I am a genius!
... but since one or two of you are even more of a genius than I am, I'm going to let you help me with this problem (my work to follow, for the sake of humour).
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Harrington on Holdem, page 150. You're in a tournament with mostly conservative opponents. SB is shortstacked after taking a few beats, but that was an hour ago, so he's not tilting. Your stack: 2100, his 550, UTG+2: 2100. Your hand, 55, in MP3. Blinds are 50-100, no ante, button has a big stack.
PF: 2 folds, UTG+2 limps, 2 folds, you limp, two folds, SB pushes for 550, BB and UTG+2 fold. Pot 850t, 450t to you.
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Pot odds: 1.85:1
I'm going to assume that villain has either two overcards or an overpair, to make things easier.
When villain has an overpair, you are a 4.5:1 dog. When villain has overcards, you're a 0.8:1 dog, which is good. (These values are taken from page 126.)
I've been out of school way too long... I'm trying to find out how often villain has to have overcards in order to make this breakeven.
0.8x + 4.5y = 1.88
I've been out of school for a while, so for a nice laugh, I'll show you why I'm a genius.
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0.8x + 4.5y = 1.88, isolating y in terms of x
==> y = (1.88 - 0.8x) / 4.5
==> 0.8x + 4.5* ((1.88 - 0.8x)/4.5) = 1.88
==> 0.8x + 1.88 - 0.8x = 1.88
==> 1.88 = 1.88
Yep, I'm pretty smart. In fact, I'm SO smart, that I knew this even before writing down the equation. [img]/images/graemlins/smile.gif[/img]
Could someone please show me their "rough work" that shows how they figured out how often villain has to have overcards?
--Dave.
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