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Flop comes out 2 suits. The air is thick with anticipation. All is quiet, even as the chatters and shouts and excitement intensify across the table. My face is flushed as I await my flush.
First, the technicalities: What are the odds of me making my flush?
The intuitive answer I gave myself was: 50%.
4 suits. Each suit has 1/4 chance of flopping.
2 cards. Add them up.
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We have 3 independent scenarios:
2 more cards of my suit + 1 card of my suit and 1 non suit + 1 non-suit and 1 card of my suit =
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A better, and mathematically similar approach is:
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There is a trick called the Rule of 4 (analogous to the Rule of 72):
We have 9 outs. To get the odds of making your outs come true, simply multiply the number of outs by 4, hence 9*4 = 36% (marvellous approximation!)
(out is defined as undealt cards that are favourable to your cause. The more outs the better!)
Now comes the game theory aspect of poker (with absolute zero knowledge of game theory, I substitute it with the rudimentary knowledge of calculating expectations)
Of course, the concept of "giving a free card" is also very crucial here--not for me, but for my opponents. I have yet to make my hand. I am waiting for a "free card"-- either on the turn or the river. Those who have me beat at the flop should make me pay (by betting heavily before the turn) to get the free card. In case, they don't, I will check quietly, but in case they do, I may just go all-in.
(A case of damned if they don't, and damned if they do, why else would I elect this hand to be my favourite poker moment.)
Why so?
Considering all possibilities when you go all-in. There are 3 and only 3 scenarios:
1) you win the pot outright. Everyone else folds ( the probability here varies. Say, each person has a 10% chance of calling)
P(0 people call) = 0.9^6 = 0.53 (assume 6 people are in the game)
P(at least 1 person call) = 0.47.
We can assume the probability of at least one person calling your all-in to be 50%.
Conversely, the probability of winning the hand outright is also 50% (everybody folds to your all-in)
2) you get called, and win. (0.5*35%=0.175)
3) you get called, and lose. (0.5*65%=0.325)
Calculate expectation.
Let PotMoney = P.
Let your current stack = x.
Let n = total number of opponents who call your all-in
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Let's calculate how high P should be for a favourable all-in.
For expectation > 0 (for n=1),
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Factors to consider before making the quantum leap:
1) The number of people playing
2) The pot
3) How much money I have
4) Chances of anyone making a full house
5) I better be holding the Ace or King of that suit, giving me a nuts flush.
6) Stages of the game (Opening?Endgame?)
I favour amassing chips at the beginning stages of the game. With a large stack, I can proceed to bully the table for the rest of the game, which is something I highly recommend to do once just for the experience. Pointing at my short-stacked opponents, I can either:
1) laugh at them
2) ask, "How much have you?", and throw the amount of bet equal to what they have, forcing them to go all-in for every card they play. And then repeat Step 1.
The gist is: for an all-in, your upside is unlimited (you never know how many "by-catch" are caught on the trailing hook), but your downside is limited to your all-in. When the gambling blood in me goes on full boil, calling all-in is something I relish!