Hello peoples,
The way golfing handicaps are determined a golfer is expected to equal or better his/her handicap roughly 1 in every 5 rounds. To my mind this assigns a 20% probability of achieving this feat each round. In a golfing group of 4 people, each playing individually, what is the probability that someone from the group will shoot to their handicap or better?
I’m quite sure it is not cumulative i.e 4/5 or 80%. Could you show the correct workings.
Cheers.
The chance of it not happening is 0.8^4 = 0.4096, so the chance of it happening is 1-0.4096 = 0.5904
If events X and Y are independent, then P(X and Y) = P(X) P(Y). (Here P(E) = probability of E occurring.) Also, for any event X, P(not X) = 1 - P(X). The probability you’re asking for is therefore (sorry about the bad markup here):
P(golfer 1 succeeds or … or golfer 4 succeeds)
= P(not (golfer 1 fails and…and golfer 4 fails)
= 1 - P(golfer 1 fails and…and golfer 4 fails)
= 1 - P(golfer 1 fails)…P(golfer 4 fails)
= 1 - (4/5)^4
= 0.5904.
Thanks for that.
Could it also be done by working out the combinations that will allow it to happen, figuring out the probability of each of those combinations happening and summing those probabilities??
Yes; in a sense, that’s what the binomial probability distribution formula does. In this case, n=4 (the 4 people), p=.20 (the 1/5 chance for each person to shoot their handicap); and if you want the probability that one or more of them does so, you’d have to add up the probabilities that the formula gives you for r=1, 2, 3, and 4.
And in this instance it happens to be the case that only one combination doesn’t allow it to happen, so it saves leg-work to calculate that probability and subtract it from 1.
…which doesn’t seem like a big deal if there are four people in the round, but if there’s a tournament with 50, that’s a lot of leg-work.
If there are 100 contestants, and they each have a 20% probability of making their handicap, then surely the same percentage would apply to the whole group. So, on average, 20 of them should, and 80 of them shouldn’t.
Yes, and this will tend to apply with increasing accuracy as the sample size grows. But if you want to know what’s the probability of a specific number making it - such as, say, exactly 17 out of 100 - then there’s some working-out to be done. Obviously 20’s the likeliest number, but 19 or 21 should be nearly as likely, and so on.
Even though 20 is the most likely number, it’s far more likely that the result won’t be 20. That’s why there are other calculations to do.