comment_law_12

# Σύγκριση εκδόσεων

Εδώ βλέπετε τις διαφορές μεταξύ της επιλεγμένης έκδοσης και της τρέχουσας έκδοσης της σελίδας.

 comment_law_12 [2019/03/05 12:13]pournaras comment_law_12 [2019/03/05 12:17] (τρέχουσα)pournaras 2019/03/05 12:17 pournaras 2019/03/05 12:13 pournaras 2019/03/05 12:02 pournaras δημιουργήθηκε 2019/03/05 12:17 pournaras 2019/03/05 12:13 pournaras 2019/03/05 12:02 pournaras δημιουργήθηκε Γραμμή 267: Γραμμή 267: Example 11: Example 11: - A pair is misinformed and ends up in 3NT going down one instead of playing in a making 6♦contract. If the polling shows that it is easy to get to 6♦with correct information, then the TDshould assign 100% of 6♦. + A pair is misinformed and ends up in 3NT going down one instead of playing in a making 6♦ contract. If the polling shows that it is easy to get to 6♦ with correct information, then the TD should assign 100% of 6♦. - If however the polling shows only a ~50% chance of the pair getting to 6♦, then the TD should assign a percentage of 6♦making together with a proportion of the various (successful and nonsuccessful) game-level contracts. After factoring in the associated uncertainty the TD obtains an adjusted frequency of: 0.50 x 1.2 = 0.60 (which equates to 6♦making ~60% of the time). + If however the polling shows only a ~50% chance of the pair getting to 6♦, then the TD should assign a percentage of 6♦ making together with a proportion of the various (successful and nonsuccessful) game-level contracts. After factoring in the associated uncertainty the TD obtains an adjusted frequency of: 0.50 x 1.2 = 0.60 (which equates to 6♦ making ~60% of the time). - Now let us suppose that for 6♦to make the declarer has to find a queen and it is a pure guess. We therefore don’t know if he would get it right or not, so it is now normal to include a proportion of both 6♦making and 6♦going down as part of the final weighted result (again giving some consideration to the margin of doubt associated with the process). Hence, if it seems that getting to 6♦is 100% certain and making it is only a ~50% chance; the assigned score would be 6♦making ~60% of the time (0.50 x 1.2 = 0.60) and going down ~40% of the time. + Now let us suppose that for 6♦ to make the declarer has to find a queen and it is a pure guess. We therefore don’t know if he would get it right or not, so it is now normal to include a proportion of both 6♦ making and 6♦ going down as part of the final weighted result (again giving some consideration to the margin of doubt associated with the process). Hence, if it seems that getting to 6♦ is 100% certain and making it is only a ~50% chance; the assigned score would be 6♦ making ~60% of the time (0.50 x 1.2 = 0.60) and going down ~40% of the time. - If the TD discovers that only ~50% of the players polled would get to 6♦, and that those in 6♦would only make it ~50% of the time then, based upon the raw percentages, we would expect thenon-offenders to get the score for 6♦making ~25% of the time. But since they are the non-offendingside, it is entirely appropriate to give them some benefit of doubt and assign 6♦making ~30% of the time (0.25 x 1.2 = 0.30). This means that the remaining ~70% would need toinclude those occasions when 6♦fails, as well some proportion of 5♦making and 3NT failing. + If the TD discovers that only ~50% of the players polled would get to 6♦, and that those in 6♦ would only make it ~50% of the time then, based upon the raw percentages, we would expect the non-offenders to get the score for 6♦ making ~25% of the time. But since they are the non-offending side, it is entirely appropriate to give them some benefit of doubt and assign 6♦ making ~30% of the time (0.25 x 1.2 = 0.30). This means that the remaining ~70% would need to include those occasions when 6♦ fails, as well some proportion of 5♦ making and 3NT failing. === Serious error === === Serious error ===
• comment_law_12.1551780821.txt.gz
• Τελευταία τροποποίηση: 2019/03/05 12:13
• από pournaras