A question for Wizard: MS stud basic strategy borderline cases
May 23rd, 2013 at 5:55:59 AM
permalink
Hi, dear Wizard,
I just did my own combinatorial analysis on the basic strategy of MS stud (with the usual rules and pay table). I am able to reproduce the results shown in the webpage (verify the optimality of the strategy, tally the # of combination for different bets, the ev for given player 2 cards etc), however, there's a tiny detail I disagree with you: when knowing 4 cards, I think one should fold to reduce variance when having 1 high and 2 mid cards (but nothing else).
I'm talking about hands like:
A2-6-T (off suit)
68-Q-3 (off suit)
J5-9-7 (off suit)
KT-6-4 (off suit)
etc
These have a conditional unit-less ev (per bet) of exactly -3/4 if the river bet is played to see the last card. This ev = -3/4 is the same for all regardless of the actual card composition, as long as it's 1 hi 2 mid but nothing. For these cases, there are already 3 units (Ante + 1x on flop(3rd street) + 1x on turn(4th street)) hence
folding EV = -3 unit * 1 (lose 100%) = -3
raise 1x on river EV = 4 unit * (-3/4) = -3
That is, folding and playing have exactly the same ev. I'm not sure if the risk-averse analysis (with certainty equivalence) applies to this high skewness game, but clearly with the same ev one should go with the lower variance (in fact, zero variance) of folding here.
Obviously this 'discrepancy' I have with the published results doesn't show in the overall EV, since playing and folding having the same ev. It's only a tiny difference in the strategy. Incidentally, the improvement in the overall variance is pretty much negligible: from 96.7595070 to 96.1939146.
yeah....so for what it's worth, just a theoretical discussion I'd like to have.
Thanks.
If one knows ANY of other players cards, then it's a totally different story......'obviously' one should play when the low cards exceeds a certain proportion among all the cards seen (dealt).
(btw, the similar situation when knowing 3 cards, 1 hi 1 mid, has a EV = -1.9xx, which is higher than -2 so one should play)
I just did my own combinatorial analysis on the basic strategy of MS stud (with the usual rules and pay table). I am able to reproduce the results shown in the webpage (verify the optimality of the strategy, tally the # of combination for different bets, the ev for given player 2 cards etc), however, there's a tiny detail I disagree with you: when knowing 4 cards, I think one should fold to reduce variance when having 1 high and 2 mid cards (but nothing else).
I'm talking about hands like:
A2-6-T (off suit)
68-Q-3 (off suit)
J5-9-7 (off suit)
KT-6-4 (off suit)
etc
These have a conditional unit-less ev (per bet) of exactly -3/4 if the river bet is played to see the last card. This ev = -3/4 is the same for all regardless of the actual card composition, as long as it's 1 hi 2 mid but nothing. For these cases, there are already 3 units (Ante + 1x on flop(3rd street) + 1x on turn(4th street)) hence
folding EV = -3 unit * 1 (lose 100%) = -3
raise 1x on river EV = 4 unit * (-3/4) = -3
That is, folding and playing have exactly the same ev. I'm not sure if the risk-averse analysis (with certainty equivalence) applies to this high skewness game, but clearly with the same ev one should go with the lower variance (in fact, zero variance) of folding here.
Obviously this 'discrepancy' I have with the published results doesn't show in the overall EV, since playing and folding having the same ev. It's only a tiny difference in the strategy. Incidentally, the improvement in the overall variance is pretty much negligible: from 96.7595070 to 96.1939146.
yeah....so for what it's worth, just a theoretical discussion I'd like to have.
Thanks.
If one knows ANY of other players cards, then it's a totally different story......'obviously' one should play when the low cards exceeds a certain proportion among all the cards seen (dealt).
(btw, the similar situation when knowing 3 cards, 1 hi 1 mid, has a EV = -1.9xx, which is higher than -2 so one should play)
