SwingsitlikeHogan
Major Champion
Out one evening by myself playing two balls I eagled our 551 yd par 5 14th hole with both balls. Probability?
Probability question for Maths experts
- Thread starter 3offTheTee
- Start date
Out one evening by myself playing two balls I eagled our 551 yd par 5 14th hole with both balls. Probability?
3offTheTee
Tour Rookie
OK Idid contact a friend with the information I posted originally who is a bit of a maths guru and he did not come up with 0.73 nor 42! His reply is partly quoted below:
What we need to know is
- how the hole was playing ie of the last 1000 players (or more), what was the average net score, after adjusting for handicap.
- what the standard deviation (or variance) was for each player.
We would have to make assumptions
- that weather conditions etc did not have any impact
- standard deviation assumes a normal distribution, ie an even bell curve. This not the case as the times a player will score an 3 above net handicap is infinitely bigger than the chances of him having a net -3on a hole. The distribution in reality would have a long tail.
- that the standard deviation is the same for all handicap players. Again, this would not be the case, higher handicap players are going to be erratic - ie a 36 handicap player will have more net eagles (gross pars) than a scratch player. likewise he will have more net double bogeys.
- there was no matchplay pressure, hangovers, tiredness, bad backs.......
Without knowing any of the above, let's assume (I am inventing)
handicap 11 has a 30% chance of getting a 4 (50% of getting a 5, 15% of a 6 and 5% of 7 or more).
handicap 12 a 25% of getting a 4......
16 a 15%...
21 a 10%
We'd then multiply these percentages together to get the probability so 1 in 889 with my invented figures.
What we need to know is
- how the hole was playing ie of the last 1000 players (or more), what was the average net score, after adjusting for handicap.
- what the standard deviation (or variance) was for each player.
We would have to make assumptions
- that weather conditions etc did not have any impact
- standard deviation assumes a normal distribution, ie an even bell curve. This not the case as the times a player will score an 3 above net handicap is infinitely bigger than the chances of him having a net -3on a hole. The distribution in reality would have a long tail.
- that the standard deviation is the same for all handicap players. Again, this would not be the case, higher handicap players are going to be erratic - ie a 36 handicap player will have more net eagles (gross pars) than a scratch player. likewise he will have more net double bogeys.
- there was no matchplay pressure, hangovers, tiredness, bad backs.......
Without knowing any of the above, let's assume (I am inventing)
handicap 11 has a 30% chance of getting a 4 (50% of getting a 5, 15% of a 6 and 5% of 7 or more).
handicap 12 a 25% of getting a 4......
16 a 15%...
21 a 10%
We'd then multiply these percentages together to get the probability so 1 in 889 with my invented figures.
chrisd
Major Champion
Unlikely ðŸ˜
Grant85
Head Pro
I made similar assumptions to get to my estimate of 2000/1.
My thinking was that the stroke index 3 is probably a long par 4 and so a 21 handicapper (and the others for that matter) are less likely to make par than your estimates.
Sure an 11 or 12 handicap may make par 25% to 30% of the time, but less on the 3 and 4 hardest holes. Especially if they are longer holes, which typically add disproportionate levels of difficulty for handicap golfers.
shortgame
Tour Rookie
SwingsitlikeHogan
Major Champion
shortgame
Tour Rookie
ExRabbit
Club Champion
The actual answer is 50:50 - it either happened or it didn't happen.
Maths doesn't have to be hard!
Maths doesn't have to be hard!
SwingsitlikeHogan
Major Champion
50:50 only works if there are, or were, only two possible outcomes...and both outcomes are or were equally likely.
drdel
Tour Rookie
IF you really wanted to compute a probability here's some of the things to consider...
https://www.myscorecard.com/cgi-bin...file=coursehandicaps&article=Course-Handicaps
Personally I'd just run it on a simple discrete simulation with it set for each player to hit 250 balls and see the number of pars. You'd need some pretty global assumptions on the course index etc.
https://www.myscorecard.com/cgi-bin...file=coursehandicaps&article=Course-Handicaps
Personally I'd just run it on a simple discrete simulation with it set for each player to hit 250 balls and see the number of pars. You'd need some pretty global assumptions on the course index etc.
D
Deleted Member 1156
Guest
I don't know the probability but I would have cashed out after the 3rd guy made his par 
ExRabbit
Club Champion
I was joking!
It is a kind of internet meme for poker players regarding the chances of the outcome of a hand.
rksquire
Head Pro
Ranges from 16-1 (simple probability as in each guy either can or cannot par the hole (50/50)) to about 1,485 to 1 if you start to introduce some but not all variables!
