First Gun Confiscation Killing...

[Roy]

I'm pretty sure if someone had a 5%-20% chance of catching a disease every year, then that would accumulate so that the odds of them catching it at some point in their lifetime would indeed increase, but would never actually reach 100%. Maybe when Roy said 95%-100% chance he meant it exclusive of 100%.

No, just rounding up from 99.9998%

[Roy]

Or cancer. Maybe cancer would be a better example. Or getting killed in an auto accident. Or having a heart attack. Or being struck by lightning...

or winning the lottery.

None of which changes the fact that the number of mass shootings in the US is increasing. - or that this whole flu discussion is an attempt to avoid that fact.

P.S. You're wrong about lightning. More USans die in mass shootings than from lightning strikes.

[Roy]

Sorry Zym, but I'm going to disagree with you too. The chances of being exposed to the flu may increase every year but that still doesn't mean you contract it. I don't think it reaches anywhere near 95% even. That formula is too simplistic to define all the variables that make up the 5% - 20% yearly number. I'd be interested if anyone has the actual number, not one posited via a formula.

it seems like a silly argument to be having anyway.

It's an extremely silly argument. A non-mathematician, who not only hasn't done the calculation but doesn't know how, is trying to tell a mathematician who has done the calculation what the answer should be.

This whole thing is just another Roy nitpick to avoid the topic by attacking an analogy with minutia.

I'm not avoiding the topic at all.

A flu vaccine protects the person who gets it. If some-one doesn't get a flu vaccine, that won't remove the protection from some-one that does.
A gun confiscation protects other people. If some-one doesn't give up their guns that [u]does[/u[ remove the protection from some-one that does.
They are not comparable.

And again, flu was only introduced to avoid talking about the increased frequency of mass shootings.

As for "nit-picking," that's become your excuse to admitting you're wrong.

[Roy]

If
1) Each year 5-20% of USans get the flu.
And
2) You get the flu every year without a flu shot
Then
3) 80%-95% of Usans must be getting flu shots each year

But that's false, so one of your premises is wrong.

No, correlation does not equal causation...as I know you know.

I do, but the above doesn't involve either correlation or causation...as I now know you don't know.

[Roy]

Nope, because I'm not wrong in this instance. You saying it over and over doesn't make you right either. The equation you posted to support Roy's number stated that there was statistically 100% chance that you would get the flu over your lifetime. That's just wrong. Both of you are wrong. There's a 5% - 20% chance every year that you will contract the flu. That doesn't accumulate because of the years you live. You'll have to do more to convince me than just your say so.

Jim showed you the calculation. The reason you aren't convinced isn't because it's just his say so, it's because you think your inexperienced gut-feeling is more reliable than actually doing the calculation.

[Littlejoe]

I do, but the above doesn't involve either correlation or causation...as I now know you don't know.

Explain to us then how your example doesn't involve correlation.


noun

• a mutual relationship or connection between two or more things.

Everything looks connected to me. an IF, IF, THEN statements are inherently correlation. I agree by the way that your example's conclusion is correct. But, these do not represent my premises...so, hence my statement. You continue to burn a straw man IMO.

[Littlejoe]

Jim showed you the calculation. The reason you aren't convinced isn't because it's just his say so, it's because you think your inexperienced gut-feeling is more reliable than actually doing the calculation.

No, it's wrong because the world is not an enclosed environment. Your simply formula doesn't account for any variables. It might hold true is some fantasy world but not the real world.

[Littlejoe]

A flu vaccine protects the person who gets it. If some-one doesn't get a flu vaccine, that won't remove the protection from some-one that does.
A gun confiscation protects other people. If some-one doesn't give up their guns that [u]does[/u[ remove the protection from some-one that does.
They are not comparable.

No, that's where you make the mistake. If I don't give up my gun, no one will be harmed because of me. Unless they pull a gun and start shooting, in which case I would be their protection, so your conclusion is false. You're not going to get rid of all the guns in the US. Heck, even the UK hasn't managed that with...how many years of strict gun control? OR even Australia, where this year they had the worst mass shooting in that country since 1996. Gun bans and gun confiscations will not stop the illegal gun deaths that are the majority of the deaths in the US. And IF somehow you could take them all away, there's going to be an increase in beating deaths, knife deaths, etc. The problem isn't the guns. If it was, then it would have been more of an issue 30 years ago when guns were much easier to get.

[Sparko]

Guys - this is basic stuff. The probabiliy of an event stays the same, per event, yes. But the probability there will HAVE BEEN a hit goes up with each event.

So, just like if I toss a coin, the probability I'll have seen at least one heads is 50% on the first toss, 75% on the second, 87.5% on the third and so on. The way you calculate that is 1 - (probability of a miss)^{<number of tries>}

So if I take a single die with probabilty 1/6 that any of the 6 numbers pops up, then the pobability I get a 1 the first time is 1/6, the probability I will have seen a 1 after two tries is 1-(5/6)² and so on.

When I play the lottery it's the same. And it doesn't matter if I play the same set of numbers or a random set of numbers each time, though it's more intuitive to think that sooner or later my numbers will come up. So if I play the lottery 14,000,000 times with a probability of winning of 1/14000000, then the probability is about 63.2%* I will have won at least 1 time.


Jim

*btw, anybody recognize what number we are running into here with that 63.2%? Very famous. Not pi

I am not a mathematician - but I think you are wrong on the lottery thing. Each time would be the same odds. The current drawing is not related to the previous ones. You are not taking a second guess at the same number each week. You are trying to guess a new number each week. So each week you have a 1/14M chance.

I get the idea of if you are trying to get a heads, the more flips you make the better your chances. But what if rather than trying to get heads, you were trying to match my flips? I flip a coin and you flip trying to match it. Would your odds go up with every flip then? If so, explain it to me.

[oxmixmudd]

I am not a mathematician - but I think you are wrong on the lottery thing. Each time would be the same odds. The current drawing is not related to the previous ones. You are not taking a second guess at the same number each week. You are trying to guess a new number each week. So each week you have a 1/14M chance.

I get the idea of if you are trying to get a heads, the more flips you make the better your chances. But what if rather than trying to get heads, you were trying to match my flips? I flip a coin and you flip trying to match it. Would your odds go up with every flip then? If so, explain it to me.

It's the same because we are dealing with the probability of a single event over the number of events and determining the odds of at least one instance of the event. The chance of the match, each time, is 1/14M. That defines the odds of the match. What you are concerned about is how we get that initial probability of a hit, which is a bit more complicated in the case of selecting some subset of numbers from a larger set and defining that as a 'hit'.

But to clarify the problem when two elements vary: consider If you are tossing a coin, and I am tossing a coin, and we want to know the probability of a match. Then the odds of the match is still 1/2. We can both get heads. or we can both get tails. Same with two die. will they match? there are 6 possible matches and 36 possible combinations. 1/6, same as the toss of a single die. So in this case, adding in the fact we need a match doesn't change the probability of a hit at all.

Now we are not taking all possible numbers in the lottery, so it is different in that we have to factor in number of possible combinations taken 6 at a time or whatever the number of numbers vs the number of picks is - as the case may be. But again, this is the problem of determining the probability of a single hit, a single event. We can assume the 1/14M is the probability of the match calculated under those circumstances, otherwise it would be false advertising.

Once the probability of a match is established( a hit ), then the probability of a hit over a given number of tries is determined by the same equation:

1 - (probability of a miss)^{<number of tries>}


Jim

[lilpixieofterror]

Stop responding to LPOT? You've run away.
Keep playing whack-a-mole? You've got OCD.

[Sparko]

It's the same because we are dealing with the probability of a single event over the number of events and determining the odds of at least one instance of the event. The chance of the match, each time, is 1/14M. That defines the odds of the match. What you are concerned about is how we get that initial probability of a hit, which is a bit more complicated in the case of selecting some subset of numbers from a larger set and defining that as a 'hit'.

But to clarify the problem when two elements vary: consider If you are tossing a coin, and I am tossing a coin, and we want to know the probability of a match. Then the odds of the match is still 1/2. We can both get heads. or we can both get tails. Same with two die. will they match? there are 6 possible matches and 36 possible combinations. 1/6, same as the toss of a single die. So in this case, adding in the fact we need a match doesn't change the probability of a hit at all.

OK the odds are still 1/2 that they will match. But how do you get that the odds they will match is higher than that if you flip several times? Each time is a discrete event not related to the events before it. That is what I am not getting.

Now we are not taking all possible numbers in the lottery, so it is different in that we have to factor in number of possible combinations taken 6 at a time or whatever the number of numbers vs the number of picks is - as the case may be. But again, this is the problem of determining the probability of a single hit, a single event. We can assume the 1/14M is the probability of the match calculated under those circumstances, otherwise it would be false advertising.

Once the probability of a match is established( a hit ), then the probability of a hit over a given number of tries is determined by the same equation:

1 - (probability of a miss)^{<number of tries>}


Jim

So back to my other joke comment... if the odds of you NOT catching the flu each year is 80%, then why doesn't THAT add up and after a few years the odds of you NOT catching the flu become close to 100%?

[oxmixmudd]

OK the odds are still 1/2 that they will match. But how do you get that the odds they will match is higher than that if you flip several times? Each time is a discrete event not related to the events before it. That is what I am not getting.

Careful. We are not saying the odds change on the second flip. We are looking at the odds it matched on the first flip OR the second flip OR both flips. The probability that after two tries, you will have seen at least 1 match.

So back to my other joke comment... if the odds of you NOT catching the flu each year is 80%, then why doesn't THAT add up and after a few years the odds of you NOT catching the flu become close to 100%?

Because it's the complementary problem. That is, this problem is just the complement of the 'will catch it' problem, which means it is defined not by 1 - x, but x itself

{ remembering x = (probability of a miss)^{<number of tries>} }

Consider a die with 6 faces, so saying I won't get a one on the first toss is similar to saying you won't get the flu this year, and has a similar probability - 5/6. But what about not getting a 1 on either of two tosses? Both event probabilities are the same, but the odds of not getting a one on either toss is (5/6)², which is 25/36<5/6.

Can we validate this? Can I show I'm not just making this up? Sure. here are all the possible pairs of tosses.

1,1 1,2 1,3 1,4 1,5 1,6
2,1 2,2 2,3 2,4 2,5 2,6
3,1 3,2 3,3 3,4 3,5 3,6
4,1 4,2 4,3 4,4 4,5 4,6
5,1 5,2 5,3 5,4 5,5 5,6
6,1 6,2 6,3 6,4 6,5 6,6

So how many are there with 1's in them? 11. And 36-11 = 25. So there are 25 that don't have a 1. There you go - 25 out of 36 is your odds after 2 tries of not getting a 1. And that is what we get from the formula, 25/36 :).

So back to your scenario. If you have an 80% chance you won't catch the flu this year, and next, and the next, over your entire life, then the probability you will NEVER catch the flu (assuming you live 80 years) becomes .8⁸⁰, which is .000002%, or very UNlikely, which incidentally when added to the probability from the other formula you WILL get the flu for a 20% chance of catching it, is 1, as it should be.

Jim

[Sparko]

Careful. We are not saying the odds change on the second flip. We are looking at the odds it matched on the first flip OR the second flip OR both flips. The probability that after two tries, you will have seen at least 1 match.



Because it's the complementary problem. That is, this problem is just the complement of the 'will catch it' problem, which means it is defined not by 1 - x, but x itself

{ remembering x = (probability of a miss)^{<number of tries>} }

Consider a die with 6 faces, so saying I won't get a one on the first toss is similar to saying you won't get the flu this year, and has a similar probability - 5/6. But what about not getting a 1 on either of two tosses? Both event probabilities are the same, but the odds of not getting a one on either toss is (5/6)², which is 25/36<5/6.

Can we validate this? Can I show I'm not just making this up? Sure. here are all the possible pairs of tosses.

1,1 1,2 1,3 1,4 1,5 1,6
2,1 2,2 2,3 2,4 2,5 2,6
3,1 3,2 3,3 3,4 3,5 3,6
4,1 4,2 4,3 4,4 4,5 4,6
5,1 5,2 5,3 5,4 5,5 5,6
6,1 6,2 6,3 6,4 6,5 6,6

So how many are there with 1's in them? 11. And 36-11 = 25. So there are 25 that don't have a 1. There you go - 25 out of 36 is your odds after 2 tries of not getting a 1. And that is what we get from the formula, 25/36 :).

So back to your scenario. If you have an 80% chance you won't catch the flu this year, and next, and the next, over your entire life, then the probability you will NEVER catch the flu (assuming you live 80 years) becomes .8⁸⁰, which is .000002%, or very UNlikely, which incidentally when added to the probability from the other formula you WILL get the flu for a 20% chance of catching it, is 1, as it should be.

Jim

PS there are 12 1's.

[oxmixmudd]

PS there are 12 1's.

You are counting the event 1,1 twice*. That still counts as just one event, not two. We are looking for the events that include at least one 1. We don't care if there is an event with more than one 1. Translating to your 80% won't catch the flu case, the scenarios where you catch the flu several times over your lifetime still count just a single example of a lifetime where a person with an 80% chance of catching the flu caught it.


Jim

* each comma separated pair is 'an event' where an event is a possible sequence of 2 tosses of the die. So perhaps it would have been clearer to say "How many events have at least one 1 in them".