obrona przeciwko nietypowym konwencjom

W poni?szych pytaniach nale?y scharakteryzowa? odzywki, podaj?c: naturalne
(liczba kart), b?d? sztuczne – wtedy znaczenie oraz zakres forsingu: GF,
F1, NF, SO, I; przy naturalnych odzywkach bezatutowych nale?y poda?
przedzia? si?y.
PO 2 MULTI
(blok na starszym plus ewentualnie jakie? silne r?ce)
1.(2) 2BA = ___________ 4/(do koloru) – 4BA=_____________
4=_____________
ktr.=____________
pas – forsuje
nie forsuje
2.(2) ktr. = __________ 2/(do koloru)
ktr.=____________
2=______________
2BA=______________
3/=____________
3/=____________
3/=____________
3BA=______________
4/=____________
4BA=______________
3.(2) pas 2/ (do koloru) ktr.=____________________________
2BA=_____________________________
pas, a potem ktr.=_______________
pas, a potem 2BA=________________
pas, a potem 3/=______________
3/=___________________________
3BA=_______________potem_________
4/=___________________________
4BA=_____________________________
4.(2) 2 pas 2BA=______________________
3/=_____________________
3=_______________________
3=_______________________
4/=_____________________
4BA=_______________________
5.(2)
pas 2 pas
(pas/2) ktr.=__________________________ dalej _____________
2BA=____________________________ dalej _____________
3=____________________________
2=____________________________
3/=__________________________
3BA=____________________________
4/=___________________________
4BA=____________________________
6.(2) ktr. 2/(do koloru) ktr.= ______+ kart w kolorze
skontrowanym
PO 2 FLANNERY
str. 1
Test na zgodność – obrona przeciwko różnym otwarciom
 Marek
Wójcicki
(11-16 PC, 5+ I 4)
1.(2) – ktr.=_________ 2 - 2BA=____________________________
3/=___________________________
3=_____________________________
3=_____________________________
3BA=____________________________
4BA=____________________________
4/=___________________________
4=_____________________________
2.(2) - 2=____________ pas – 2BA=__________________________
2=___________________________
3/=_________________________
3=___________________________
3=___________________________
3BA=__________________________
4/=________________________
4BA=__________________________
3.(2) - 2=_____________ pas -2BA=__________________________
2=___________________________
3/=_________________________
3=___________________________
3=___________________________
3BA=__________________________
4/=________________________
4BA=__________________________
4.(2) - 3=_____________ pas -3=___________________________
3BA=__________________________
4/=________________________
4=___________________________
4=___________________________
4BA=__________________________
5.(2) – pas - 2 - ktr.=____________________________________
2BA=__________________dalej______________
2=___________________________
3/=_________________________
3=___________________________
3=___________________________
3BA=__________________________
4/=________________________
4BA=__________________________
6.(2) – pas - 2 - ktr.=____________________________________
2BA=__________________dalej______________
3/=_________________________
3=___________________________
3=___________________________
3BA=__________________________
str. 2
Test na zgodność – obrona przeciwko różnym otwarciom
 Marek
Wójcicki
4/=________________________
4BA=__________________________
Po 2 s?aba dwukolorówka na dwóch starszych
1.(2) – ktr.=____________ pas - 2/=__________________
2BA=___________________
3/=__________________
3=____________________
3=____________________
4/=__________________
4=____________________
4=____________________
4BA=___________________
pas=___________________
2.(2) - 2=____________ 2 - 2BA=______________________
3/=______________________
ktr.=_______________________
3/=_______________________
3BA=________________________
4/=_______________________
4BA=_________________________
3.(2) – 2BA=__________ - pas - 3=_______________________
3=_______________________
3=_______________________
3=_______________________
4/=_____________________
Po 2 FLANNERY
(11-16 PC, 5+ I 4)
1.(2) – ktr.=____________ pas - 2=__________________________
2BA=___________dalej__________
3/=________________________
3=__________________________
3=__________________________
2.(2)- 2=___________ pas – 2BA=____________________________
3/=__________________________
4/=__________________________
4/=__________________________
4BA=___________________________
3.(2) – pas - 4 - 4=___________________________________
4BA=___________________________________
ktr.=__________________________________
4.(2) – pas - 2 - ktr.=__________________dalej____________
2BA=____________________dalej____________
3/=____________________________________
str. 3
Test na zgodność – obrona przeciwko różnym otwarciom
 Marek
Wójcicki
4/=____________________________________
4BA=_____________________________________
Po 3BA GAMBLING
(pe?ny kolor m?odszy, bez bocznego stopera)
1.(3BA) ktr.=_________________ dalej_________________________
4=___________________ dalej_________________________
4=___________________ dalej_________________________
4/=______________________ co z pozycj? sforsowan??
___________________________
4BA=________________________
5/=_______________________
2.(3BA) pas – 4/5(do koloru) – ktr.=_______________________
4BA=_________________________
5BA=_________________________
4=_________________________
5=_________________________
3.(3BA) pas – 4/5(do koloru) – pas
4/5 - ktr.=____________________________________
4/=___________________________________
4/5BA=___________________________________
str. 4
Test na zgodność – obrona przeciwko różnym otwarciom
 Marek
Wójcicki