PokerTracker

When I filter this column, hands should appear that won the SD but I also get several losses

what is the reason ?


r_probe_wtsd_won

sum(if[cash_hand_player_statistics.flg_r_bet AND lookup_actions_t.action LIKE 'X' AND lookup_actions_p.action NOT LIKE '%R' AND cash_hand_player_statistics.flg_p_face_raise AND cash_hand_player_statistics.enum_f_cbet_action = 'C' AND ((cash_hand_summary.cnt_players >= 3 and cash_hand_player_statistics.val_p_raise_aggressor_pos < cash_hand_player_statistics.position) OR (cash_hand_summary.cnt_players = 2 AND cash_hand_player_statistics.flg_showdown AND cash_hand_player_statistics.flg_won_hand AND cash_hand_player_statistics.flg_blind_b)), 1, 0])

The original cnt_r_probe column has separate parts for counting river probes in multiway hands and heads up hands but in your expression the cash_hand_player_statistics.flg_showdown AND cash_hand_player_statistics.flg_won_hand part is within the final OR condition so it's only being used for heads up hands. If you want to count showdown/winning hands for all river probes you can just place the edit at the beginning of the expression instead:

r_probe_wtsd_won

Code: Select all

sum(if[cash_hand_player_statistics.flg_showdown AND cash_hand_player_statistics.flg_won_hand AND cash_hand_player_statistics.flg_r_bet AND lookup_actions_t.action LIKE 'X' AND lookup_actions_p.action NOT LIKE '%R' AND cash_hand_player_statistics.flg_p_face_raise AND cash_hand_player_statistics.enum_f_cbet_action = 'C' AND ((cash_hand_summary.cnt_players >= 3 and cash_hand_player_statistics.val_p_raise_aggressor_pos < cash_hand_player_statistics.position) OR (cash_hand_summary.cnt_players = 2 and cash_hand_player_statistics.flg_blind_b)), 1, 0])

Great thank you very much

I had understood that the order of the variables did not affect the final filter, I will take that into account

The order wouldn't matter if only AND was used but since there is an OR condition then you can't just put the extra code within that condition if you want it applied everywhere:

z AND a AND b AND c AND ((e) OR (f))

b AND c AND d AND ((e) OR (f)) AND z

both of the above will give the same result however

b AND c AND d AND ((e) OR (f AND z))

will not.