rpls

Got it, thanks

I tried a different approach and got a different result. I will count the number of favorable cases and divide it by the total numbers of possible cases. Aa means first finesse works Af means first finesse fails Same for B e C finesses All possible cases are: Aa Ba Ca Aa Bf Ca Aa Ba Cf Aa Bf Cf Af Ba Ca Af Bf Ca Af Ba Cf Af Bf Cf So I have 8 possible cases All favorable cases are: Aa Ba Ca Aa Bf Ca Aa Ba Cf Af Ba Ca So the quotient of favorable cases by the number of possible cases equals to one half –50%

I dont mind looking dumb, so here I go: Yes, I agree Humm, dont get it Could you please detail it a bit more?

Hi I have this slam and in order to win I need to win two out of 3 finesses Are the odds 50% or 25%? One can reason like this: imagine declarer loses the first finesse; then he needs the second finesse to work AND the third too, which means 50*50% equals 25%. Do you agree? If needed I can post the hands Regards

Nowadays transfers are more and more on the scene Anyway I appreciated your scheme; thanks!

Playing at IMPs, you are dealt: [hv=pc=n&w=s5h872dajt953ckj4&d=e&v=e&b=6&a=1h2s]133|200[/hv] What are my options? 2NT invite+, with 3+ cards2NT Lebensohl type Usually responder has room to distinguish between 3 and 4+ card support, but not here; what do you recommend?

Hi i wonder if it is possible to generate a hand with 12-14 HCP if non vulnerable and in case its vulnerable then the hand must be in the 15-17 HCP range Thanks in advance

problem solved distribuicaoRespondente = (sS >= 5) ? (sSn >= 4) : (sHn >= 4)

Hi, i read the manual about this and cant have it working here's my code: //teste para Gladiator - 7 nov 2019 sS = spades (south) sH = hearts (south) sD = diamonds (south) sC = clubs (south) sSe = spades(east) sHe = hearts (east) sSn = spades(north) sHn = hearts (north) sDn = diamonds (north) sCn = clubs (north) zona = 15 < hcp (south) and hcp (south) < 19 distribuicaoST = shape (south, any 4333 + any 4432 + any 5332) ricoQuinto = sS == 5 or sH == 5 menorQuarto = sD == 4 or sC == 4 restricaoRicoMenor = not (ricoQuinto and menorQuarto) restricaoRicos = not ((sH > 4 and sS > 3) or (sH > 3 and sS > 4)) restricoes = restricaoRicoMenor and restricaoRicos distribuicao = distribuicaoST and restricoes interventor = zona and distribuicao zonaAbridor = 5 < hcp (east) and hcp (east) < 11 distribuicaoAbridor = sSe ==6 or sHe == 6 abridor = zonaAbridor and distribuicaoAbridor if (sSe == 6) distribuicaoAdvancer = (sHn >= 5 and sCn >= 5); else distribuicaoAdvancer = (sSn >= 5 and sCn >= 5); zonaAdvancer = hcp (north) > 8 advancer = distribuicaoAdvancer and zonaAdvancer advancer = zonaAdvancer condition abridor and interventor and advancer what i want is to know how to use a simple if/else statement can anyone please help me? Thanks in advance NB my code works perfectly if i exclude the "if"

I must confess i started reading the manual but that looks like "mandarin" for me :) my goal is to be able to generate hands for training purposes Thanks, nige1

Great! The code is very elegant: are you a pro? I dont understand the last 2 lines, mostly the last one :(

in the meanwhile i got it: condition shape(south, any xxxx - any 4333 - 4432 - 5332) and hcp(south)>=16

Hi Thanks for answering What i'm doing here is testing this precision style openings :) As an aside, how can i code something like: i dont want a 4333 or 4432 or 5332 shape? I'm trying hcp(south)>=16 and condition shape(south, ! (any 4333 + any 4423+ any 5332)) but it wont work Regards

Hi basically i want to generate hands that would open in one of a major; this implies opener cant have a longer minor suit i have the code for opening 1 M: ( spades(south)>=5 and hcp(south)>=10 and hcp(south)<=15 ) || ( hearts(south)>=5 and hcp(south)>=10 and hcp(south)<=15 ) but i dont know how to code the restriction i mentioned above Thanks in advance for your help

I just tried the last code and it worked perfectly I wanna thank both of you for your help Regards