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Problem of the Week #91: January 9 Problem of the Week #91: January 9

01-14-2011 , 02:34 PM
As a general skill-gambling rule, play higher variance against stronger players and lower variance against weaker players (if things are otherwise close). Longer matches/games favor stronger players because it gives weaker players more opportunities to make mistakes.
Problem of the Week #91: January 9 Quote
01-14-2011 , 09:34 PM
A-Let’s look a simple situation: no cube and cash game.
Black’s chances = 85,65% (= 5/36 + (1-5/36)*(1-6/36))
White’s chances = 14,35%
Equity = 71,30% (=85,65%-14,35%)
Equity
(A) No Double: 71,30%
(B) Double / Pass: 100%
(C) Double/Take: 142,60%
So: Double / Pass

B-The problem is more complicated because of the cube and the length of the match.
We note MWC(1,7) the match winning chance if Black 1- away and White 7- away, MWC(3,5) the match winning chance if Black 3- away and White 5-away, etc.
We have:

(A) No Double: MWC(A) = 85,65%*MWC(1,7)+ 14,35%*MWC(3,5)

(B) Double / Pass: MWC(B) = MWC(1,7)

(C) Double/Take: MWC(C) = 85,65%*MWC(0,7) + 14,35%*MWC(4,4)
= 85,65% + 14,35%*MWC(4,4)
= 92,82% if same level
< 92,82% if white strong
> 92,82% if black strong

(1) Black should double <=> the minimum of [MW(B), MWC(C)]>MWC(A)

I-If White is a very strong player MW(B)= MWC(1,7) is probably less than 85,65%
so
(1) Black should double <=> MW(B)>MWC(A)
<=> MWC(1,7)> 85,65%*MWC(1,7)+ 14,35%*MWC(3,5) True because MWC(1,7)>MWC(3,) so
Black should double

If Black doubles :
(2) White should take <=> MWC(C)<MWC(B)
<=> 85,65% + 14,35%*MWC(4,4)<MWC(1,7) False because MWC(1,7) < 85,65% so :
Double/Pass

II-If White is a weak player MWC(1,7) is probably near 100%
so :
(1) Black should double <=> MWC(C) >MWC(A)
<=> 85,65% + 14,35%*MWC(4,4) > 85,65%*MWC(1,7)+ 14,35%*MWC(3,5) probably False
So no double.

If Black doubles
(2) White should take <=> MWC(C)<MWC(B)
<=> 85,65% + 14,35%*MWC(4,4)<MWC(1,7) probably True so :
No double/Take
Problem of the Week #91: January 9 Quote
01-15-2011 , 02:18 AM
Well this has gotten interesting re: bill's post on the beaver and crawford. As i have read the whole thread.
............
(a) Assume you are Black and playing a very strong player. Should Black double? If Black doubles, should White take or drop?
Black should double and White should drop.
If white takes his only chance is if black does not roll doubles, white will have 1 chance to roll doubles, 6 out of 36 and would have to double before the roll. It would be a take for black
and a win of the tourney for black if white missed and a win for white if he hit doubles.
.............
(b) Same question, but now you are Black playing a weak player.
Black should double, the odds are just too good to not go for it all.
Problem of the Week #91: January 9 Quote
01-16-2011 , 10:13 AM
This concept was hotly debated in the poker community for quite a while, whether you should accept or push small edges even if you are a superior player.


The general consensus is that the skill edge is never quite as big as one thinks in poker, and it is usually a good idea to just push all of the edges.

However, in my work on the subject, I am fairly certain that there is a very important tweak. There are particular edges you should be more likely to forego, in a game of part skill and part luck such as poker (and bg)...those being edges which are:

1.small
2.increase the luck to skill ratio the most
3.Are match/game/hand/tournament dependent and of high variance.

For example, in poker, if someone opens utg and you as a superior player are deciding whether to flat call the open on the button with some speculative hand such as 22, 67s etc, lets say that the math dictates it is a marginal call by .01%. In this case you should push that edge and make the call, as subsequent action is still skill based, and not automatically determinative of the hand/game/tourney etc.

On the other hand, if someone open shoves allin and has you covered, and you determine that it is a call by .01% then this is an edge you should forego, as the luck to skill ratio would be pushed right to the maximum and would be instantly determinative of the final result.

It SEEMS to me that a similar analysis to BG problems such as this is called for. If everyone's calculations are correct such that this is a marginally decent double, i imagine that it might not be a bad idea to forego the edge for similar reasons discussed above. Interestingly though, if the opponenet is that inferior such that he will not know to reship....well then.

Obviously the danger in always playing like this is that you will OVERestimate your skill edge and UNDERestimate your edge you are passing up, but as a good player you should be able to avoid ethat trap as well.

Hope this made sense.
Problem of the Week #91: January 9 Quote
01-16-2011 , 11:12 PM
Let's first say that, hypothetically, the cube would never be turned again from the posted position.

White' winning chances are 31/36 x 1/6 = 31/216 = 14,35%.
Therefore, Black's winning chances are 85,65%.

Now, if I refer to Woolsey's MET (1994, http://www.bkgm.com/rgb/rgb.cgi?view+801), here are the leader-trailer chances:

At 1-away 7-away Crawford = 91% - 9%
At 3-away 5-away = 66% - 34%

So, right now, if we would not factor in the cube handling:

White has a 14,35% X 34% + 85,65% x 9% = 12,59% of match winning chances.
Black has a 14,35% x 66% + 85,65% x 91% = 87,41% of match winning chances.

Now, let's say Black doubles.

If White drops, it becomes 6-0 Crawford (1-away 7-away) and White has 9% of Winning the match.

If White takes, he will theorically re-double, which means that when White wins the game, he wins the match as well. Therefore, that comes down to the initial chances of 14,35% for White.

Clearly, in (a), Black should hold the cube and White will take if he's doubled.

Now, what about in (b) ?

I guess we may have to assume that White, being a weak player, will miss the re-double (or even drop Black's cube!)

If that's the case, When White wins the game, the match becomes 4-4 and it's theorically 50-50 (I say theorically, because if White is in fact weak, it's a bit more than 50% for Black). But when he loses the game, he loses the match.

Therefore, White's winning chances would go like this:

14,35% x 50% + 85,65% x 0% = 7,18%

So if Black suspects that White will either drop the cube or forget to re-double, than Black should double.


(a) No double, take
(b) Double (suspecting either drop or take with no re-double)
Problem of the Week #91: January 9 Quote
01-17-2011 , 10:10 AM
I think you could lose your market by not doubling. LOL
No double/take for both
Problem of the Week #91: January 9 Quote

      
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