Ah, Dr. Dave; Ever the theorist, eh?
Our group of about 16 - 18 guys has been playing the way I mentioned for about 25 years. We have a plus 1, a couple of 5's and 7's, some 10 to 13's and at least 5 guys in the 16 to 18 hdcp range. Over time group winnings have pretty much balanced out even. When I was a single digit paired up with say, an 18, I won about as many times as I lost. Ever the empiricist, TFlan ;-) --- Dave Tutelman <[EMAIL PROTECTED]> wrote: > At 06:28 PM 8/6/2007, TOM FLANAGAN wrote: > >Of course this only applies if you're playing a > 4-some > >event. > > That's why I was asking what the game was. > > >But still, within groups we have team bets, > >Generally the high and low vs the 2 in the middle, > or > >something like that. Again, we try to match up > total > >handicap strokes. > > OK, let's assume we're talking about 4-ball better > ball match play. > Then matching up total (or average) handicap strokes > seems like a > natural way to do it. You should be able to play > level that way. But > experience (or a little thought) says that it's not, > if there is more > than a few strokes difference within the foursome. > > That's because, in a better-ball event, one golfer > who is > substantially better than the others will dominate. > For instance, > suppose you have two 15-hcp guys playing a 10 and a > 20. The 10 is > enough better than both the members of the other > team that his play > will completely dominate, and the 20 will only > occasionally be "in > the hole". Even though the average of the two teams > is the same, the > 10-20 team will beat 15-15 a disproportionate > fraction of the time. > > I remember seeing an article (probably a decade ago) > that made a lot > more sense. Match the average of the INVERSE of the > handicaps. That > way, a lower handicap carries more weight than a > higher handicap of > the same difference. Put another way: > > Effective handicap = 1 / (1/A + 1/B) > > where > A = handicap of the A-player > B = handicap of the B-player > Effective handicap = the handicap to use > when comparing > teams for fairness. > > For you electrical guys out there, it's the parallel > resistance > formula. Tom is saying, if the handicap is a > resistor, then matched > teams should have the same series resistance. The > article would argue > (and I would agree) that it makes more sense for > matched teams to > have the same parallel resistance. > > If you want Tom's formula (match by average > handicap) to make sense, > you should play a high-low game rather than simple > best ball. In that > game, each player is always in the hole. The best > scores of each team > is matched, AS IS THE WORST. For instance, suppose a > hole goes: > Red: player A=4, player B=7 > Blue: player A=5, player B=6 > > In simple best ball, Red wins the hole. > In hi-lo, Red gets a point (4 beats 5) and Blue gets > a point (6 beats 7). > > Hope this is more helpful than confusing. > > Cheers! > DaveT > > > -- > No virus found in this outgoing message. > Checked by AVG Free Edition. > Version: 7.5.476 / Virus Database: 269.11.6/938 - > Release Date: 8/5/2007 4:16 PM > > >
