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merenovice or some other bright spark!!
pod1
Member Posts: 4,377
help please. i understand the basics of %/ odds/ outs. but i got a tad stumped last night in a live game. playing in tourny and we got down to final 3. chip leaders sat on 32k 2nd 28k and im on 22k. blinds are 500k and 1k. i am in the bb with k8 off, dealer folds sb flat calls. flop comes kc jh 8c ( i have no clubs). agressive sb raises to 6k. knowing the player well ,i put him on either flush draw or is on the rob. thinking he is not pot committed and can get away from this hand i push all in. after a min he says" i know im behind but i got outs" and turns over j c 10c. hits club on river. a very new player behind m,e says " unlucky" " cant believed he called". i tried to explain to him that he had 2js +9 +3 10s to improve his hand. which i made about 50%.. so i wasnt that upset he called. he then turned round and said but you got 4 cards to improve your hand to a full house. "how do those odds equate into that". and i was stumped. could some one explain to me 1 was it a bad push or call. 2 who was actually ahead at flop. 3 how do cards that i can hit effect the odds. brain hurtin now, hope this isnt too all over the place for someone to help. phil
Comments
Your equity 57.67%
Their equity 42.33%
21-6=15 to call
pot=44
15/44*100= villian needs 34.1% equity to call which he has.(plus he might put you on a tp instead of two pair.)
I suspect that you're more interested in the general theory here than the specifics of this example so I'll add my tuppence worth.
Firstly, in this example, he has less outs than you give him credit for since the 10's are no good (your KK88 would beat his JJTT).
However, when he assessed his odds he would not know that you had flopped two pairs so he would expect his 10's to be outs and it is quite reasonable for him to think this.
Generally flopping a pair and a flush draw is a very strong hand and, unless your opponent has a set or two pairs it is reasonable to assume that you are at least level (odds wise).
Strictly from a mathematical point of view:
a) the probability of two events both happening (so long as they are "independent") is obtained by mutiplying the probabilities of the individual events.
b) the odds of either of two ("mutually exclusive") things happening is obtained by adding the probability of the individual events
However, the events here are not mutually exclusive or independent so the maths is a little more complicated. I can go into more depth if you wish!
If your opponent has a flush draw and you have two pairs after the flop, your opponent's odds of winning the hand would (roughly) be reduced proportionally to the number of outs that you have for the re-draw.
If you have 2 pairs, there are 4 cards that would make you a full house.
When we use "the rule of 4 and 2", we are actually factoring in a little of these re-draws since the odds of hitting a card on the turn (or river) is close to 2.3% not 2% so multiplying your outs by 4 (or 2) to give a percentage for your chances of winning a hand is generally reasonable although there are examples where it does not work too well - I can quote examples if you're interested.
I'll use an example of tossing a "fair" coin, i.e one that will come down either heads or tails with an equal chance.
The probability of a head is 0.5 and the probability of a tail is 0.5.
The coin coming down heads or coming down tails are "mutually exclusive" events, i.e. they cannot both happen on a single toss of the coin.
Therefore the probability of a coin coming down either heads or tails is 0.5 + 0.5 = 1.
If the coin is tossed twice then the two tosses are "independent" events, i.e. it doesn't matter what happens on the first toss, the odds of the second toss coming heads or tails are not affected.
Therefore the probability of two tails being tossed are 0.5 * 0.5 = 0.25.
You can see that this is correct if you consider the four possible outcomes of two coin tosses, i.e. HH, HT, TT & TH.
A pack of cards is different if we don't replace the cards after choosing the first one.
Imagine we had a bag which contained two black balls and two white balls.
If we pick a ball out at random, then the probabilities of picking a white ball or a black ball are each 0.5.
However, if we select a black ball and don't replace the ball in the bag then if we pick a random second ball the probability of picking a black ball is 0.33 (i.e. 1 in 3) and the probability of a white ball is now 0.67 (2 in 3).
This is an example of events that are not "independent".
If we pick a card from a pack of cards at random, the probability of picking a heart is 1/4 (i.e. 13/52) and the probability of picking a jack is 1/13 (i.e. 4/52).
However, these two events are not "mutually exclusive" since it is possible to pick the jack of hearts which is both a jack and a heart. :-)))
It all seems like a right load of
MATHEMATICS to me.
There are a huge number of variables to be considered and the consequent calculations are very complicated.
After much hard work, I came up with a probability of 0, i.e. your and your wife's gen1talia will be "mutually exclusive" this evening.
However, as stated earlier, the rule of "4 and 2" is generally a good guideline since it accounts for re-draws.
When it comes to a 15 second decision on the flop for all your chips, the best that you can generally do is have a stab at the number of outs that you think you have and multiply it by 4 for a percentage of winning the hand.
If you're not all-in (and your opponent is not all-in) it becomes much more complicated.
On top of all this, you need to factor in your opinion of your opponent, e.g. how likely is he to make moves with drawing hands.