11% for 1 player (HeadsUP) holding KT.
so multiply by 3 (3 oponents) to get 3.33% = 1/30 that somone better than u. ( its ok to multiply low odds..)
however, against 3 oponents ur odds to win in the showdown is 85%, they can straight or even get flush or full house until the river.
*2)there is 1 nut straight flush (QhJh)=0.1%
and there r 43 other straight flush with Jh (Jh and all other 43 cards remaining (excluding Qh))) = 4.34%

so there is 4.44% chance that 1 oponent is better than u.

/images/graemlins/wink.gif

[hagelito]

[ QUOTE ]
*1) there r 12 hands with KT (4K*3T) from 1082 possible hands in the flop give u 1.11% for 1 player (HeadsUP) holding KT.
so multiply by 3 (3 oponents) to get 3.33% = 1/30 that somone better than u. ( its ok to multiply low odds..)
however, against 3 oponents ur odds to win in the showdown is 85%, they can straight or even get flush or full house until the river.
*2)there is 1 nut straight flush (QhJh)=0.1%
and there r 43 other straight flush with Jh (Jh and all other 43 cards remaining (excluding Qh))) = 4.34%

so there is 4.44% chance that 1 oponent is better than u.

/images/graemlins/wink.gif

[/ QUOTE ]

1) Ok, that seems right. But the 85% really didn't apply to my scenario (which I didn't explain fully) because it was early in a $5000 NL freeroll (start 1000 chips) and I had about 1500 and average was about 1100 and no drawing hands could call with the right odds there. There where only a few 100s in the pot (don't remember exactly), and I didn't want to split or lose the pot to a T or K or any other backdoor draws for a cheep price.

2) Yes I see it now (how to calculate), but that's just in theory, in real money games I guess it isn't likely that someone would play all those combinations /images/graemlins/wink.gif

Thank you very much for enlighten me! *two thumbs up*
/Hagel.

[MtDon]

RE: Question 1

I think you are making a mistake by multiplying the chances having a KT by 3 to get the probability you are beat.

You need to consider the total number of people at the table who would call with KT before the flop.

If the table is 10 handed and you are 100% certain that all the other players would call before the flop with KT, then you multiply by 9 instead of 3.

If you think, for example, that the players in 3rd and 4th position wouldn't call with KT, but everyone else would, then multiply by 7 instead of 3. Unless, of course, the only other players left in are the Big Blind and the players in 3rd and 4th positions. Then the only player who might have a KT is the Big Blind, so multiply by 1.

This can be refined by assigning probabilities to how likely each player would call before the flop with KT.

The basic idea is that the probability of someone holding a given hand after the flop is based on the probability of any one person having that hand multiplied by the number of players who would have called with that hand before the flop, rather than how many people are left in the hand.


-- Don