ain't it grand

[gwnn]

[hv=pc=n&s=skj974ha9532da2cj&n=saq2hk7dkt7caq532&d=s&v=0&b=11&a=1sp2cp2hp2sp3hp4cp4dp4np5c(0%2F3)p5dp5sp7sppp]400|300[/hv]

 

W leads the ♣9.

[Phil]

Just try ruffing the heart. You'll make when they are 33 or RHO has the length and they can't uppercut you. For ruffing hearts high you still need to deal with the sT

 

The club lead could be from anything so I wouldn't put much stock in trying to figure out if LHO has heart length.

[gwnn]

Just try ruffing the heart. You'll make when they are 33 or RHO has the length and they can't uppercut you. For ruffing hearts high you still need to deal with the sT

 

The club lead could be from anything so I wouldn't put much stock in trying to figure out if LHO has heart length.

I mean by the critical moment (K of hearts, A of hearts, low heart, LHO following) I know RHO doesn't have the length. So isn't it between:

 

a) (hearts 3-3 and spades 3-2) + (hearts 4-2 and spades 3-2 with the T onside) + (some change for spades 4-1 with the stiff T)

b) hearts 3-3

 

I'm mainly just posting this because I'm afraid I'll get something wrong if I try multiplying the percentages, and to share some boards.

[gwnn]

OK first effort at multiplying percentages (I'm always a bit unsure when I do it because I suspect that in actual fact they are far from independent):

 

a) = 68%*(36%+1/2*24%)+1/5*28%*36%=30.6% (=spades 3-2 and hearts 3-3 or half of the 4-2's OR one-fifth of the 4-1's with hearts 3-3)

b) = 36%*96% = 34.6% (hearts 3-3 but spades not 5-0)

 

there's some rounding going on but it wouldn't change a 4% edge. or did I miss something for a?

 

(note: these are not the actual final %'s of the grand's chances as there is an approximately 24% chance that LHO had a doubleton heart which both a and b picks up)

Edited May 21, 2016 by gwnn

[Phil]

Won't check your math now but remember when at the time you play the 3rd heart that hearts being 33 is much higher than 36%.

[gwnn]

Of course I know that... But it's higher than 36% for both cases. I'm not biased towards a) or b) when I take a priori probabilities.

 

As a simple example, if we compare the probabilities for HHH and HHT of a biased coin, 60% heads, 40% tails, you could either compare just the last coin toss (60-40) or the a priori probabilities of all of it: 0.36*0.6=0.216 vs 0.36*0.4=0.144. In both cases the ratio is the same: 1.5.

 

It was just a shortcut I used to use known percentages instead of percentages I would have to justify.

[gwnn]

OK I just re-entered my equations in the calculator and now they are 34.66% for a and 34.56% for b. hmmm

 

 

For more exact values (3-3=35.520%, half of 4-2's=24.225%, 3-2=67.8%, 4-1=28.26%) I get:

 

a=34.30%

b=34.12%

 

So basically the two are equally good/bad. But I still think I might be missing some case.

[1eyedjack]

Cashing trump ace before ruffing heart may pick up another minor combination at no obvious cost.

[gwnn]

Cashing trump ace before ruffing heart may pick up another minor combination at no obvious cost.

Hmm you mean like for case b, hearts 4-2 along with stiff T of trumps I guess? But we still can't ruff two hearts in dummy. Or erm hearts 4-1 with specifically east having the stiff T so that he can't overruff anything. That is 0.68% if I'm still using these unconditional probabilities but I really think it would be a different result given that I'm giving someone 4-4 in the majors and way too few minors*.

 

Case a) would suffer a lot from cashing a trump early.

 

*-ok what the heck let's just do it:

 

we assume east's hand to have 2 hearts (from 6) and the stiff T of spades (from 5) and know nothing about the rest of his hand (10 random cards out of 15). The probability is:

 

C(6,2)*C(1,1)*C(15,10)/C(26,13)=45045/10400600=0.43%

 

Sanity check, what if I am looking at stiff T of spades with the long hearts (I can't pick it up but let's just look at it, in the "naive multiplication" case it would have the same probability)? It should be higher than 0.68% by about the same amount that 0.43% is lower, so about 0.93%?

 

C(6,4)*C(1,1)*C(15,8)/C(26,13)=96525/10400600=0.93% -- yes, 0.93%!

(of course C(6,4)=C(6,2), but it is easier to write these equations when you know that the second terms in the parantheses add up to 13).

 

TL; DR: OK so case b has an additional edge of about 0.43% - making it better than a by a whisker instead of worse by half a whisker.

[Phil]

Curious what the non spade lead suggests.

[gwnn]

Yes I know. Not sure I know how strong of an inference that is - I don't know that much about my opponents. I'm not really trying to find excuses for my line. At the table I thought ruffing high was going to be significantly better than ruffing low (the number of possibilities is longer but each of them is smaller), but clearly I was wrong. I agree that if the two lines are approximately equal, I should take the lead into account. For the moment I was just trying to figure out whether they were equal and it seems they are close enough.