# Difference between revisions of "Talk:1252: Increased Risk"

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Saying that unfortunately Cueball is mistaken in his calculations because he said 50% instead of 49.99999992% is a bit of an exaggeration. [[User:Xhfz|Xhfz]] ([[User talk:Xhfz|talk]]) 20:19, 16 August 2013 (UTC) | Saying that unfortunately Cueball is mistaken in his calculations because he said 50% instead of 49.99999992% is a bit of an exaggeration. [[User:Xhfz|Xhfz]] ([[User talk:Xhfz|talk]]) 20:19, 16 August 2013 (UTC) | ||

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+ | ;Chaos at explain section | ||

+ | Please stop adding this, it does not explain the comic, it only belongs to this discussion page: | ||

+ | |||

+ | :Note that the 50% figure is an approximation. Assuming the odds of being attacked by a dog is ''x'', the odds of being attacked by a dog at least once in two visits is 1 - (1-''x'')<sup>2</sup>. The odds of being attacked at least once in three visits is 1 - (1-''x'')<sup>3</sup>. Therefore, if one visit has one in a billion probability of attack, then two visits have not 2 in a billion, but 1.999999999 in a billion. Similarly, three visits have a probability of 2.999999997 in a billion. Saying 50% instead of 49.99999992% is a reasonable approximation. | ||

+ | |||

+ | :Unfortunately, [[Cueball]] is mistaken in his calculations. This is easier to see with an event that has greater probability, such as a coin toss. Assuming the odds of getting heads in one flip is .5, the odds of getting heads at least once in two flips is .75 (i.e., 1 minus [.5 X .5], the odds of getting tails both times), and the odds of getting heads at least once in three flips is .875 (1 minus [.5 X .5 X .5], the odds of getting three tails in a row). Getting heads in three flips is not 50% more likely than getting heads in two flips. With very low probabilities (such as the probability of attack by a dog swimming with a handgun), Cueball's calculation gives an extremely close approximation of the actual probability, but one can't apply the same logic to events of just any probability. |

## Revision as of 22:13, 16 August 2013

I think this is to address the old chestnut of "<something> will *double* your risk of getting cancer!", or the like, where the risk of getting that cancer (in this example) is maybe 1 in 10,000, so doubling the risk across a population wouldmake that a 1 in 5,000 risk to your health... which you may still consider to be an acceptable gamble if it's something nice (like cheese!) that's apaprently to blame and you'd find abstinence from it gives a barely marginal benefit for a far greater loss of life enjoyment. Also, this sort of figure almost always applies towards a *specific form* of cancer, or whatever risk is being discussed, meaning you aren't vastly changing your life expectancy at all. In fact, the likes of opposing "red wine is good/bad for you" studies can be mutually true by this same principle (gain a little risk of one condition, lose a little risk from another). (Note: I don't know of any particular "cheese gives you cancer!" stories doing the rounds, at the moment. I bet they have done, but I only mention it because I actually quite like cheese. And I probably *wouldn't* give it up under the above conditions.)

It's also possible that this covers the likes of "<foo> in <country> is 10 times more dangerous than it is <other country>" statements. Perhaps *only* ten incidents happened in the former, and a single instance in the latter, out the *whole* of each respective country. Or a single incident occured in both, but the second country is ten times the size, so gets 'adjusted for population' in the tables. And, besides which, that was just for one year and was just a statistical blip that will probably revert-towards-the-mean next year.

Finally, for a given risk of some incident happening on the first two trips, with no 'memory' or build-up involved, it pretty much is half-as-likely-again for the incident to have happened (some time!) in three separate trips. (Not quite, if those that lose against the odds and get caught by the incident the first or second trip never get to *have* a (second or) third trip... but for negligable odds like thegiven example, of the dog with the handgun, it's near-as-damnit so.) 178.104.103.140 11:12, 16 August 2013 (UTC)

Where did "dogs with shotguns" come from? I only saw "handgun" in the comic. Besides, I interpreted the risk as being hit by a negligent discharge from the handgun, not being deliberately attacked by the dog. Also, since probabilities are the set of real numbers between 0 and 1 inclusive, there are an uncountable number of them. "A x% increase in a tiny risk is still tiny" is an inductive statement, which means it could only be used to argue that a countable set of numbers is tiny. 76.64.65.200 12:24, 16 August 2013 (UTC)

- If induction base is uncountable, you can prove it for the whole [0; 1]. For example your induction base may be "every risk under 0.00000000000000000001% is tiny". --DiEvAl (talk) 12:38, 16 August 2013 (UTC)

I think it's worth mentioning that this comic doesn't distinguish between percentages and percentage points. --DiEvAl (talk) 12:35, 16 August 2013 (UTC)

Is it the case that doing something three times increases risk by 50% over two times inherently? I feel like this is the case, but it's early, here. Also, I'm not sure Randall is attacked by a dog, he may be using it as a diversion. I think that he's done this before. Theo (talk) 12:56, 16 August 2013 (UTC)

- (First, good point, DiEvAl, about the percentages/percentage-points. I
*knew*I'd missed something out in my first thoughts. I actually tend to assume*against*percentage points, which is somewhat the opposite from what I've seen in the general public.) - Actually, depends on how you count it. But I was using the "encounter 'n' incidents per trip", "encounter '2n' incidents per two trips", "encoutner '3n' incidents per three trips" measure, where 3n==2n+50%. But that works best with a baseline of >>1 incidents per trip assumed. In reality, if the chance is a fractional 'p' for an occurance in one instance, it's (1-p) that it
*didn't*occur thus (1-p)^{n}that it didn't occur in any of 'n' instances and 1-(1-p)^{n}that it did (at least once, possible several times or even all). Not so simple, but for p tending to zero it 'does' converge on 1.5 times for across three what you'd expect for two (albeit because 0*1.5=0). Like they say, "Lies, Damn Lies...", etc. ;) 178.104.103.140 14:22, 16 August 2013 (UTC)

I don't think Randall is being attacked by a dog at all. What he's saying is that if you are going to think getting attacked by a shark is so likely, then you better be watching out for that never-gonna-happen dog scenario too. Jillysky (talk) 13:56, 16 August 2013 (UTC)

Is 0.000001% really "one in a million"?

- If 1% = 1 in 100, then
- 0.1% = 1 in a 1,000
- 0.01% = 1 in a 10,000
- 0.001% = 1 in a 100,000
- 0.0001% = 1 in a 1,000,000
- 0.00001% = 1 in a 10,000,000
**0.000001% = 1 in a 100,000,000**

Would it be more accurate to leave off the % sign? Assuming I'm right, I think it'd be less confusing to leave it and reduce the numbers by a couple orders of magnitude. --Clayton 12.202.74.87 14:36, 16 August 2013 (UTC)

*If the chance of the dog attack is 0.000000001% (one in a billion) on each visit to the beach, then the chance of attack over two visits is 0.000000002% whereas in three visits it becomes 0.000000003%*

Um, no. Following that logic, if I go to the beach a billion times then I **will** get shot by a dog that is packing. Rather, each visit to the beach has it's own odds, like the rolling of dice? On any particular visit there's a one-in-a-billion chance. And that's true on each subsequent visit as well. Tuesday's visit to the beach isn't twice as dangerous just because I was at the beach on Monday. CFoxx (talk) 16:26, 16 August 2013 (UTC)

- For each visit that is the case. Because it's one visit, that's true. However, if (time not being a factor) one were to have a billion visits planned, the odds over all would be increased. Pretty sure that overall this means that you got the joke faster than I did. Thanks for the clarification! Theo (talk) 17:06, 16 August 2013 (UTC)
- The odds overall may increase with multiple visits. But not, at least, at the rate listed. Otherwise that billionth trip (if one survived that long as one is likely to do) would be certain death. CFoxx (talk) 17:30, 16 August 2013 (UTC)
- Correct. Technically, the odds we are worried about are the "probability of being shot one or more times by a dog". So if the probability is 1/10^9 for any given day, than the odds of not being shot are (10^9-1)/10^9 for any given day, and the odds of not being shot over three days are (10^9-1)^3/10^27, and then the odds of being shot one or more times are 1-((10^9-1)^3/10^27), which is roughly 2.999999997000000001/10^9. That is close, but slightly less, than 3/10^9. 206.174.12.203 18:01, 16 August 2013 (UTC)Toby Ovod-Everett
- Absolute incorrect: You always have to look at the single event. More events do not belong together, you always have the same probability at each single event. So, even 10 billion events may or may NOT result in a disaster. Math isn't easy.--Dgbrt (talk) 19:17, 16 August 2013 (UTC)
- I believe what CFoxx was saying is that if the odds of something happening on any given day are one in three, then the odds of that thing happening at least once during a four day period is NOT 4/3rds! I was pointing out that the proper way to calculate the odds for a four day period is to say that the odds of it not happening on any given day are two in three. You take that probability and raise it to the fourth power, giving the odds that it won't happen at all during a four day period of 16/81, thus the odds that it will happen during that four day period is 65/81. I then did that same calculation for the 1 in a billion chance per day and applied it to the three day period, and recognized that he was correct that the true probability of the event happening one or more times over a three day period was not three times the probability of it happening on any given day, but also noted that the difference for a 1 in a billion chance over a small period is pretty close to the simplistic (but incorrect) approach. My rough estimate for the "one in a billion per day" event happening one or more times during a billion day period is 63.21%.206.174.12.203 21:33, 16 August 2013 (UTC)Toby Ovod-Everett

- Absolute incorrect: You always have to look at the single event. More events do not belong together, you always have the same probability at each single event. So, even 10 billion events may or may NOT result in a disaster. Math isn't easy.--Dgbrt (talk) 19:17, 16 August 2013 (UTC)

- Correct. Technically, the odds we are worried about are the "probability of being shot one or more times by a dog". So if the probability is 1/10^9 for any given day, than the odds of not being shot are (10^9-1)/10^9 for any given day, and the odds of not being shot over three days are (10^9-1)^3/10^27, and then the odds of being shot one or more times are 1-((10^9-1)^3/10^27), which is roughly 2.999999997000000001/10^9. That is close, but slightly less, than 3/10^9. 206.174.12.203 18:01, 16 August 2013 (UTC)Toby Ovod-Everett

- The odds overall may increase with multiple visits. But not, at least, at the rate listed. Otherwise that billionth trip (if one survived that long as one is likely to do) would be certain death. CFoxx (talk) 17:30, 16 August 2013 (UTC)

Just a thought: is the title text a reference to the Sorites paradox? --AJ 80.42.221.105 17:25, 16 August 2013 (UTC)

Rats! I made the newbie mistake of editing something before I found the discussion page. I looked for it, honest I did! I see that UTC has already brought up what I referred to as "Cueball's error" in my (pre-log-in) edit. I did find it hard to believe I'd be the first xkcd fan to notice this error. I think this is worth addressing in the explanation, though I of course won't take offense if someone wants to obliterate my edit and start over. (CLSI) -- CLSI (talk) *(please sign your comments with ~~~~)*

Maybe he means this: Florida man shot by his dog, police say http://usnews.nbcnews.com/_news/2013/02/26/17107343-florida-man-shot-by-his-dog-police-say?lite -- Jb (talk) *(please sign your comments with ~~~~)*

Saying that unfortunately Cueball is mistaken in his calculations because he said 50% instead of 49.99999992% is a bit of an exaggeration. Xhfz (talk) 20:19, 16 August 2013 (UTC)

- Chaos at explain section

Please stop adding this, it does not explain the comic, it only belongs to this discussion page:

- Note that the 50% figure is an approximation. Assuming the odds of being attacked by a dog is
*x*, the odds of being attacked by a dog at least once in two visits is 1 - (1-*x*)^{2}. The odds of being attacked at least once in three visits is 1 - (1-*x*)^{3}. Therefore, if one visit has one in a billion probability of attack, then two visits have not 2 in a billion, but 1.999999999 in a billion. Similarly, three visits have a probability of 2.999999997 in a billion. Saying 50% instead of 49.99999992% is a reasonable approximation.

- Unfortunately, Cueball is mistaken in his calculations. This is easier to see with an event that has greater probability, such as a coin toss. Assuming the odds of getting heads in one flip is .5, the odds of getting heads at least once in two flips is .75 (i.e., 1 minus [.5 X .5], the odds of getting tails both times), and the odds of getting heads at least once in three flips is .875 (1 minus [.5 X .5 X .5], the odds of getting three tails in a row). Getting heads in three flips is not 50% more likely than getting heads in two flips. With very low probabilities (such as the probability of attack by a dog swimming with a handgun), Cueball's calculation gives an extremely close approximation of the actual probability, but one can't apply the same logic to events of just any probability.