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On 31 Oct, 11:32, 1LK wrote:
On Oct 31, 5:36 am, Ian wrote: How old are you? A 1 in 80 chance of dying today means that you have a 79 in 80 chance of making it to tomorrow, which is a (79/80)^365 = 1% chance of making it through a year. I'll bet even 100 year olds have a better survival rate than that ... Ian Well under 100. Ian (although not as far under as I'd like). It's a multifactorial analysis and some of the factors that go into it are personal, but, given the assumptions, it's probably a decent characterization. The risk isn't additive, BTW; it remains about the same day to day as long as the factors used to calculate it are stable. So you really do only have a 1% chance of being here next year? What are you doing posting to Usenet - go flying. It's probably not worth buying a flarm, by the way. Or a parachute. Ian |
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On Oct 31, 8:09 am, Ian wrote:
On 31 Oct, 11:32, 1LK wrote: On Oct 31, 5:36 am, Ian wrote: How old are you? A 1 in 80 chance of dying today means that you have a 79 in 80 chance of making it to tomorrow, which is a (79/80)^365 = 1% chance of making it through a year. I'll bet even 100 year olds have a better survival rate than that ... Ian Well under 100. Ian (although not as far under as I'd like). It's a Not exactly. The odds ratio applies to any point in time; it's neither cumulative or additive. I have a 98.75% chance of being here tomorrow; on a day a year from now I'll have roughly the same odds of being here a day after that. I do wear a parachute, BTW, but again, given my age and condition, it's not certain that I could extract myself successfully from a lawn dart. I wear it because I'd rather be busy trying to use it, than sitting on my hands waiting to hit the ground. As you know, FLARM isn't usable here (US), but I'm planning to add a transponder and an ELT this year. I take aspirin as well. Not sure why I post to Usinet, perhaps the riskiest activity of all. I control what I can and accept the rest and, yes, I think I'll go flying. Ray Warshaw 1LK multifactorial analysis and some of the factors that go into it are personal, but, given the assumptions, it's probably a decent characterization. The risk isn't additive, BTW; it remains about the same day to day as long as the factors used to calculate it are stable. So you really do only have a 1% chance of being here next year? What are you doing posting to Usenet - go flying. It's probably not worth buying a flarm, by the way. Or a parachute. Ian- Hide quoted text - - Show quoted text - |
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On 31 Oct, 19:11, 1LK wrote:
Not exactly. The odds ratio applies to any point in time; it's neither cumulative or additive. I have a 98.75% chance of being here tomorrow; on a day a year from now I'll have roughly the same odds of being here a day after that. If you have a 98.75% chance of being here tomorrow, then you have a 98.75% x 98.75% chance of being here a day after that, a 98.75% x 98.75% x 98.75% chance of seeing Sunday and so on. Sure, /if/ you make it to 1st November 2008 you have a 98.75% chance of making it to 2nd ... but there's only a 1% chance, on these figures, that you'll put it to the test. Ian |
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The calculation which yields the 80/1 is only true for the single
instance; my odds of being here next year are another thing entirely. For that you need to do mortality computations. Ray |
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On 1 Nov, 14:07, 1LK wrote:
The calculation which yields the 80/1 is only true for the single instance; my odds of being here next year are another thing entirely. For that you need to do mortality computations. I'm having real difficulty understanding this. Are you saying that ... 1) There is a 1/80 chance you'll be dead by this time tomorrow and that 2) There is a 1/80 chance you'll be dead by this time the day after? Ian |
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On Oct 31, 7:11 pm, 1LK wrote:
Not exactly. The odds ratio applies to any point in time; it's neither cumulative or additive. I have a 98.75% chance of being here tomorrow; on a day a year from now I'll have roughly the same odds of being here a day after that. Think of it like tossing a coin to see whether you live. The probability of each toss being "heads" is 0.5, and is independent of the previous results. To stay alive for a week, you have to toss "heads" 7 times in a row, and the probability of that is 0.5 ^ 7 = 0.078125 = 1 in 128 Personally, I hope you have more than a 0.9875 ^ 365 chance of being alive in a year. (i.e. 0.01014 or 1 in 98). Of course, if you are still alive in a year, there would be the same chance (0.9875) that you would be alive in a year and a day. |
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results.
To stay alive for a week, you have to toss "heads" 7 times in a row, and the probability of that is 0.5 ^ 7 = 0.078125 = 1 in 128 It's not binary, it's multifactorial. Ray |
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On Nov 1, 2:11 pm, 1LK wrote:
The calculation which yields the 80/1 is only true for the single instance; Single instance of what? If it is the "single instance of a day", then the calculations are correct. my odds of being here next year are another thing entirely. For that you need to do mortality computations. To stay alive for a week, you have to toss "heads" 7 times in a row, and the probability of that is 0.5 ^ 7 = 0.078125 = 1 in 128 It's not binary, it's multifactorial. I don't understand: what do you mean by "it" and "multifactorial"? Binary? Well yes, flipping a coin is binary; that's why I subsequently used your figures (that you didn't bother to include). It might help if you could explain the reasons (based on an equivalent example, if you prefer) why you believe that the calculations are wrong. Examples I can think of are - it is not a 1.25% chance of dying on every day, only on some days - each day shouldn't be treated as independent from the preceding days (but that doesn't fit with your original statement) Anyway, I am glad that your mortality isn't as imminent as it at first appeared. |
#9
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Tom's calculations indicate how hard it is for the human mind to grasp
probability, and thus why we cannot calculate risk properly. Tom's coin analogy fails because he is looking for an unbroken sequence of survival, which therefore takes into account the past in predicting the future. His calculations are cumulative. Even with coin tosses, we can see that once we ignore the past and stop cumulating results, the calculation changes. Thus, at the start of the week, the chance of survival for a week at coin toss levels is 1 in 128. The chance of surviving for 8 days is worse, at 1 in 256. However, if our subject survives day 1, his chance of making day 8 increases to 1 in 128, and by the end of day 7 it has risen to 50:50. The older he gets, the longer his chances of living forever! I think (but as a European writing after what UK government-sponsored has recently described as a "hazardous" level of wine consumption I cannot be sure) this may be related to Zeno's paradox (in Tom Stoppard's words, "... thus proving that the arrow never reaches it's target and Saint Sebastian died of fright"). If we ignore the past, however, each day's chance is the same at 0.5. Thus Ray (may he live forever) is able to state that next year his chances will be pretty much the same, if he makes it that far. Cumulation of probabilities is what the human brain does automatically. Suppose the chance of being killed on a glider flight is 1 in 1,000. The mind (without extensive training) deals with this in a number of ways: 1. I can fly safely 999 times, then have to give up or I will certainly die on the 1,000th. If I'm already dead, I was "statistically" unlucky. 2. I've had 500 flights, so my risk level has risen to 50:50. 3. At my club we fly 1,000 flights a year between us, so one of us is sure to die flying. Unless I'm badly mistaken, none of these are true statements. I try to think as follows: a. In the UK where I fly, gliding fatalities are on average around 2.5 per annum out of 5,000 pilots, so my "statistical" risk is around 1 in 2,000 of dying through gliding each year. b. I can do a number of things to reduce my personal risk to less than 1 in 2,000, so I'll try to do those things. c. This is, to me, an acceptable level of risk for the pleasure I get from gliding. The good thing is that these probabilities are not cumulative. I've been flying for 11 years, so if they were cumulative my "statistical" risk might be down to under 1 in 20. It ain't. What can be cumulative are personal mistakes - careless rigging, no positive check, lack of sleep, etc. etc. These are the things I worry about. Tom Gardner wrote: On Nov 1, 2:11 pm, 1LK wrote: The calculation which yields the 80/1 is only true for the single instance; Single instance of what? If it is the "single instance of a day", then the calculations are correct. my odds of being here next year are another thing entirely. For that you need to do mortality computations. To stay alive for a week, you have to toss "heads" 7 times in a row, and the probability of that is 0.5 ^ 7 = 0.078125 = 1 in 128 It's not binary, it's multifactorial. I don't understand: what do you mean by "it" and "multifactorial"? Binary? Well yes, flipping a coin is binary; that's why I subsequently used your figures (that you didn't bother to include). It might help if you could explain the reasons (based on an equivalent example, if you prefer) why you believe that the calculations are wrong. Examples I can think of are - it is not a 1.25% chance of dying on every day, only on some days - each day shouldn't be treated as independent from the preceding days (but that doesn't fit with your original statement) Anyway, I am glad that your mortality isn't as imminent as it at first appeared. |
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On Nov 1, 10:46 pm, Chris Reed wrote:
Tom's calculations indicate how hard it is for the human mind to grasp probability, and thus why we cannot calculate risk properly. Tom's coin analogy fails because he is looking for an unbroken sequence of survival, Well, yes. If I die on Tuesday, it would seem somewhat optimistic to assume that I'll be alive on Wednesday! Am I missing something? which therefore takes into account the past in predicting the future. Er, no. Conditional probability and all that! His calculations are cumulative. Even with coin tosses, we can see that once we ignore the past and stop cumulating results, the calculation changes. Er, for the conditions I stated, no. Thus, at the start of the week, the chance of survival for a week at coin toss levels is 1 in 128. The chance of surviving for 8 days is worse, at 1 in 256. However, if our subject survives day 1, his chance of making day 8 increases to 1 in 128, and by the end of day 7 it has risen to 50:50. True, but missing the point. The chance of getting to day 7 from day 1 is 1 in 128, so the chance of getting to day 8 from day 1 is still 1 in 256. No change. No one is disputing if you've reached day 7 then the chance of getting to day 8 is 1 in 2. Conditional probability, etc. But on day N the chance of getting to day N+356 is vanishingly small. The older he gets, the longer his chances of living forever! I think (but as a European writing after what UK government-sponsored has recently described as a "hazardous" level of wine consumption I cannot be sure) this may be related to Zeno's paradox (in Tom Stoppard's words, "... thus proving that the arrow never reaches it's target and Saint Sebastian died of fright"). Rats. You took the words right out my mouth! Leibnitz and Newton also had a few things to say in this area ![]() If we ignore the past, however, each day's chance is the same at 0.5. Thus Ray (may he live forever) is able to state that next year his chances will be pretty much the same, if he makes it that far. Quite correct. But of course we are actually talking about the chance of him getting there (which would seem to be unfortunately small based on his statements). Cumulation of probabilities is what the human brain does automatically. Suppose the chance of being killed on a glider flight is 1 in 1,000. The mind (without extensive training) deals with this in a number of ways: 1. I can fly safely 999 times, then have to give up or I will certainly die on the 1,000th. If I'm already dead, I was "statistically" unlucky. 2. I've had 500 flights, so my risk level has risen to 50:50. 3. At my club we fly 1,000 flights a year between us, so one of us is sure to die flying. Unless I'm badly mistaken, none of these are true statements. Correct (except under pathologically perverse circumstances ![]() I try to think as follows: a. In the UK where I fly, gliding fatalities are on average around 2.5 per annum out of 5,000 pilots, so my "statistical" risk is around 1 in 2,000 of dying through gliding each year. b. I can do a number of things to reduce my personal risk to less than 1 in 2,000, so I'll try to do those things. c. This is, to me, an acceptable level of risk for the pleasure I get from gliding. The good thing is that these probabilities are not cumulative. I've been flying for 11 years, so if they were cumulative my "statistical" risk might be down to under 1 in 20. It ain't. I think the concept of "cumulative" is seriously misleading in this context. I think what you're really trying to say is that the probability of dying on day X from cause Y is *not independent* of the probability of dying on day X+1 from cause Y. Under such conditions the "1 in P^N" calculation is clearly and simply invalid. In the absence of other information, I chose to presume "independent" and you have chosen "not independent". |
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