benlessard

Club is out because declarer choose 6Nt instead of 6H wich make it likely hes got 2 C stoppers. D is out because partner could have X 5D (wich show D not lightner or clubs because otherwise you ll risk playing 5D xx) I think a H is unlikely to cost a trick while i see many scenario where a S cost a trick. Its true that partner may have the K of S and a red stopper but he easily might have just the A fo S and leading S = no more guess for declarer. I ll prefer a H to a S but im far from sure.

Enough shape and enogh values to make a neg X. The chance i get a plus score at 2H isnt good enough to my taste.

Clear 3Nt by opener if i dont have a way to show super-acceptance. Vul at imps you dont pull 1Nt to play 3C unless you have real hope of making 3C if responder has real hope of making 3C then 3Nt by opener is a wise gamble. With a no big hope to make 3C hand responder should just wait for the double and run to 2C.

You will find a lot of very good players (mostly over 50) that for them rebidding 1Nt without a stopper or with a stiff is unthinkable. All im saying was from the "classic standard" point of view. Me i dont remember having some terrible boards with opening 1C with 4-4 in the m and rebidding D after a neg X but at the same time i see that playing 2D as a reverse is probably superior.

I want to declare. [hv=n=sxxhkqxdxc&w=sj9xxhd9xc&e=sxxxhjtxdc&s=sakxhxdxcx]399|300|[/hv]

Partner would lead K with AKQ of S. So Declarer has the Q of S. I think Qxx and AKQxxx of H is a bit heavy for a non-vul 2H preempt. So im going to play partner for a H trick. So Im asking for a switch and even if partner dont cash his S king im not afraid that hes going to lose it.

I think its the opposite. If you are opening 1C with 4-4m than 2D as non-forcing make sense. If you open 1D with 44m than playing 2D as non-reverse make little sense (only useful with 5C+4D weakish hands)

X WTP. If you are happy with +100 on this deal than my feeling is that you are getting robbed way too often or that you are expecting partner to reopen with all kinds of crap. If partner pull to 5H we are going to make it.

In "classic" standard its 100% not forcing. The reason was that a 1Nt rebid showed a stopper or xxxx. 2H with 3 and 2C with any 5 suit were possible. However i believe that 2D forcing might be better than a 2D non-reverse. Especially if you open 1D all 4-4 minors hands.

pass easy. All red in IMPs the double of 2S is more a competitive double than a lead directing double. So partner may easily have wasted S values. Its just wishfull thinking to want to try for low HCP games because we think all of partner hand is working.

Pass WTP

I think knowing if it MP or imps is pretty important here. Obviously at Mp you should strive to play 1Nt instead of 2m. 1m----1S-----1Nt------??? In Mp its clear that rebidding 2S with 5S is a winner but in imps 1Nt is probably safer if partner can have a stiff.

For me a 2 over 1 isnt forcing so 2C is a general force up to 2nt and its the only bid for me. I see no reason to lose the first 5/6 tricks in 2Nt/3Nt when you have a much safer alternative. (1C)---1S-----(P)------2C (P)-----2S-----(P)------2Nt isnt forcing.

If opener often raise with 3♠ support then responder will often pass 1Nt holding 5S . So this make rebidding 1Nt with a singleton more attractive. PS If you play some unbalanced diamond system the responses i suggest are. 1D---1S ??? 1Nt = clubs 2C= 6D or D+H reverse 2D = D+H NF non-reverse strenght 2H = weak or GF S raise 2S = intermediate raise. 2Nt + GF Agree and i think 2D is a beginner bid.

I really like 2H as 2nd negative and i think that playing Kokish over it isnt really pertinent. I have no problems for 2C----2D (GF)----2H (here to be artificial) but to play it that it completly force 2S isnt to my taste. The unbalanced hand should strive to describe not the big balanced hand. Showing 2 extra point in a balanced hand when your are GF is less important than allowing responder to make a 2nd natural bid. a setup like 2C----2D(Gf)------???? 2H all balanced hands doesnt force 2S 2S S 2Nt H 3C C 3D D is very superior to 2D(GF) coupled with 2H-Kokish.

This is false if there is 50% of getting 2Y you should swap the enveloppe. Saying that the envelope already contain Y and 2Y where by switching you are gaining Y or losing Y is a fallacy. Since Y wont have the same value in both cases. The correct calculation is assuming envelope X with x/2 or 2x as the other envelope. Switching will provide. 1/2*2x + 1/2*1/2x = 5/4x wich is higher then x that you get if you do not switch. The answer of the problem is that in an infinity if you have Y in the envelope the other envelope is more likely to have y/2 than to have 2y. If you have a predertermined set of enveloppe you should always switch unless you are getting the highest envelope. 1,2,4,8,16,32,64. Even without looking at the envelope but knowing that if you get the highest number (64) you keep 64 even if you switch you should always switch.

He can but with this hand he will go on anyway.

A 4H Red vs W should show 8.5--9 tricks. This hand is a trick too strong but the prospect of opps sacrifice at 4S is so high that i prefer a direct 4H. If the opps compete further I double to show a great hand. If im W the X should suggest a save at 5H. If im red the X is extra strenght.

Ive passed 1Nt and partner made 5 he had a 3334 with a fairly good 14. Its was worth 0/7 MP :) My regular partner did invite with the same hand.

I just bid 4S. Partner didnt redouble 4C so we got 1 club loser. Partner will need AKQ of spades or AKxxxxx to have slam significatly higher than 50%. With those holding and the K of H i expect him to keep going after 4S anyway.

Ive noticed a fallacy in the wikipedia page about the 2 envelope problems. This is complete BS, If you lose Y but you had the 2Y envelope you lose a Y that is worth half of the amount in your envelope. But if you hold the 1st envelope with Y and you switch you win Y when Y is the full amount of the envelope. If you have 50$ and switch. If the pair was Y=50$ & 2Y=100$ you win Y but that Y is worth 50$. If the pair was Y=25$ and 2Y=50$ you still lose Y but y here is worth 25$. Its clear that in infinity, real number dont have the same probability. Otherwise by adding them you would get more than 1. With the wallet problem its obvious that the more money you have into your wallet= the more likely the other wallet is smaller than yours. So the probability of having the highest wallet is 50% only if you dont look at the amount you have. Once you know the amount you have in your own wallet the probability depend on the amount you have even if you are in infinity.

2H WTP and i think 3D is a mistake. 1- Partner will pass 3D with hands that make 3Nt a good bet and we are in IMPs. 2- partner will bid 3Nt with the K of H and a shaky club stopper. Axxxx Kxx xx Qxx Better to be in 6D than to be in 3Nt 3- Our hand is good for slam.

Wasnt able to find the thread about the envelope problem. Can you post the link ?thanks.

X followed by 3D over 3C show this hand. With a 2 suiter you would bid 2Nt. So double implies 4H + longer suit or a 3 suited hand.

Belgian mathematician Maurice Kraitchik proposed this puzzle: Two people, equally rich, meet to compare the contents of their wallets. Each is ignorant of the contents of the two wallets. The game is as follows: whoever has the least money receives the contents of the wallet of the other (in the case where the amounts are equal, nothing happens). One of the two men can reason: "Suppose that I have the amount A in my wallet. That's the maximum that I could lose. If I win (probability 0.5), the amount that I'll have in my possession at the end of the game will be more than 2A. Therefore the game is favourable to me." The other man can reason in exactly the same way. In fact, by symmetry, the game is fair. Where is the mistake in the reasoning of each man?