Re: Poll Results for Euchrescience
- Hi Ace,
> "and if your P has Ac, the opps may ruff it."Yes, but I'm not why my statement needs correcting. 1st chair, after
> With all due respect, pardon the correction, Rob.
> Only 1st might ruff your partner Ace, 3rd is void of trump in the
> polled situation.
all, is a member of the set called "opps." So if he ruffs it, then it
is also true the "opps" ruffed it. ;)
- --- In EuchreScience@yahoogroups.com, "bimbert84" <bimbert84@y...>
> Hi Ace,
> > "and if your P has Ac, the opps may ruff it."
> > With all due respect, pardon the correction, Rob.
> > Only 1st might ruff your partner Ace, 3rd is void of trump in the
> > polled situation.
> Yes, but I'm not why my statement needs correcting. 1st chair,
> all, is a member of the set called "opps." So if he ruffs it, thenit
> is also true the "opps" ruffed it. ;)- - - - - - - - -
> Damn engineers.
> -- Rob
Dang me! And here I thought you were applying definition 2. of the
unofficial jargon term, "opps," but you were using definition 3.
Somebody in my home last night must have spoken the key
word "obvious," which induces me into a temporary hypnotic state.
Clarified and out! (10-4)
- The mathematical analysis
by "bimbert84" is, as usual,
unassailable. But here's
another analysis, from
a slightly different
perspective, on which I
invite his critique (and
After three tricks, each
player besides the dealer
has a 1/4 (2/8) probability
of holding the ace of clubs.
Each of the opponents has, in
addition, a 1/7 probability
of holding the king of clubs
(so does dealer's partner;
but, it is agreed, dealer's
partner's holding the king of clubs does not help the dealer, except
in keeping it out of the opponents' hands, which is already a factor
in assessing the opponents' probabilities).
So, the probability the opponents will take the fourth trick if the
queen of clubs is led is: 50 per cent, the probability one of them
holds the ace (1/4 x 2), PLUS 3.5 per cent, the probability they hold
a king less the probability the dealer's partner holds the ace (2/7
minus 1/4), EQUALS 53.5 per cent; and your probability of taking the
fourth trick leading the queen of spades is the complement, 46.5 per
The probability the opponents will take the FIFTH trick if the queen
of clubs is saved until then is: 80 per cent, their probability of
holding either ace or king of clubs (2/5 x 2), MINUS 20 per cent, your
partner's probability of holding the ace (1/5), EQUALS 60 per cent;
and your probability of taking the FIFTH trick leading the queen of
spades is the complement, 40 per cent.
So, even before consideration of the "squeeze play," probability
favors leading the queen at fourth trick. But these are only
probabilities, ignoring what might be sluffed on the fourth trick. I
realize that the opponents' initial 80 per cent probability on fifth
trick is highly inflated because it equates a king with an ace. Now
is the time to apply the "squeeze" probabilities. I await the
(Hovering throughout all these calculations is the 1/28 probability
the queen of clubs is good "all day" -- i.e., the 2/8 x 1/7 [1/4 x
1/7] probability that both the ace and king of clubs are in the deck.
This probability is included in the calculations above and, therefore,
not mentioned individually.)
You can inflate the opponents' "likelihood" (but not their
probability) of holding a winner of a fifth trick queen of clubs lead
by their ability to communicate -- e.g., the third-chair opponent
could be induced to save a king of clubs instead of an ace of diamonds
by seeing his partner toss the king of diamonds on the fourth trick.
Yet another argument can be made for the precept "Don't squeeze your
partner"; but it is sociopolitical, not mathematical: Let's say your
partner does not and will not understand the math, even if it favors
holding the queen of clubs for the fifth trick; and let's say that
this partner is someone important to you, or with power over you --
your tournament partner, say, or maybe your boss, or even your wife.
How are you going to explain squeezing that ace of clubs from her hand
if an opponent's king takes the last trick? Is making the correct
play worth the point you didn't make?
I think that the moral to this story may be: "If you're the dealer's
partner, save the ace of clubs."
Hindsight? Not entirely. Diamonds are "next," and a bit less likely
to be led to the fifth trick.
And such a policy might free the dealer for the "squeeze play" after
> . . . Yes, but if you lead the queen of clubs on trick 4, they [the
> opponents] don't need communication. It all boils down to who holds
> what. If you lead trump, you at least force them to make the right
> play. . . .
> Unless I'm missing something, a trump lead on trick 4 fails to a Qc
> lead only when:
> 1) Your P holds Ac + A; AND
> 2) He throws the Ac away on trick 4; AND
> 3) Opps hold Kc; AND
> 4) Opps save Kc on trick 4.
> Odds? For #1, there is only one holding of interest, namely Ac + Ad
> (note that we know As is out-of-play since it did not surface on
> trick 1). There are 8 cards unknown, 2 in each hand plus two in the
> kitty (plus As in kitty), for a total of 28 possible 2-card
> combinations. So that's 1-in-28 for Ac + Ad, or about 4%.
> For #2, let's just call it 50%.
> So we're already down to 2% and we haven't even addressed factors
> #3 and #4 (which I'm not going to bother to do here). In all other
> cases, the trump lead fares AT LEAST as well, and in many cases, is
> the only way to win.
> Note that does NOT mean we'll march 98+% of the time if we lead
> trump! It simply means that 98+% of the time the trump lead is at
> least as good as the Qc lead.
> POLL QUESTION: Score 8-8 You pick up 9h to hold J9h, Jd Ks Qc.
> 1st leads Qs, 2nd plays Ts, 3rd plays Js, you win trick with Ks.
> You lead Jh, 1st plays Ah, 2nd plays Th, 3rd plays 9d.
> What do you lead now?
> CHOICES AND RESULTS
> Jack of diamonds, 9 votes, 69.23%
> Nine of hearts, 1 votes, 7.69%
> Queen of clubs, 3 votes, 23.08%