Simple math problem from Europen Ch

[Statto]

1)5-3-2-3: there are 6C3 = 20 ways to deal 3 clubs.

2)5-3-3-2: there are 6C2 = 15 ways to deal 2 clubs.

You already pointed out that all combinations where ♦J drops have been eliminated, and as Phil K correctly stated, that leaves just 10 combinations for each.

 

If we assume the ♠ lead to be from a 5-card suit, East has shown 5 ♠, 3 ♥ and 2 ♦, and has 3 spaces left for the remaining ♦. West has shown the same number of cards in red suits, but only 3 ♠, so has 5 vacant spaces for the last 2 ♦. This makes it much more likely the finesse will succeed. A quick check at http://www.automaton.../en/OddsTbl.htm makes the finesse 63% and the drop 53%.

 

Of course it's different if the lead was from a 3-card suit, now just 37% finesse, still 53% drop. But enough more often than not it will be from a 5-card suit to favour the finesse, I think.

 

Edit: if you think the lead will be from a 3 card suit rather than 5 card suit more than 40% of the time, you should play for the drop...

[PhilKing]

A quick check at http://www.automaton.../en/OddsTbl.htm makes the finesse 63% and the drop 53%.

 

Of course it's different if the lead was from a 3-card suit, now just 37% finesse, still 53% drop. But enough more often than not it will be from a 5-card suit to favour the finesse, I think.

 

 

Any tool that can reach a total chance of 116% for two almost mutually exclusive lines has my admiration. I don't think the chance of AQxxx xxx Jxx AK fully accounts for this, somehow. How it can then drop to a cumulative 90% shows a worrying drop in class.

 

Did you program in that the non-leader followed to three diamonds?

 

Anyway, I will do an extra simulation later today, fully expecting a result of 4/3 against, although it will not be for some time due to having just finished a monster poker session. B-)

[Statto]

Any tool that can reach a total chance of 116% for two almost mutually exclusive lines has my admiration.

They're not mutually exclusive at all. You always make if it started with Jxx onside (27%, assuming lead was 5th), and never make if it started with Jxxx offside (11%). Finesse succeeds if it started with Jxxx onside (36%), and drop if it started with Jxx offside (27%). 27% + 36% ~= 63%. 27% + 27% ~= 53% (subject to rounding errors).

[helene_t]

sorry nonsense argument

[PhilKing]

They're not mutually exclusive at all. You always make if it started with Jxx onside (27%, assuming lead was 5th), and never make if it started with Jxxx offside (11%). Finesse succeeds if it started with Jxxx onside (36%), and drop if it started with Jxx offside (27%). 27% + 36% ~= 63%. 27% + 27% ~= 53% (subject to rounding errors).

 

But the question is whether to finesse or play for the drop when second hand plays low on the third round - not whether to win the trick when he plays the jack.

[bluecalm]

You already pointed out that all combinations where ♦J drops have been eliminated, and as Phil K correctly stated, that leaves just 10 combinations for each.

 

Paraphrasing previous poster: I don't know why you quoted me and then wrote this comment so I take it as you quoted a part you agree with... :)

 

You always make if it started with Jxx onside (27%, assuming lead was 5th), and never make if it started with Jxxx offside (11%). Finesse succeeds if it started with Jxxx onside (36%), and drop if it started with Jxx offside (27%). 27% + 36% ~= 63%. 27% + 27% ~= 53% (subject to rounding errors).

 

All those percentages are wrong.

You can try build in BBO calculator to see this, it will show you exact combinations with % or you can calculate it by hand which should be easy if you follow my posts in this thread.

EDIT: unless you mean after playing exactly two round of diamonds but not 3rd. Then the percentages are almost correct (it should be 10.71%, 26,78%, 35,7%) but we really should eliminate Jxx onside if we talk about what decision to make here.

[PhilKing]

Simulation 2 was way in favour of the finesse, running off 6 anaswered winners at one point. Added to sim one the results are pretty close to the 57% (4/3) expected.

[bidule5]

No need for simulation, it is really simple math.

I am with wyman with different notation:

West 5332 with DJ = C(2,5)*C(2,6)=10*15

West 5323 without DJ =C(2,5)*C(3,6)=10*20

so finesse =4/7

[Statto]

EDIT: unless you mean after playing exactly two round of diamonds but not 3rd. Then the percentages are almost correct (it should be 10.71%, 26,78%, 35,7%) but we really should eliminate Jxx onside if we talk about what decision to make here.

Sorry all, confusion, yes I did mean before embarking on the 3rd round of ♦. And yes those numbers are very familiar, I just rounded them.

 

But it's far simpler. There are 4 vacant spaces against 3 after the half round of ♦, assuming lead from 5. If the ♠ lead is from 3 it's 5:2 against. Happily, 26.78:35.7 is also 3:4, so we're all right. Apart from those who were wrong.