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Picture of sigmoid
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I guess all that training I learned from Clint was a waste?
Tap, rack, and go


________,_____________________________
Guns don't kill people - Alec Baldwin kills people.
 
Posts: 1309 | Location: Idaho | Registered: July 07, 2010Reply With QuoteReport This Post
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If your ball ammo is anything like the army's ball ammo you will get a lot of practice with a 320.


“So in war, the way is to avoid what is strong, and strike at what is weak.”
 
Posts: 11009 | Registered: October 14, 2004Reply With QuoteReport This Post
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Picture of Angel King
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Thank you all for your replies which I enjoy reading very much. Let's go back to the M&P for a second since it was brought up as an alternative here. I purchased my own and as I normally do with my duty guns, Kept track of how many rounds I put through it. One afternoon training behind my in-laws house in the desert, the gun had a catastrophic failure because the conical trigger return spring broke. At about the 2000 round count.

The M&P did not and does not in the current 2.0 version allow for a manual forward reset of the trigger. Apex tactical offered a part for the 1.0 which allowed for a manual forward reset so you could still shoot the gun one round at a time. They were very helpful and send me two trigger return springs of a modified version with hard cotton inside to obsorb vibration and extend life.

When the 2.0 came out, I contacted Apex and was informed they did not offer a forward trigger reset fix for the 2.0. The gentleman I spoke to said a broken trigger return spring would cause the 2.0 to be a paperweight.

I had lost confidence in my M&P 40 back then and had switched to a Glock 22 GEN4 which I could not shoot as well.

Although I like to M&P a lot, I just can't get myself to use it at work. To be fair, when I Google incidence of the trigger return springs breaking, most documents are back in the 2011 to 13 erra. I think they're probably just fine today, but based on my own personal experience I won't use one for work.


An eagle does not capture flies.
 
Posts: 1080 | Location: Between heaven and hell | Registered: March 24, 2004Reply With QuoteReport This Post
Freethinker
Picture of sigfreund
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quote:
Originally posted by hrcjon:
Your exact question about the probability of failure of your gun I don't think can be answered unless you have a direct connection to a deity.


That’s what I thought I was doing by appealing to statisticians. Wink

Thanks for your insights and discussion. I have not read it all closely, but will do so. Smile




6.4/93.6

“Wise men talk because they have something to say; fools, because they have to say something.”
— Plato
 
Posts: 47410 | Location: 10,150 Feet Above Sea Level in Colorado | Registered: April 04, 2002Reply With QuoteReport This Post
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I own a P320 Compact and a G17. In the past 25 years I've owned a dozen or more Glocks in various calibers. None ever failed in any major capacity; I did do mag spring replacements. My P320 Compact had to go back to Exeter for a out of spec extractor, and then a short time later for "the fix."

I really want to love the P320, but I just can't trust it yet. The Glock is so utterly proven, and keeps being proven, that as much as I love the P320's trigger, reset, and accuracy, it's just going to be a while before I can get comfortable with the system. It's just too new for me. In 2-3 years, when the P320 has been truly vetted, I'll most likely be on board.

Until then, I'm staying with Glock and PPQ.
 
Posts: 1094 | Location: Arkansas | Registered: September 25, 2008Reply With QuoteReport This Post
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Picture of maladat
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quote:
Originally posted by sigfreund:
Is there way to calculate the probability that after X number of rounds without a failure of some sort that the next shot will not result in a failure? I.e., if I fire 200 rounds without a failure, what is the probability that shot 201 will not result in a failure? And keep in mind that I’m not asking about sampling a large number of firearms as if they were light bulbs and I was trying to determine what their average life was. I’m referring to a single firearm and a specific number of shots fired. To be more specific, I have fired my 357/40 P320 exactly 2653 times without a malfunction. What is the probability that if I must defend myself with it tonight that I will have a malfunction with the first shot I fire? (The answer is obviously not 1 ÷ 2653 or 0.000377, because if it were that simple, the probability that there would be a malfunction after two shots were fired would be 1 ÷ 2, or 50%—or sort of like having a 50/50 chance of winning the lottery: I’ll either win or I won’t.)


I'll take a shot at this question. I'm not a statistician, but I am a computer scientist that has had to study a whole bunch of probability theory (which is the mathematical basis of statistics).

Note that all probabilities used here are given from 0 to 1 rather than from 0% to 100% (with 0 equivalent to 0%, and 1 equivalent to 100%) unless specifically noted.

First, it seems obvious that the per-shot failure probability of a given firearm is probably not constant - things like how clean the firearm is, parts breaking in, springs wearing out, etc., might affect it.

Given a large population of firearms and detailed data, we could model this, but we're talking about single firearms here, so on this front, we have to just give up and say we're going to keep the firearm well maintained and assume the per-shot failure rate is a constant. Let's call it "F."

Second, given a number of shots taken, "N," with no failures, we don't have good information about how frequently failures occur, so we can't estimate F directly.

What we CAN do, is estimate an upper bound value for F - an estimate we are pretty sure is larger than the real value for F.

Specifically, we are looking for a confidence interval for the value of F. What that means is that, given our observation (N shots fired without a failure), we believe with some probability "C" that F is less than some specific value.

There is an a fairly involved proof of the following that I won't reproduce here. If you want to read about it, look up the confidence interval of a binomial distribution with no successes.

It turns out that in cases like this, given 0 failures in N trials, if you want confidence C that F is less than some value F0, F0 is just the value of F that gives you a probability (1-C) of getting 0 failures in N trials (in general, this is not necessarily the case).

So, say, if you want to know with 95% confidence that F is less than some specific value F0, then F0 is the value of F which gives you a 5% chance of getting 0 failures. The reason the probabilities flip like that is essentially because you're saying "I might have gotten really lucky this time, so the failure probability might be worse than it looks." The more confident you want to be about your estimate of the worst possible failure probability, the more lucky you have to be worried about having gotten in your test.

The probability C of getting 0 failures in N trials is trivial to calculate:

(1-F)^N = 1-C

Now we can rearrange:

(1-F) = (1-C)^(1/N)

F = 1 - (1-C)^(1/N)

So, let's say we shot 200 rounds without a failure, and we want to be 95% sure about our estimate of the gun's failure probability.

F = 1 - (1-0.95)^(1/200) = 0.0149, or 1.49%.

So, basically, shooting 200 rounds without a failure lets us be 95% sure that the gun's per-shot failure rate is less than 1.5%, but beyond that, we can't really say anything. It could be 1 in 100 (1%) or 1 in 10,000 (0.01%) and we have no idea.

There's an approximation you can use that reduces this formula to:

F ~ -ln(1-C)/N.

Using the same numbers from above, we get:

F ~ -ln(1-.95)/200 = 2.996/200 = 0.015, or 1.5%.

This leads to the so-called "rule of three" - for a 95% confidence interval, you can just use F = 3/N. There's an article on Wikipedia about the rule of three. https://en.wikipedia.org/wiki/...of_three_(statistics)

You can generalize the "rule of three" to other confidence values. For 99% confidence, -ln(1-.99) = 4.61, so you can use F = 4.61/N. For 99.9% confidence, -ln(1-.999) = 6.9, so you can use F = 6.9/N. And so on.

So with 99.9% confidence, shooting 200 rounds without a failure means your per-shot failure rate is no worse than 6.9/200 = 0.0345, or 3.45%.

Being 99.9% sure instead of 95% sure makes the maximum failure rate worse because with 95% confidence, you're only worried that you might have gotten really lucky when you shot 200 rounds without a failure. To be 99.9% sure, you have to be worried that you might have gotten really, really, REALLY lucky when you shot 200 rounds without a failure.

This sort of analysis is pretty common, e.g., in medical studies - if they test a new drug on 1000 people and have no serious adverse reactions, they still want to be pretty confident about the maximum rate of serious adverse reactions.
 
Posts: 6319 | Location: CA | Registered: January 24, 2011Reply With QuoteReport This Post
Freethinker
Picture of sigfreund
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maladat, thank you very much for that discussion. After my post above that got me to thinking about the question again, it occurred to me that what I was interested in wasn’t exactly some probability of an event’s happening, but what you described as confidence. I didn’t perceive it just as you described, but that’s really what I was seeking. It may be that the other people I asked my question of didn’t understand what I was getting at, so thanks for your insight.

I also understand that when evaluating something like the reliability of a firearm, if anything changes from the original conditions, such as using different ammunition, that can affect the conclusions, but at least now I have at least a vague idea of a starting point.




6.4/93.6

“Wise men talk because they have something to say; fools, because they have to say something.”
— Plato
 
Posts: 47410 | Location: 10,150 Feet Above Sea Level in Colorado | Registered: April 04, 2002Reply With QuoteReport This Post
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Picture of sigmoid
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quote:
Originally posted by jbourneidentity:
I own a P320 Compact and a G17. In the past 25 years I've owned a dozen or more Glocks in various calibers. None ever failed in any major capacity; I did do mag spring replacements. My P320 Compact had to go back to Exeter for a out of spec extractor, and then a short time later for "the fix."

I really want to love the P320, but I just can't trust it yet. The Glock is so utterly proven, and keeps being proven, that as much as I love the P320's trigger, reset, and accuracy, it's just going to be a while before I can get comfortable with the system. It's just too new for me. In 2-3 years, when the P320 has been truly vetted, I'll most likely be on board.

Until then, I'm staying with Glock and PPQ.


I ran a G19 about 15 years ago, after a few hundred rounds of factory ammmo, during a training session, the recoil rod came shooting out in all its glory
Nope, never had that kind of failure in all 8 of my Sig’s
You can keep the Glocks


________,_____________________________
Guns don't kill people - Alec Baldwin kills people.
 
Posts: 1309 | Location: Idaho | Registered: July 07, 2010Reply With QuoteReport This Post
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Picture of Angel King
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So today I went to a local law enforcement gunshop and handled a few 320's. Here is my initial opinion:

The best part of the 320 is #1 The grip ergonomics and #2 looks if the slide. The trigger was short and crisp, but heavy.

The aspects I did not care for are: It felt a bit 'Fragile' compared to the robustness of a Glock. For example, looking at the striker assembly underneath of a removed slide, the parts appeared less than overbuilt. I also did not care for the fact that the slide moved noticeably when the striker was released.

I would love to have one for target, but not work. The 320X was pretty cool.

I did however handle a G17 Gen5. I had been under the impression there were not enough changes to go away from the proven Gen4. That trigger was amazing compared to a stock Gen4 which is terrible. The grip felt better too without the finger grooves. Mostly, it felt rugged.

I was very surprized a G5 17 was my top pick.

The above is just my impressions of the 320 and not intended to offend anyone. It would be my second choice. Thank you for all your time and replies.


An eagle does not capture flies.
 
Posts: 1080 | Location: Between heaven and hell | Registered: March 24, 2004Reply With QuoteReport This Post
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Are the VTAC 320’s raising any eyebrows? Got fondle one for a while at my LGS, and I was intrigued. Good ergo’s, looked nice, and a nice trigger, but it’d be nice to shoot one first.
 
Posts: 198 | Location: SE Louisiana  | Registered: August 29, 2016Reply With QuoteReport This Post
Told cops where to go for over 29 years…
Picture of 911Boss
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I was a Glock fan for well over 20 years, had as many as 5 Glocks at a time. After picking up a VP9 and P320, I started selling them off and now I haven’t had a Glock for going on two years.

Nothing against them, but I prefer the P320 and have no worries about it’s reliability or dependability.






What part of "...Shall not be infringed" don't you understand???


 
Posts: 10947 | Location: Western WA state for just a few more years... | Registered: February 17, 2006Reply With QuoteReport This Post
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posted Hide Post
quote:
Originally posted by maladat:
quote:
Originally posted by sigfreund:
Is there way to calculate the probability that after X number of rounds without a failure of some sort that the next shot will not result in a failure? I.e., if I fire 200 rounds without a failure, what is the probability that shot 201 will not result in a failure? And keep in mind that I’m not asking about sampling a large number of firearms as if they were light bulbs and I was trying to determine what their average life was. I’m referring to a single firearm and a specific number of shots fired. To be more specific, I have fired my 357/40 P320 exactly 2653 times without a malfunction. What is the probability that if I must defend myself with it tonight that I will have a malfunction with the first shot I fire? (The answer is obviously not 1 ÷ 2653 or 0.000377, because if it were that simple, the probability that there would be a malfunction after two shots were fired would be 1 ÷ 2, or 50%—or sort of like having a 50/50 chance of winning the lottery: I’ll either win or I won’t.)


I'll take a shot at this question. I'm not a statistician, but I am a computer scientist that has had to study a whole bunch of probability theory (which is the mathematical basis of statistics).

Note that all probabilities used here are given from 0 to 1 rather than from 0% to 100% (with 0 equivalent to 0%, and 1 equivalent to 100%) unless specifically noted.

First, it seems obvious that the per-shot failure probability of a given firearm is probably not constant - things like how clean the firearm is, parts breaking in, springs wearing out, etc., might affect it.

Given a large population of firearms and detailed data, we could model this, but we're talking about single firearms here, so on this front, we have to just give up and say we're going to keep the firearm well maintained and assume the per-shot failure rate is a constant. Let's call it "F."

Second, given a number of shots taken, "N," with no failures, we don't have good information about how frequently failures occur, so we can't estimate F directly.

What we CAN do, is estimate an upper bound value for F - an estimate we are pretty sure is larger than the real value for F.

Specifically, we are looking for a confidence interval for the value of F. What that means is that, given our observation (N shots fired without a failure), we believe with some probability "C" that F is less than some specific value.

There is an a fairly involved proof of the following that I won't reproduce here. If you want to read about it, look up the confidence interval of a binomial distribution with no successes.

It turns out that in cases like this, given 0 failures in N trials, if you want confidence C that F is less than some value F0, F0 is just the value of F that gives you a probability (1-C) of getting 0 failures in N trials (in general, this is not necessarily the case).

So, say, if you want to know with 95% confidence that F is less than some specific value F0, then F0 is the value of F which gives you a 5% chance of getting 0 failures. The reason the probabilities flip like that is essentially because you're saying "I might have gotten really lucky this time, so the failure probability might be worse than it looks." The more confident you want to be about your estimate of the worst possible failure probability, the more lucky you have to be worried about having gotten in your test.

The probability C of getting 0 failures in N trials is trivial to calculate:

(1-F)^N = 1-C

Now we can rearrange:

(1-F) = (1-C)^(1/N)

F = 1 - (1-C)^(1/N)

So, let's say we shot 200 rounds without a failure, and we want to be 95% sure about our estimate of the gun's failure probability.

F = 1 - (1-0.95)^(1/200) = 0.0149, or 1.49%.

So, basically, shooting 200 rounds without a failure lets us be 95% sure that the gun's per-shot failure rate is less than 1.5%, but beyond that, we can't really say anything. It could be 1 in 100 (1%) or 1 in 10,000 (0.01%) and we have no idea.

There's an approximation you can use that reduces this formula to:

F ~ -ln(1-C)/N.

Using the same numbers from above, we get:

F ~ -ln(1-.95)/200 = 2.996/200 = 0.015, or 1.5%.

This leads to the so-called "rule of three" - for a 95% confidence interval, you can just use F = 3/N. There's an article on Wikipedia about the rule of three. https://en.wikipedia.org/wiki/...of_three_(statistics)

You can generalize the "rule of three" to other confidence values. For 99% confidence, -ln(1-.99) = 4.61, so you can use F = 4.61/N. For 99.9% confidence, -ln(1-.999) = 6.9, so you can use F = 6.9/N. And so on.

So with 99.9% confidence, shooting 200 rounds without a failure means your per-shot failure rate is no worse than 6.9/200 = 0.0345, or 3.45%.

Being 99.9% sure instead of 95% sure makes the maximum failure rate worse because with 95% confidence, you're only worried that you might have gotten really lucky when you shot 200 rounds without a failure. To be 99.9% sure, you have to be worried that you might have gotten really, really, REALLY lucky when you shot 200 rounds without a failure.

This sort of analysis is pretty common, e.g., in medical studies - if they test a new drug on 1000 people and have no serious adverse reactions, they still want to be pretty confident about the maximum rate of serious adverse reactions.


Absolutely excellent response. I wish everyone here could be as comprehensive and yet actually quite concise and precise with their answers. Thanks again.
 
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