Handicapping an Uneven Scramble

[Trivia Jockey]

I am organizing a golf trip coming up in a couple weeks where one of the events is a team scramble.  We usually do 4 vs. 4 which is fairly easy to handicap, but this year we had a cancellation so the teams will be 4 vs. 3.  I'm trying to find a way to handicap it so it will be a fair competition.

I know a popular way of calculating a scramble team's handicap is (Golfer A index x .25) + (Golfer B index x .20) + (Golfer C index x .15) + (Golfer D index x .10), where A is the best golfer, B is second-best, etc.  (And for the team of 3 you just drop the Golfer D piece.)

But is that enough?  Should you still somehow account for the fact that the team of 4 gets 4 chances at each shot?

Any suggestions? 

P.S. I am trying to avoid the need for the team of 3 to play a fourth ball on every shot, every hole, if possible.  Maybe just add an extra tee shot on each hole?  Or an extra putt?

[iacas]

No, you can't just drop the D golfer. That fourth shot is helpful. The more shots you get, the better the score you're going to shoot.

If you imagine a handicap of 5, 10, 15, and 20, 0.25 * 5 + 0.2 * 10 + 0.15 * 15 + 0.1 * 20 = 7.5. If you dropped the 20, that's 0.25 * 5 + 0.2 * 10 + 0.15 * 15 = 5.5 - So that team would be getting two fewer strokes despite losing a guy. They should get MORE strokes.

Perhaps something like… 0.35 * 5 + 0.3 * 10 + 0.25 * 15 = 8.5. But even that feels low to me.

[Ty_Webb]

Personally I'd have them rotate the player who hits the fourth shot and have them hit four shots each time. Then handicap I'd make it 0.25 * A + 0.2 * B + 0.15 * C + 0.1 * (A + B + C)/3. 

If you don't want to have someone hit twice every shot, then rather than saying they can for one tee shot, I'd give the team of three one time per hole that they can have one of their players (at their choice) hit a second shot. If they use it on the tee, then they can't use it anymore. I'd think putting is where it could be most valuable since it has the starkest contrast between good outcome and bad outcome, but off the tee if you have nothing in play would obviously be where you'd want it. Do the handicap the same way - the difference then would be that they could always pick their best player to hit the extra shot, but they still get the extra handicap strokes for the worse players so it somewhat evens out.

If you're set on only having them hit three shots at a time, then you'd need to figure out a way to handicap it fairly. I agree with Erik in his example that 8.5 doesn't feel like enough. Only one extra stroke for 50-60 extra shots on the round (I'm assuming there will be quite a few putts where the D player wouldn't be needed and they're not going to be shooting 15 under) seems like not a lot. My initial reaction was to say you'd want to double the allowances. If you lose the 5 handicap, then you'd get 0.5 x 10 + 0.4 x 15 + 0.3 x 20 = 17 shots. If you lose the 20 handicap then you'd get 11. 3.5 strokes for losing the 20 handicap seems like quite a lot to me, so maybe split the difference? Something like 0.5 x A + 0.35 x B + 0.2 x C could work. That would be 9 shots if you lose the 20 handicap and 14.25 if you lose the 5 handicap. 

If the handicaps are more even, let's say they're all 5 handicaps, 0.5 x A + 0.35 x B + 0.2 x C would be 5.25 shots, while the four 5 handicaps would get 3.5. 1.75 shots is probably not enough for losing the 4th go at every shot. If they're all scratch then both teams get 0. More I think about it, I think multipliers might not be the best option. Instead I'd say the four man team gets the 0.25 x A + 0.2 x B + 0.15 x C + 0.1 x D. Three man team gets 0.25 x A + 0.2 x B + 0.15 x C + 4. I think 4 scratch players would beat 3 scratch players at a scramble by somewhere in the 3-5 strokes range on average. The extra man is worth less the worse they get, but that's getting offset by the shots they would be giving the team, so if you lose a 20, you only get 2 extra strokes. Put another way, this is equivalent to assuming you have a fourth player who is never good enough to hit a ball in play, but has a 40 handicap in return. That seems reasonable to me. 

[iacas]

14 minutes ago, Ty_Webb said:

Personally I'd have them rotate the player who hits the fourth shot and have them hit four shots each time. Then handicap I'd make it 0.25 * A + 0.2 * B + 0.15 * C + 0.1 * (A + B + C)/3. 

That'd be too high - the A, B, and C players are all better than the D player would be, but you're giving them the same weight. I'd be more inclined to let them hit a fourth shot, but not give any bump to the index: just do 0.25A + 0.2B + 0.15C. An A player getting to hit every third shot a second time is a big bonus.

[Ty_Webb]

38 minutes ago, iacas said:

That'd be too high - the A, B, and C players are all better than the D player would be, but you're giving them the same weight. I'd be more inclined to let them hit a fourth shot, but not give any bump to the index: just do 0.25A + 0.2B + 0.15C. An A player getting to hit every third shot a second time is a big bonus.

Yep - that's a fair point. I guess it somewhat depends on what the handicaps are though. If you've got eight 20 handicaps in total and one drops out, then both teams have 4 players hitting each shot. All shots are being hit by 20 handicappers, but you'd be giving two extra strokes to the team with 4 players rather than 3. On the other hand, if you've got the ranges from your example, and one of the 20s drops out, then the three player team will have better players hitting the fourth shot every time, for which they get only 1 less stroke. (0.1 x 20 vs 0.1 x (5+10+15)/3). If one of the 5s drops out, then the 3-player team's handicap would be 0.25 x 10 + 0.2 x 15 + 0.15 x 20 = 8.5. Where the 5, 10, 15, 20 would be 7.5. Seems a little mean spirited to lose the best player and only get 1 extra shot. My way would make that 10 vs 7.5, which seems a little fairer. I think it's too difficult to come up with an always fair way to do it that works regardless of the playing ability of whoever is missing.

[Trivia Jockey]

This is good stuff, thanks guys.

[iacas]

3 hours ago, Ty_Webb said:

Yep - that's a fair point. I guess it somewhat depends on what the handicaps are though. If you've got eight 20 handicaps in total and one drops out, then both teams have 4 players hitting each shot. All shots are being hit by 20 handicappers, but you'd be giving two extra strokes to the team with 4 players rather than 3. On the other hand, if you've got the ranges from your example, and one of the 20s drops out, then the three player team will have better players hitting the fourth shot every time, for which they get only 1 less stroke. (0.1 x 20 vs 0.1 x (5+10+15)/3). If one of the 5s drops out, then the 3-player team's handicap would be 0.25 x 10 + 0.2 x 15 + 0.15 x 20 = 8.5. Where the 5, 10, 15, 20 would be 7.5. Seems a little mean spirited to lose the best player and only get 1 extra shot. My way would make that 10 vs 7.5, which seems a little fairer. I think it's too difficult to come up with an always fair way to do it that works regardless of the playing ability of whoever is missing.

The only time it would work to just let another player hit a shot is if everyone is the same handicap… and even then, having already hit the shot, that team would still be at an advantage because the same player is always getting a second chance to re-hit the shot.