I read in a paper that the chance of getting a hole in one are 12,500 to 1. The article continued that to get 4 holes in one is (1/125000) to the power of four.

Why does the chance decrease per attempt? The universe doesn’t know you’ve had one or two or three results already so why are increasing holes in one less likely?

I see that intuitively this might be the case, but I’m not fully convinced. Part of me believes that the chance of getting any number of holes in on should be 12,500 to 1.

In: 1

The chance you’re getting a heads in a coin flip is 1/2. The chance you’re getting a heads in a SECOND coin flip is ALSO 1/2.

But the chance of getting heads in both flips is 1/4.

This is because with one coin flip, there are only two outcomes: heads and tails.

With BOTH coin flips, there are four outcomes:

Heads/heads

Heads/tails

Tails/heads

Tails/tails

Think of it like a combination lock. If the lock will open if you select the correct number out of 125,000 numbers, you have a 1/125,000 chance of guessing it. If you needed to select two separate numbers as the combination, it would be 1/(125,000)^2.

For each digit you picked for the first number, you would have to go through every digit for the second number.

1/1, 1/2, 1/3….3452/1, 3452/2…

Your chance of getting a hole in one is 1/12500

Your chance of getting a second hole in one on the next hole is also 1/12500 because the two events are independent.

But the odds of both happening in a row are 1/12500 *times* 1/12500 since you’re asking about the odds of it happening twice in a row.

Four times in a row is (1/12500)^4 – a very small number that only the world’s greatest golfer (Kim Jong-Il) could ever hope to get.

Now getting 4/18 in a typical round of golf is a slightly more complex equation than 4 in a row, since you’re allowed multiple different permutations of 4/18. It’s been too long since my last stats class to work that one out.

Because it’s a series. You cannot get 2 holes in one without first getting a hole in one on the first hole. And can’t get 3 without first getting the first two. The dependency changes the odds. The odds of getting one on any given hole is the same, but the odds of doing it on the holes consecutively is the more difficult aspect.

Probability and statistics are fun. It’s honestly the coolest math. And you’re right OP, every single attempt at a hole in one is a separate event, with its own probability of happening regardless of any other events. So if you sink the very first hole, the second hole your chance will still be 1/125000 regardless of the first attempt. And it goes on like that. If you take each hole as a separate instance, your chance will always be the same. But if you are looking at the set of holes, and your chance of sinking any amount of them above 1, then your odds go way down, essentially 1:125000^n n being the number of holes you want to get in one.

To reiterate, each event has the same odds, but calculating the probability of multiple events is as you listed, which is 1:125000^4. As you complete holes in one, your odds aren’t going up or down. It’s just that the entire set of events happening have the overarching odds of 1:125000^4.

>Why does the chance decrease per attempt?

* Because there is only one way to succeed (everyone gets a hole-in-one) but there are lots of ways you can fail (any single person misses their hole-in-one, then they whole things fails).

* If you only consider a single hole-in-one, how many changes are for it to fail?

* Just 1.

* But if you consider three, let’s list all the ways it can go wrong.

* 1 – All three golfers fail

* 2 – First golfer gets it but second, and third fail.

* 3 – First golfer gets it, second golfer gets it, but third fails.

* 4 -First golfer gets it, second fails, but third gets it.

* 5 -First golfer fails, second fails, but third gets it.

* 6 – First golfer fails, second gets it, third gets it.

* And how many ways can we succeed?

* All three get it

* So just one.

* So you can see for a single hole-in-one, it’s 1 chance to succeed and 1 chance to fail, so it’s 50/50

* But with three holes in a row, it’s 1 chance to succeed and 6 to fail.

I think you get the general concept but you’re misunderstanding what is being calculated.

>Why does the chance decrease per attempt?

It doesn’t. Each individual attempt has the same chance of being a hole in one. Your shot on the first hole has a 1 in 12,500 chance of going in. Your first shot on the second hole also has a 1 in 12,500 chance of going in, regardless of what happened on the first hole. The first shot of the third and fourth holes are also a 1 in 12,500 chance each.

The figure being calculated (the fraction to the 4th power) represents the odds of you getting a hole in one on the first hole AND getting a hole in one on the second hole AND getting a hole in one on the third hole AND getting a hole in one on the 4th hole. Just getting a hole in one on the fourth hole is your standard 1 in 12,500 odds, but the chances of getting 4 holes in one in 4 attempts are lower than any individual attempt.

The chance per attempt doesn’t decrease, they stay independent events and this is the chance of a single hole-in-one.

If you want to know chance of 4 hole-in-ones, they all 4 must happen. And the chance of multiple things happening is always less as a single event.

If doesn’t matter if it is the same event multiple times, or multiple events.

There is a chance of your phone ringing this hour. There is a chance of a mail delivery tomorrow.

The chance of a mail delivery AND your phone rining is always less than the individual events. while the chance of a mail delivery OR your phone ringing are always higher than the individual events.

I’ll answer this from an intuitive perspective rather than mathematical.

Let’s say you’re at a party with a bunch of strangers. You get drunk and you begin betting people that if they could guess your name correctly by whispering it in your ear, that you’ll give them ten dollars. It’s against the rules for others to share your name with anyone else.

In one scenario, the first person guesses correctly but all subsequent guesses are wrong. One would be reasonable to assume the first attempt was a lucky guess.

In the next scenario, all guesses are correct. One would be reasonable to assume that the participants were cheating.

You could say that any given guess would be independent of one another, but intuitively the pattern of multiple correct guesses might indicate unfair play.

From a mathematically minded perspective, we can look at this a different way. Basically, successive attempts at something increases the total number of outcomes, therefore decreasing the chance of any specific outcome. For example, if you flip a coin once , you can either get heads or tails. Flip it twice? Well now you can have h/h, h/t, t/h, or t/t. There is only one outcome out of four that you have heads twice in a row.

Applying this your example probability, only one outcome out of 12500 is a hole in one. When attempting four? Well, that’s only one outcome out of 12500*12500*12500*12500. this is because we expand the set of possible outcomes, all of which except one include a failure to achieve a hole in one.

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