http://iowacomicbookclub.com/ups1.php When Tony returned from Atlanta, where he and his wife hide every winter, he brought us a good new golf game, which he learned from his friends down there. Here’s how it works: Let’s say you have four guys, playing two against two. Each pair counts just twenty of the thirty-six hole scores they collectively make during the round, and the winning pair is the one with the lower twenty-score total. After finishing a hole, each pair has to decide, before teeing off on the next hole, whether to count one, both, or neither of the scores they just made. The effect is similar to that of Stableford scoring, since disasters can usually be ignored.

Let’s say that you make a net birdie the first hole, while your partner makes a net double-bogey. You presumably want to forget all about the double-bogey, so you announce (before hitting your tee shots on No. 2) that you’re going to count only the birdie.That makes you one under par after one hole, and leaves you with nineteen scores still to tally before the end of the round. Let’s also say that your opponents, on that first hole, both make net pars. They have to decide whether to count one or both—or to ignore them in order to leave plenty of room for birdies and eagles later on. The truly tough decisions come toward the end of the round, as opportunities dwindle. You don’t want to arrive on seventeenth tee knowing that, no matter what, you and your partner will have to count all four of your scores on the final two holes.

The game can be played with any number of players, in any combination. If you have four-man teams, each team counts forty scores; if you are playing as individuals, each player counts ten. If you have a threesome versus a foursome, the threesome counts thirty balls, then divides by three, while the foursome counts forty balls, then divides by four.

My friends and I reckon all hole scores only in relation to par, so that a five on a par 5 is the same as a three on a par 3. But you could make the game both more interesting and more complicated by counting absolute scores instead.

I love any game that can help eliminate my blow-up holes. Playing a Peoria Handicap event on Friday, so I just hope my blow-up holes are on the lucky 6.

We play a close variation of that game with some of my regular gang. One thing we do to make it a touch more interesting is we have a “landmine bet” or “worst bet” in addition to the 20 ball bet. Usually it is half of the value of the 20 ball bet. To calculate the landmine bet you take a predetermined number of worst scores and add them together. The only time we use this feature is when we are playing a round where pace of play doesn’t really matter (golf trips where you are going to have endure long waits) and when we are playing with a group of folks who are all reasonably low handicappers (we made the mistake of playing landmine with a 26 handicapper only once—forcing him to play out every shot was complete torture—especially since he was really probably a +40).

Our lingo is “Best and Worst”. We would set the bet up by saying do you want to play 25 & 5 for $20 bucks. Its implied we would be using 25 best scores & 5 worst scores and the bet would be $20 bucks for the best and $10 for the worst (generally, if one person is the contributor of all the worst scores they have to pay everyone on the team’s landmine bets– similarly, anyone who manages to get all 18 of their scores on the card sweeps the winnings–I’m not sure this has happened more than 2 times ever in our game– if none of your scores gets counted you are excluded from the winnings– *see below how this could happen). The landmine game is great when you are trying to improve by eliminating your blow-up holes. My favorite variation is to play 25 & 2. I find it really keeps me focused on keeping doubles or worse off the card. When it is only 2 landmines—you get the question… “Who got hit with the landmine? How did you make a triple on #3?” Since I hate rehashing my worst holes and never want to have to pay for the whole team, it keeps me dialed in after bad shots.

*We don’t specify the number of holes per player so if you pick a 25/2 game and have 4 players per team with 48 scores –ties count as contributing so it is possible but unlikely that someone might not contribute.

Thanks for the idea for a new game. Always looking for something different. We put a small twist on it. The 18th hole has got to be included, basically making the game 9-Ball plus 18. Gives the final hole more of a chance to mean something in the match for all players.

Great idea. We will adopt this immediately.

Quick question. If playing in teams, does each team member have to count 10 holes? Or could one person count, for example, 13 holes while the partner counts only 7?

You could make each team member count ten, but we don’t. Theoretically, all the balls could come from one player, although that doesn’t happen. One adjustment we’ve made recently: on the eighteenth hole, every ball has to count. That means that if you’re playing 20-Ball you need to count 18 balls over the first seventeen holes, because you know that you’re going to have to count two on the last. The amazing thing to me is that, almost every time we’ve played this game, the match has come down not just to the last hole but to the last putt.

We played today with three four-man teams, each of which counted twenty balls for the first nine and twenty for the second. Rick had to leave after nine holes, so my team was down to three guys. The only adjustment we could think of was to let us count fifteen balls for the second nine, instead of twenty. We came in last, but it seemed to work OK.

Is it possible to play foursomes against threesomes?

Yes. The 3-some should count 15 balls each side rather than 20. No adjustment is necessary. There is a falsehood that the 3-some score should be adjusted by multiplying it by 4/3rds, but that is completely incorrect. That would only work if the 3-some was required to also count 20 balls each side.

Take the total threesome’s 30 scores, divide by three and multiply that times 4.

This would equate to 40 scores score.

This answer is completely incorrect. The threesome would have to take 20 scores per side, not 15 per side, if you intend to apply this adjustment.

We played this game with two threesomes and one foursome using net scores. Which turned out -10, -11, and -12, however when we divided the threesome’s net score (103) by three they ended up with the lowest score even though their net score placed them in third place, and the foursome, which was -12 ended up with the highest score when we did the math. Something doesn’t sound right when you do the dividing in addition to making the foursome score 10 more balls which I initially thought would level the game between threesomes and foursomes. Tell me where I went wrong. Thanks,

I agree that something is a little screwy. The solution we have adopted is to play it only when we have teams with the same number of guys. The solution we use in our regular Sunday Morning games–allowing threesomes to start off -2–works well with regular best-ball but doesn’t work with 20-ball. I don’t know what the best solution is. Let me know if you come up with something.

I thought about this for a day so see if this approach works. If groups are equal then playing low net (-11 or -12) score works. However, if you use the total # of strokes then the team that appears to win based on low net may not win when counting total strokes. Why? because teams may choose more par 3s than par 5s which could throw off the scoring which is why low net works better, although I could probably make the argument that part of the strategy of trying to come in with the lowest total number of strokes forces you to concentrate on trying to maximize the par threes and only take one from each par 5. When you have threesomes vs foursomes merely making the foursome play the additional 10 balls doesn’t level the playing field. It does in terms of the % of balls; however the foursome has 10 more balls in play which gives them more chances to have a lower net score. So using some of the advice from above-dividing the threesomes net score by 3 and THEN multiplying by 4 would equate to adding the fourth player and that helps for comparison. Another way to calculate is to take their total stroke count and divide by either 30 or forty and you’ll come up with a number that can be used to differentiate the winning team. The challenge is that it isn’t as straight forward as using the low net number to ID the winning team and players think that there is some “pencil whipping” going on to figure out the winner. Let me know if this makes sense or I’m just making this harder and more confusing. thanks,

I’m not a hundred percent sure I understand–but I should say first that we count balls only in relation to par, so that a net par on a par 3 counts the same as a net par on a par 5, etc.–it’s just a par, or a zero. We played 11-ball today with five guys (the eleventh ball was the score on the eighteenth, which everyone had to count), and ended with two of us tied for the win at plus-4. In my case, that was seven pars and four bogeys; in Tim’s, it was an eagle, five pars, and six bogeys.

This is a great discussion and one that we have had many times when playing with 3 and 4-somes. We have each team take 10 x the # of players (ie. 30 or 40 balls). To have one match for all teams, we have ended up with taking the net score relative to par for each team and divide it into the number of players in that group, to arrive at the lowest score “per person”. The team with the lowest score per person wins.

Example:

3-some (30 balls) scores 9-under – divide by 3 – final score is -3.0

4-some (40 balls) scores 13-under – divide by 4 – final score is -3.25

4-some wins as they have the lowest net score to par per person.

Just bring a calculator and see if you can do the math after a few beers :)!

I have been lobbying my group here to adopt dividing by the number of persons on each team, when we have 3-somes and 4-somes, and while it works well when everyone is under par, I don’t see how the math works out when teams are over par. Can someone explain?

A 3 man team at +6 would be +2 per player. A 4 team at +12 would be +3 per player. 3 man team wins!