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Short-handed

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Short-handed - Mon Dec 09, 2002, 11:32 PM
(#1)
Deleted user
From the Poker Pages article: Short-handed Hold'Em: Preflop Play (Part I), By Jason Pohl

Example 3:

Player 1 has 6s 5s {small blind}
Player 2 has Kh 9h {big blind}
Flop: Jh 3d 8c {4 small bets in the pot}

Player 1 bets. It's a complete bluff. If Player 2 plays back, Player 1 is in trouble and will likely muck. If there is even a 25% chance of an immediate fold, Player 2 would be correct to raise. (Note: This is true regardless of what Player 2 holds, but the next scenario will illustrate why holding big cards makes this raise even more profitable.)

Can someone explain the math behind this statement?
 
Old
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Tue Dec 10, 2002, 05:52 PM
(#2)
Deleted user
The way I see it Player 2 is effectively getting a raise in 4 times for the price of 1 small bet. 1 time in 4 Player one folds giving Player 2 $5. 3 times in 4 Player 2 pays $2 for a total of $6. So Player 2 has made the pot larger (which she still may win) 4 times for the cost of 1 small bet.
8)

Den
 
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Wed Dec 11, 2002, 12:15 AM
(#3)
Deleted user
If I read this correctly.....P2 is risking 1 extra small bet (his raise) to win 5 small bets already in the pot (after P1's flop bet) ......if P1 will fold 1 in 4 times to the raise that makes the play profitable.


Does that work?
 

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