- I'd actually like to see some strategy discussion. I have never been

able to discuss strategy with anyone, so everything I do is based on

my own experience.

Let's see if I agree with the world on the following strategy I have:

If a bower is the upcard, there is only one condition under which I

would refuse to take it -- if I had zero cards in that suit to begin

with. In all other cases (i.e., I have at least one other card in

that suie) I take it, regardless of the rest of my hand.

I have had (online) players who seem to refuse bowers in more

situations than that. But the above has always worked for me.

Is there a prevalent view on this subject?

Thanks.

Kyle. - Hi Joe,

> > Of the times where it makes a difference, here

Note I said "Of the times where it makes a difference." A guarded L

> > are the requirements for success of the two leads:

> >

> > 1) A lead: L in 4th (50%); AND 4th ducks (< 100%).

> > 2) R lead: L in 2nd (50%).

>

> I would say that #2 should read:

> 2) R lead: unguarded L in 2nd (50%).

is not one of those times.

> Your line of play addresses trying to take the first

Sort of. I was simply examining the relative merits of leading the A

> 3 tricks, not necessarily the whole hand.

vs. leading the R.

> However, as I have said, I'd be grateful I cashed my

Depending on the rest of my hand, I might do that, too.

> off A and I'd be leading offsuit away from my RA combo

> and hoping for endplay.

> What are the chances you lead the R and get 2nd's

16%.

> unguarded L when you hold just RA? Didn't you put

> that at less than 15% when you hold RAK?

> And the chance that 4th has Lxx OR 2nd has Lx

22%.

> was very high.

> With only 2 trump between 1st & 3rd, Lx in 2nd

Yes, and that's what's interesting. A guarded L in 2nd is actually

> is even more likely.

more likely than a lone L. This is due to the mass of non-trump in

1st's hand.

Nevertheless, whatever the odds of the R winning, they will always be

better than the odds of the A winning.

For the R lead to be successful, it requires only the proper card

distribution. There is no luck or skill involved. Call the

likelihood of this distribution P(R).

For the A lead to be successful, it requires two things: the proper

card distribution, call this P(A); and a duck by the dealer, call

this P(D).

Since P(R) = P(A), it is always true that P(R) > P(A) * P(D).

What does this mean? It means if you lead the R a bunch of times and

lead the A a bunch of times, the R lead will net more success. How

much more? That depends on how often the dealer ducks.

This is all I've been saying. This does not address whether either R

or A is the best lead, nor does it address the score, nor the

opponent's playing style, nor anything else. Just the logic that in

the long run, the A lead can never be more successful than the R lead.

-- Rob