Quote:
Originally Posted by rescued
Strat question. How many streets can we pull with a 3 flush in back before we gotta settle for a pair type hand?
The math behind this is going to be beyond complicated. Assuming the question is "If I draw a card that will make a pair on the bottom and keep me from mis-setting, at what number of remaining draws should you place the card on the bottom."
To do this(I think) you would have to calculate your expected value for both plays:
(1) EV(place on bottom)
(2) EV(place elsewhere)
For simplicity let's say that your mid and top hands will split, meaning there is no chance of sweeping or being swept;
EV(1)= -1 (this is assuming, again for simplicity, you will lose the bottom hand if you make this play)
EV(2)= [RoyaltyValue*(%chance of making royalty)] – [MisSetPenalty*(%chance of MisSet)]
For flush(assuming 4 pts for Royalty and 1 pt for winning the bottom)
EV(2)=5*(%make flush) - 6*(1-%makeFlush)
Set the EV's equal and you have your break even point.
-1=5*(%MakeFlush) - 6*(1-%MakeFlush)
-1=5*(%MakeFlush+6*(%MakeFlush)-6
5=11(%makeFlush)
%makeFlush=5/11=45.4%
So you would need to be able to make your flush 45.4% of the time.
If all of your outs (10) are live then the breakeven point is crossed on 8th street where with 5 draws left your odds are *42.9%* via
OFC Odds Calculator
/shameless self-marketing
Note: There are quite a few factors left out:
-You could possibly win the bottom without the flush
-Even if you don’t hit the flush you might not mis-set
-If sweeping/being swept comes into play these number change significantly.
-I now want to kms