Of horseshoes and probability

[Just Plain Old]

Hello all, thank you in advance for your help!
Probabilty was never a strong subject for me and it was over 30 years ago so here I am looking for help.

Recently I have started playing horseshoes, the object of which is to throw a horseshoe at a metal pin 40 feet away and get it to hook on - a ringer.

Now my experimental probabilty over a long term (200 attempts) is roughly 5%. OK, I am not good -yet.

When I practice I will throw 4 shoes in each inning. I have observed, and logic dictates, that I am most likely to get, with greatly decreasing probability, the following:

1 - no ringers and none close
2 - no ringers but close
3 - 1 ringer and none close
4- 1 ringer and some close
5 - 2 ringers and none close


And there it ends. I rarely get 2 ringers and some close and have never thrown 3 of the 4 for ringers.

So my question: Given my 5% chance of throwing a ringer and throwing 4 shoes per inning, how many innings, in theory, would I need to throw to get 3 of 4 as ringers?

I know this is more than a simple probabilty question and that is where my old brain fails me.

Clearly getting better would increase my chnces but still, how do I figure it?

Thanks for your thoughts and theories.

 

[stapel]

Recently I have started playing horseshoes....

My Granddad used to love playing horseshoes. He'd often play when he and Gramma were at the lake during the summer (they had a trailer home).

Now my experimental probabilty over a long term (200 attempts) is roughly 5%. OK, I am not good -yet.

When I practice I will throw 4 shoes in each inning. I have observed, and logic dictates, that I am most likely to get, with greatly decreasing probability, the following:

1 - no ringers and none close
2 - no ringers but close
3 - 1 ringer and none close
4- 1 ringer and some close
5 - 2 ringers and none close

And there it ends. I rarely get 2 ringers and some close and have never thrown 3 of the 4 for ringers.

So my question: Given my 5% chance of throwing a ringer and throwing 4 shoes per inning, how many innings, in theory, would I need to throw to get 3 of 4 as ringers?

That will depend upon additional information. If you're assuming that your future results will match the current results, then possibly you will never do what you've never done. But are you allowing for improvement over time? And, if so, at what rate have you been improving so far?

Thank you!

 

[Just Plain Old]

More thoughts:

If I look at each throw of the shoe as a single isolated event then it stands to reason that I have a 1/20 chance of getting the ringer with each throw. Right? Of course I am looking to consider the events that occur in groups of 4. with no overlap between groups of course.

Naturally my getting better would greatly improve my chances on any given event occurring. With time I will improve. However for the sake of simplicity let us assume that I remain at a measly 5% accuracy.

It might be that there is no real way to calculate / predict what i want but the clear pattern that I see for the events that have occurred indicate that while my talent may be poor and somewhat random it does appear to be predictable.

Gary Smith at Pomona College did some interesting research on the predictability of certain patterns in horseshoes that show indicators for streaks, both up and down, but never really touched on what I am looking for.

Just as random information - a world class player may shoot 75% on a good day... I've got a way to go !

Thanks

 

[jmpehrson]

The binomial distribution could be applied

Assuming a rate of success of 0.05 (a ringer) per throw, this implies a failure rate of 0.95 (for this example, a near miss is a failure). Assume the rate of success is constant (neither improves or declines throughout the 4 throws), the probability of getting a certain number X of ringers out of N throws, given a success rate (p) and failture rate (q) can be calculated using the binomial distribution P(x) = N!/(X!(N-X)! x P^(X) x Q^(N-X). So, probability of 0 ringers is P(0) = 0.81451, P(1) = 0.17148, P(2) = 0.01354, P(3) = 0.00048, P(4) = 0.00001. Good luck improving your game!