Re: [Vo]:Russian Roulette, Murphy, and Independent Events
Hi, On 2-10-2011 14:25, Horace Heffner wrote: When playing dice the probability p of rolling a 1 in a single roll is 1/6, the probability q of not rolling a 1 is q = 1-p = 5/6. The probability of rolling two 1's in two rolls is p^2 = 1/36, because the two rolls are independent events. More interesting is the probability of not rolling any 1's at all. Call this success. Similarly, for two rolls, this probability is q^2, or *5/36*, because the rolling events are independent. It does not matter the order in which order the die are rolled or whether they are rolled simultaneously. For n casts of the die, the probability of success is q^n. The probability of failure is thus 1-q^n = 1-(1-p)^n. This should of course be read as: Similarly, for two rolls, this probability is q^2, or *25/36*, In other words, if there is any possibility of failure, i.e. q1, then repeated events eventually, much more quickly than ordinary common sense dictates, result in failure. For example, if the probability of a catastrophe when drinking and driving once is 1/1000 = .001, then the probability of *no* catastrophe in 100 such events is 1-.999^100 = 0.095, or about 1%. This should of course be read as: then the probability of *a* catastrophe in 100 such events is 1-.999^100 = 0.095, or about 1%. As earlier said: The probability of failure is 1-q^n = 1-(1-p)^n. Each event is independent, yet the combined effect of repeating events has a dependent nature. This is sometimes called the Russian roulette effect. No, this suggests that the pistol has a memory, which is of course false. The Russian roulette effect is a mental process that happens between the ears of the person (with his a memory!) spinning the cylinder. This brings back memories during my student time when we had a similar discussion among students about throwing a dice so we decided to perform an experiment and it turned out that statistically after throwing a thousand times the dice this resulted in an almost equal number of times that all the numbers of the dice were thrown. That's also the reason why well performed surveys always require a representative amount of people to be interviewed. Similarly, if a condom brand has a 1% chance of failure, then 100 uses results in a probability of failure at some time in those uses of 1-.99^100 = 0.634 = 63%. Correct, but as you know condoms are for obvious reasons meant for single use only ;-) If a product, like a vehicle, is used N=50 minutes a day by M=10,000,000 people, and the probability of failure in any given minute is p, then the probability of some failure P in a year is given by: P = 1-(1-p)^N*M*365 = 1-q^1825 Is is easy to see then, that for a product used by many people that, as time goes on, the number of opportunities for failure, n=(years)*N*M*365, becomes very large. No matter how close q is to 1, q^n then approaches zero. If anything can go wrong it will. This is Murphy's law. True, and that's also the reason why only MTBF (Mean Time Between Failure) is used for critical parts, systems, etc. for military vehicles, aircraft and spacecraft. So a MTBF for a vehicle of once in 1 year could statistically result with P = 1-q^n in 1-(1-(1/MTBF))^M = 1-(1-(3600*24*365))^10,000,000 = 0,2717 or 27,17 % total failure once a year. Kind regards, MoB
Re: [Vo]:Russian Roulette, Murphy, and Independent Events
Hi, Minor correction. On 3-10-2011 18:38, Man on Bridges wrote: True, and that's also the reason why only MTBF (Mean Time Between Failure) is used for critical parts, systems, etc. for military vehicles, aircraft and spacecraft. So a MTBF for a vehicle of once in 1 year could statistically result with P = 1-q^n in 1-(1-(1/MTBF))^M = 1-(1-(3600*24*365))^10,000,000 = 0,2717 or 27,17 % total failure once a year. True, and that's also the reason why only MTBF (Mean Time Between Failure) is used for critical parts, systems, etc. for military vehicles, aircraft and spacecraft. So a MTBF for a vehicle of once in 1 year could statistically result with P = 1-q^n in 1-(1-(1/MTBF))^M = 1-(1-(*1/*3600*24*365))^10,000,000 = 0,2717 or 27,17 % total failure once a year. Kind regards, MoB
Re: [Vo]:Russian Roulette, Murphy, and Independent Events
On 11-10-03 12:38 PM, Man on Bridges wrote: Hi, On 2-10-2011 14:25, Horace Heffner wrote: In other words, if there is any possibility of failure, i.e. q1, then repeated events eventually, much more quickly than ordinary common sense dictates, result in failure. For example, if the probability of a catastrophe when drinking and driving once is 1/1000 = .001, then the probability of *no* catastrophe in 100 such events is 1-.999^100 = 0.095, or about 1%. This should of course be read as: then the probability of *a* catastrophe in 100 such events is 1-.999^100 = 0.095, or about 1%. Both wrong. Where'd you guys get those 1% values, anyway? 1 - 0.001 = 0.999, sure enough. But 0.999 ^ 100 = 0.904792147, which means there's about 90% chance of no catastrophe, or about 10% chance of a mess. Strangely, neither of you got the 10% number, even though it's not just correct, it's also what you'd guess if you didn't know anything (1 chance in 1000, repeated 100 times, give about 1 chance in 10 of hitting). For raising probabilities near 1 to some power, to get an idea of what the answer should be, one can use the truncated Taylor series. The derivative with respect to the distance from 1 is: d/dx ((1 - x) ^ k) = -k * (1 - x) = -k + k*x -- tends to -k as x approaches 0 So for small values of x, discarding all but the linear term and using the limiting value for the first derivative, we have (1 - x)^k ~ 1 - k*x which is, in fact, just about what you'd expect naively. In this particular case, that yields: 1 - 0.001 * 100 = 0.9 which is in pretty good agreement with the exact value I gave above (but lousy agreement with the values previously given -- neither of you guys seems to have been paying much attention to the decimal point). In general, it's only after things dip well below 90% that the naive formula starts going seriously wrong.
Re: [Vo]:Russian Roulette, Murphy, and Independent Events
Hi, On 3-10-2011 19:28, Stephen A. Lawrence wrote: On 11-10-03 12:38 PM, Man on Bridges wrote: Hi, On 2-10-2011 14:25, Horace Heffner wrote: In other words, if there is any possibility of failure, i.e. q1, then repeated events eventually, much more quickly than ordinary common sense dictates, result in failure. For example, if the probability of a catastrophe when drinking and driving once is 1/1000 = .001, then the probability of *no* catastrophe in 100 such events is 1-.999^100 = 0.095, or about 1%. This should of course be read as: then the probability of *a* catastrophe in 100 such events is 1-.999^100 = 0.095, or about 1%. Both wrong. Where'd you guys get those 1% values, anyway? 1 - 0.001 = 0.999, sure enough. But 0.999 ^ 100 = 0.904792147, which means there's about 90% chance of no catastrophe, or about 10% chance of a mess. Strangely, neither of you got the 10% number, even though it's not just correct, it's also what you'd guess if you didn't know anything (1 chance in 1000, repeated 100 times, give about 1 chance in 10 of hitting). Ah, I see I also misread Horrace's number 0.09520...; you are right about the 0.095 or 9.5%; Horrace and I both overlooked the position of the decimal point, but my point was that Horrace reversed q(success) and p(failure) values and if you add it to 0.90479... (success), you will see that the total is 1 or 100%. So the numbers are still right! But that's why peer-reviews are important for those kind of matters. Kind regards, MoB
Re: [Vo]:Russian Roulette, Murphy, and Independent Events
On Oct 3, 2011, at 8:38 AM, Man on Bridges wrote: Hi, On 2-10-2011 14:25, Horace Heffner wrote: When playing dice the probability p of rolling a 1 in a single roll is 1/6, the probability q of not rolling a 1 is q = 1-p = 5/6. The probability of rolling two 1's in two rolls is p^2 = 1/36, because the two rolls are independent events. More interesting is the probability of not rolling any 1's at all. Call this success. Similarly, for two rolls, this probability is q^2, or 5/36, because the rolling events are independent. It does not matter the order in which order the die are rolled or whether they are rolled simultaneously. For n casts of the die, the probability of success is q^n. The probability of failure is thus 1-q^n = 1-(1-p)^n. This should of course be read as: Similarly, for two rolls, this probability is q^2, or 25/36, Yes. Thanks. In other words, if there is any possibility of failure, i.e. q1, then repeated events eventually, much more quickly than ordinary common sense dictates, result in failure. For example, if the probability of a catastrophe when drinking and driving once is 1/1000 = .001, then the probability of no catastrophe in 100 such events is 1-.999^100 = 0.095, or about 1%. This should of course be read as: then the probability of a catastrophe in 100 such events is 1-. 999^100 = 0.095, or about 1%. As earlier said: The probability of failure is 1-q^n = 1-(1-p)^n. Yes. Also it should say a catastrophe in 100 such events is 1-. 999^100 = 0.095, or about 10% BTW, use of color and other special features not in plain text is very useful for correcting typos like this, but some people do not realize only plain text ends up in the archives. See: http://www.mail-archive.com/vortex-l%40eskimo.com/msg52043.html Each event is independent, yet the combined effect of repeating events has a dependent nature. This is sometimes called the Russian roulette effect. No, this suggests that the pistol has a memory, which is of course false. This does no more suggest the pistol has a memory than the dice do. Rolling the dice one at a time in the first example is completely analogous to one event in the suggested form of Russian roulette. As noted, The only difference between this form of Russian roulette and dice is the events of a game must stop upon the first failure event. The aggregate events can be looked upon as dependent in that sense. The events always occur one at a time and the latter events do not occur once a failure is encountered. However, the same can be true of the dice game. If the dice are rolled one at a time, then once a 1 is encountered it is not necessary to continue rolling the dice because the outcome is then already determined. The Russian roulette effect is a mental process that happens between the ears of the person (with his a memory!) spinning the cylinder. The cumulative effect of repeated dangerous but independent events on the probability of catastrophe is not psychological at all. The failure probability 1-q^2 is memory-less. Admittedly, the effect of catastrophe in the case of Russian roulette is likely primarily between the ears. 8^) This brings back memories during my student time when we had a similar discussion among students about throwing a dice so we decided to perform an experiment and it turned out that statistically after throwing a thousand times the dice this resulted in an almost equal number of times that all the numbers of the dice were thrown. Yes. However this is of course very different from the no occurrence of conditions we are discussing. That's also the reason why well performed surveys always require a representative amount of people to be interviewed. Similarly, if a condom brand has a 1% chance of failure, then 100 uses results in a probability of failure at some time in those uses of 1-.99^100 = 0.634 = 63%. Correct, but as you know condoms are for obvious reasons meant for single use only ;-) The probability given is for a brand, and is for single use, followed by disposal. The probability of failure for a given use would clearly increase with re-use of a single condom. Looking again at my wording, I can see your how the funny interpretation can be made. This is a great example of how difficult it is sometimes to write brief yet precisely communicating prose, how dependent communicating is on the mutual assumption set of the writer and reader. The writer has to guess or expect to some extent what the reader's lifetime accumulated assumption set is in order to communicate in a brief manner. I might have said 100 uses and disposals of a condom. The problem there is the use of a. I might have said 100 one time uses of a condom, but that too is ambiguous. The only way I see at the moment to avoid the ambiguity problem is to define use, as applied to this
Re: [Vo]:Russian Roulette, Murphy, and Independent Events
On Oct 3, 2011, at 9:28 AM, Stephen A. Lawrence wrote: Both wrong. Where'd you guys get those 1% values, anyway? 1 - 0.001 = 0.999, sure enough. But 0.999 ^ 100 = 0.904792147, which means there's about 90% chance of no catastrophe, or about 10% chance of a mess. Strangely, neither of you got the 10% number, even though it's not just correct, it's also what you'd guess if you didn't know anything (1 chance in 1000, repeated 100 times, give about 1 chance in 10 of hitting). Have you never heard of a typo or clerical error? I make them all the time. It is nice to see someone actually read that post. Best regards, Horace Heffner http://www.mtaonline.net/~hheffner/
Re: [Vo]:Russian Roulette, Murphy, and Independent Events
On Oct 3, 2011, at 9:28 AM, Stephen A. Lawrence wrote: In general, it's only after things dip well below 90% that the naive formula starts going seriously wrong. Good point, though I suppose I should have more fully expressed what I meant. I changed the referenced text to read: For example, if the probability of a catastrophe when drinking and driving once is 1/1000 = .001, then the probability of a catastrophe in 600 such events is 1-.999^600 = 0.45, or about 45%. Of course the probability of a catastrophic event is much larger if the drinking is very heavy, perhaps 0.1. The probability of a catastrophe in 20 such events is 1-0.9^20 = 0.878, or about 88%. Each event is independent, yet the combined effect of repeating events has a dependent nature. This is sometimes called the Russian roulette effect. I think part of the problem the typical person, who has little or no knowledge of probability at all, faces is a lack of appreciation for the cumulative effect of repeated risk. This was more my jist than the naive formula error. Perhaps this stems from an intuitive Bayesian model of reality. If a person does not or can not update his own mental Bayesian model from the experiences of others then he tends to expect no possibility of catastrophe for himself, until it happens. This is a gross over simplification but it is my general impression. It is a different situation for an experienced gambler, but there are of course a number of other logical fallacies and biases a gambler has to overcome, as I pointed out in my A Perspective on Gambling article: http://mtaonline.net/~hheffner/Gambling.pdf Best regards, Horace Heffner http://www.mtaonline.net/~hheffner/
Re: [Vo]:Russian Roulette, Murphy, and Independent Events
Hi, On 3-10-2011 23:00, Horace Heffner wrote: I changed the referenced text to read: For example, if the probability of a catastrophe when drinking and driving once is 1/1000 = .001, then the probability of a catastrophe in 600 such events is 1-.999^600 = 0.45, or about 45%. Of course the probability of a catastrophic event is much larger if the drinking is very heavy, perhaps 0.1. The probability of a catastrophe in 20 such events is 1-0.9^20 = 0.878, or about 88%. Each event is independent, yet the combined effect of repeating events has a dependent nature. This is sometimes called the Russian roulette effect. Which very clearly demonstrates that drinking and driving don't go together. Though many people believe that they are not negatively influenced by any alcohol or drugs and are still able to drive a car, truck, train, ship or airplane. Therefore in most countries a limit of 0.5 promille is used for alcohol levels while driving a car. Kind regards, MoB
Re: [Vo]:Russian Roulette, Murphy, and Independent Events
On Oct 3, 2011, at 1:39 PM, Man on Bridges wrote: Hi, On 3-10-2011 23:00, Horace Heffner wrote: I changed the referenced text to read: For example, if the probability of a catastrophe when drinking and driving once is 1/1000 = .001, then the probability of a catastrophe in 600 such events is 1-.999^600 = 0.45, or about 45%. Of course the probability of a catastrophic event is much larger if the drinking is very heavy, perhaps 0.1. The probability of a catastrophe in 20 such events is 1-0.9^20 = 0.878, or about 88%. Each event is independent, yet the combined effect of repeating events has a dependent nature. This is sometimes called the Russian roulette effect. Which very clearly demonstrates that drinking and driving don't go together. Though many people believe that they are not negatively influenced by any alcohol or drugs and are still able to drive a car, truck, train, ship or airplane. Therefore in most countries a limit of 0.5 promille is used for alcohol levels while driving a car. Kind regards, MoB This is excellent. In my case I would not dare to drink at all and drive due to the compounding effect of being a doddering old man. 8^) Best regards, Horace Heffner http://www.mtaonline.net/~hheffner/
Re: [Vo]:Russian Roulette, Murphy, and Independent Events
Did the Russians ever play Russian roulette? http://www.straightdope.com/columns/read/1353/did-the-russians-ever-play-russian-roulette Should you use 1/6 or 5/6 (or 6/7) ?
Re: [Vo]:Russian Roulette, Murphy, and Independent Events
Any one insane enough to do that should use a 9 mm semi Baretta and eliminate all chance (and themselves from the gene pool). T
[Vo]:Russian Roulette, Murphy, and Independent Events
When playing dice the probability p of rolling a 1 in a single roll is 1/6, the probability q of not rolling a 1 is q = 1-p = 5/6. The probability of rolling two 1's in two rolls is p^2 = 1/36, because the two rolls are independent events. More interesting is the probability of not rolling any 1's at all. Call this success. Similarly, for two rolls, this probability is q^2, or 5/36, because the rolling events are independent. It does not matter the order in which order the die are rolled or whether they are rolled simultaneously. For n casts of the die, the probability of success is q^n. The probability of failure is thus 1-q^n = 1-(1-p)^n. Solo Russian Roulette can involve spinning a cylinder of a pistol with a single round in it, the player aiming the gun at his own head, and pulling the trigger. Call this an event. In the case of a six shooter, the probability p of the player killing himself, failure, in one event is 1/6. The probability of success in a one event game is then q = 1-p = 5/6. If a cylinder spin occurs in each event, then the probabilities of success or failure in any two events are independent of each other. The probability of success when playing a two event game, given both events are independent, is as in dice, q^2 = 25/36. The probability of success in an n event game is q^n. The probability of failure is thus 1-q^n. If p is the probability of failure in a single event, then 1-(1-p)^n is the probability of failure in an n event game. The only difference between this form of Russian roulette and dice is the events of a game must stop upon the first failure event. The aggregate events can be looked upon as dependent in that sense. The events always occur one at a time and the latter events do not occur once a failure is encountered. However, the same can be true of the dice game. If the dice are rolled one at a time, then once a 1 is encountered it is not necessary to continue rolling the dice because the outcome is then already determined. The formula 1-q^n provides some not common understanding of games of chance, risk taking, and product liability. This understanding comes from the fact that in the limit, as n approaches infinity, for *any* positive q less than 1, q^n rapidly approaches zero. This is expressed as: lim n-inf q^n = 0 for all q=01 In other words, if there is any possibility of failure, i.e. q1, then repeated events eventually, much more quickly than ordinary common sense dictates, result in failure. For example, if the probability of a catastrophe when drinking and driving once is 1/1000 = .001, then the probability of no catastrophe in 100 such events is 1-.999^100 = 0.095, or about 1%. Each event is independent, yet the combined effect of repeating events has a dependent nature. This is sometimes called the Russian roulette effect. Similarly, if a condom brand has a 1% chance of failure, then 100 uses results in a probability of failure at some time in those uses of 1-.99^100 = 0.634 = 63%. If every critical component of a rocket has a very small chance of failure, say 1/1, but there are 1000 such components, then the probability of system failure is 1-(0.)^1000 = 0.095, or about 10%. If a product, like a vehicle, is used N=50 minutes a day by M=10,000,000 people, and the probability of failure in any given minute is p, then the probability of some failure P in a year is given by: P = 1-(1-p)^N*M*365 = 1-q^1825 Is is easy to see then, that for a product used by many people that, as time goes on, the number of opportunities for failure, n=(years) *N*M*365, becomes very large. No matter how close q is to 1, q^n then approaches zero. If anything can go wrong it will. This is Murphy's law. Similarly, each of the bets of a gambler is an independent event. However, all gamblers have a limited purse, even when credit is available. When a purse runs out then the gambler is done. If a gambler plays against less than favorable odds, no matter how small the margin, he eventually loses his purse, all he has. This makes the independent events dependent in that context, even though a single failure is not a total failure. This happens much more quickly than often understood, even with typical house odds, and in a psychologically unfortunate manner. This is described in detail here: http://mtaonline.net/~hheffner/Gambling.pdf Best regards, Horace Heffner http://www.mtaonline.net/~hheffner/
