[ARCHIVED THREAD] - Probability problem. (Page 1 of 2)
Posted: 3/26/2016 8:18:31 PM EDT
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Let's say that an EMS district has a constant number of calls per year (with a random distribution), about 200. That means that the probability of a call on any given day is 200/365.35. That much I can figure out. But what if the department has gone 14 straight days without call? What is the probability that they get a call on the 15th day? |
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Quoted: w00t! If you flip a coin 20 times and get heads every time, what is the chance it will be heads on the 21st try. Quoted: Quoted: If the events are independent, the same. w00t! If you flip a coin 20 times and get heads every time, what is the chance it will be heads on the 21st try. |
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Quoted: 50% Quoted: Quoted: Quoted: If the events are independent, the same. w00t! If you flip a coin 20 times and get heads every time, what is the chance it will be heads on the 21st try. Uh, yeah...it was kind of rhetorical to drive your point home. I wasn't testing you.  |
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Quoted: Uh, yeah...it was kind of rhetorical to drive your point home. I wasn't testing you. Quoted: Quoted: Quoted: Quoted: If the events are independent, the same. w00t! If you flip a coin 20 times and get heads every time, what is the chance it will be heads on the 21st try. Uh, yeah...it was kind of rhetorical to drive your point home. I wasn't testing you. |
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Quoted: What I think confuses people about problems like this is that the odds, from the outset, of going 14 or 15 days in a row without a call are very small, but that is totally irrelevant once the 14 days without a call have already happened. Can you show that mathematically? |
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Quoted: w00t! If you flip a coin 20 times and get heads every time, what is the chance it will be heads on the 21st try. Quoted: Quoted: If the events are independent, the same. w00t! If you flip a coin 20 times and get heads every time, what is the chance it will be heads on the 21st try. If you flip a coin 20 times but you know it will land on heads 7 times, what is the chance it lands on heads on the 10th try when the previous 9 tries have all landed on tails? See, this is not simple. |
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Quoted: If you flip a coin 20 times but you know it will land on heads 7 times, what is the chance it lands on heads on the 10th try when the previous 9 tries have all landed on tails? See, this is not simple. Quoted: Quoted: Quoted: If the events are independent, the same. w00t! If you flip a coin 20 times and get heads every time, what is the chance it will be heads on the 21st try. If you flip a coin 20 times but you know it will land on heads 7 times, what is the chance it lands on heads on the 10th try when the previous 9 tries have all landed on tails? See, this is not simple. It is. It's still 50% Because you don't know it will land on either side any number of times. ETA: Or are you saying you are guaranteed to have the same number of calls this year? |
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Quoted: If you flip a coin 20 times but you know it will land on heads 7 times, what is the chance it lands on heads on the 10th try when the previous 9 tries have all landed on tails? See, this is not simple. Quoted: Quoted: Quoted: If the events are independent, the same. w00t! If you flip a coin 20 times and get heads every time, what is the chance it will be heads on the 21st try. If you flip a coin 20 times but you know it will land on heads 7 times, what is the chance it lands on heads on the 10th try when the previous 9 tries have all landed on tails? See, this is not simple. Each event has only 2 possible outcomes, and all previous outcomes have no influence on future outcomes. If a coin flipped 1000 times came out heads every time and you figure "it cant possibly be heads next time" that's a fallacious assumption. There is no reason why it couldn't be heads again more than tails. |
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Quoted:0th try when the previous 9 tries have all landed on tails? See, this is not simple. It is. It's still 50% Because you don't know it will land on either side any number of times. ETA: Or are you saying you are guaranteed to have the same number of calls this year? Yes, that is the constraint. |
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Quoted: Yes, that is the constraint. Quoted: Quoted:0th try when the previous 9 tries have all landed on tails? See, this is not simple. It is. It's still 50% Because you don't know it will land on either side any number of times. ETA: Or are you saying you are guaranteed to have the same number of calls this year? Yes, that is the constraint. How about: (200 - calls received so far) / (365.35 - days elapsed in year) |
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Can you show that mathematically? Quoted:
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What I think confuses people about problems like this is that the odds, from the outset, of going 14 or 15 days in a row without a call are very small, but that is totally irrelevant once the 14 days without a call have already happened. Can you show that mathematically? The odds of going 14 days in a row without a call are (1-200/365)^14. To understand this better, on any given day, the probability of not getting a call is about 45.2%. The probability of going 14 days in a row without getting a call is 0.00149%. The probability of going 15 days in a row without a call is 0.000673%, but the probability of not getting a call on day 15 is still just 45.2% |
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Everyday has equal odds. Would they? It would be interesting to see the previous years breakdowns. As days without calls add up, I would think that the following days odds of a call increase. A coin flipped 6 times lands on heads. The seventh flip has equal odds of being heads or tails. NO, it does not. It is 50/50 for any one SINGLE flip but the odds change dramatically that after a coin is flipped and lands on heads 6 times in a row it will continue to land on heads and not tails. The odds of coming up tails on the seventh flip are greatly enhanced in your example. 50/50 only holds true for any ONE single flip. AND, This is not flipping one single coin over and over again. This is having the entire population covered flip a coin and then deciding the 200 people out of that population that would make a call. So, lets say you have 10,000 people you are covering that would call into your EMS station. They all flip a coin and if it lands on heads, they don't need the service...... if ONE lands on tails they do...... just a rough example. It is a much more complicated question than some think. You can use google to see the equations and examples. |
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Yeah, but the probability of getting heads 21 times in a row isn't 50/50  Quoted:
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If the events are independent, the same. w00t! If you flip a coin 20 times and get heads every time, what is the chance it will be heads on the 21st try. NO it is not. |
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Quoted: Yeah, but the probability of getting heads 21 times in a row isn't 50/50 Quoted: Quoted: Quoted: Quoted: If the events are independent, the same. w00t! If you flip a coin 20 times and get heads every time, what is the chance it will be heads on the 21st try. |
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Ah, 200 mandatory calls.
Frequency would be better. 200/365 is a rate of 200cpy, or 0.548cpd. If x = the number of calls already taken prior to the 14 monotonous days, and y = the number of days already elapsed in the year, then the frequency pre-14dayperiod is x/y=__cpd Therefore, since we must not include the 14 days in the rate for the remainder of the year, the rate for the remainder of the year is (200 - x)/(365 - 14 - y)=__cpd |
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The coin doesn't take into consideration how many previous times it has landed. Each event has only 2 possible outcomes, and all previous outcomes have no influence on future outcomes. If a coin flipped 1000 times came out heads every time and you figure "it cant possibly be heads next time" that's a fallacious assumption. There is no reason why it couldn't be heads again more than tails. Quoted:
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If the events are independent, the same. w00t! If you flip a coin 20 times and get heads every time, what is the chance it will be heads on the 21st try. If you flip a coin 20 times but you know it will land on heads 7 times, what is the chance it lands on heads on the 10th try when the previous 9 tries have all landed on tails? See, this is not simple. Each event has only 2 possible outcomes, and all previous outcomes have no influence on future outcomes. If a coin flipped 1000 times came out heads every time and you figure "it cant possibly be heads next time" that's a fallacious assumption. There is no reason why it couldn't be heads again more than tails. Yes. The odds of it coming up heads 1000 times in a row is virtually zero, but in the event that it's already come up 999 times, there's a 50% chance it will make 1000. |
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Quoted: How about: (200 - calls received so far) / (365.35 - days elapsed in year) Quoted: Quoted: Quoted:0th try when the previous 9 tries have all landed on tails? See, this is not simple. It is. It's still 50% Because you don't know it will land on either side any number of times. ETA: Or are you saying you are guaranteed to have the same number of calls this year? Yes, that is the constraint. How about: (200 - calls received so far) / (365.35 - days elapsed in year) Looks good at first, but if the station goes 165 days without a call, then the probability of getting a call the next day is one. But we know it isn't. |
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Quoted: Looks good at first, but if the station goes 165 days without a call, then the probability of getting a call the next day is one. But we know it isn't. Quoted: Quoted: Quoted: Quoted:0th try when the previous 9 tries have all landed on tails? See, this is not simple. It is. It's still 50% Because you don't know it will land on either side any number of times. ETA: Or are you saying you are guaranteed to have the same number of calls this year? Yes, that is the constraint. How about: (200 - calls received so far) / (365.35 - days elapsed in year) Looks good at first, but if the station goes 165 days without a call, then the probability of getting a call the next day is one. But we know it isn't. You can have more than 1 call per day? |
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I would like to point out that I recognize it is a complicated question. 
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It is a much more complicated question than some think. You can use google to see the equations and examples. I would like to point out that I recognize it is a complicated question.
Once again, I will ask. What level course? |
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Quoted: What level stats class? As soon as I saw probability and phone calls, something else immediately came to mind... This isn't for a class. This is the result of observing an argument between a superstitious person and a non superstitious person. i.e., "Don't mention that we haven't had a call in two weeks or we'll have one in the next hour!" So I got to thinking, what is the chance, in real math terms? |
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You can use google to see the equations and examples. Quoted:
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Everyday has equal odds. Would they? It would be interesting to see the previous years breakdowns. As days without calls add up, I would think that the following days odds of a call increase. A coin flipped 6 times lands on heads. The seventh flip has equal odds of being heads or tails. NO, it does not. It is 50/50 for any one SINGLE flip but the odds change dramatically that after a coin is flipped and lands on heads 6 times in a row it will continue to land on heads and not tails. The odds of coming up tails on the seventh flip are greatly enhanced in your example. 50/50 only holds true for any ONE single flip. AND, This is not flipping one single coin over and over again. This is having the entire population covered flip a coin and then deciding the 200 people out of that population that would make a call. So, lets say you have 10,000 people you are covering that would call into your EMS station. They all flip a coin and if it lands on heads, they don't need the service...... if ONE lands on tails they do...... just a rough example. It is a much more complicated question than some think. You can use google to see the equations and examples. I hear ya bro but I think you are letting "Law of Averages " lead you down the wrong path on this one. Just my opinion. No judgments. |
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If you teach this shit for a living, did you catch he might need something involving a lambda? Quoted:
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Independent events always have the same probability. I teach this shit for a living. If you teach this shit for a living, did you catch he might need something involving a lambda? Yeah, the number of phone calls per day would be Poisson, I think. I'm taking prob theory right now. |
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Quoted: You can have more than 1 call per day? Quoted: Quoted: Looks good at first, but if the station goes 165 days without a call, then the probability of getting a call the next day is one. But we know it isn't. You can have more than 1 call per day? Yes. You can have up to five calls in one day. If this is significant I'll put it as a constraint in the OP. |
| If it is a probability problem you can't constrain how many total there will be. You can say that the average in the past has been 200 and base future probabilities on that. In that case the day after the 14 day streak has the exact same chances as any one of those days did. Events in the past like that can not materially affect events in the future. You don't need a proof for that, it is a tautology. For chance events to make exactly 200 outcomes per year every year is a statistical impossibility. Only an artificial algorithm that adjusts the probability of future events based on prior events can achieve this and that will not produce a random distribution either. |
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Yeah, the number of phone calls per day would be Poisson, I think. I'm taking prob theory right now. Quoted:
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Independent events always have the same probability. I teach this shit for a living. If you teach this shit for a living, did you catch he might need something involving a lambda? Yeah, the number of phone calls per day would be Poisson, I think. I'm taking prob theory right now. Calls to a phone number are not random events. Please don't confuse probability with unpredictability. They are not the same. |
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What you *can* do is calculate how weird it would be:
Expected Value You would need to use your assumption of an average number of calls per day. Obviously, you can think of scenarios like "all calls in a year come on the same day" and your odds would be 1 over the number of days left, etc. Game theory types mess this up all the time. |
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Say I set up a daily game show where I call up 1 contestant at a time to flip 10 coins.
If the contestant flips 10 coins and they all flip HEADS or they all flip TAILS, then the contestant wins. If the contestant flips 10 coins, but the sides don't all match, then the contestant loses. That's it. The odds for that are... 1/1024... but if I guarantee that there will be 200 winners, then the odds or probability don't really matter. I'll run out of 'losing shows' and then have to declare each contestant through the end of the year... all winners. That sounds weird because there'd likely only ever be ~1 winner per year, otherwise. But what if the odds, being the finicky thing they are... what if I got all 200 winners before the end of the year? |
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Quoted: If it is a probability problem you can't constrain how many total there will be. You can say that the average in the past has been 200 and base future probabilities on that. In that case the day after the 14 day streak has the exact same chances as any one of those days did. Events in the past like that can not materially affect events in the future. You don't need a proof for that, it is a tautology. For chance events to make exactly 200 outcomes per year every year is a statistical impossibility. Only an artificial algorithm that adjusts the probability of future events based on prior events can achieve this and that will not produce a random distribution either. Yes, dan45678 would agree with you on this. But follow me here. The station does not get exactly 200 calls/year, but the standard deviation is pretty low year to year. They're going to get about 200 calls. Shouldn't that matter? |
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Calls to a phone number are not random events. Please don't confuse probability with unpredictability. They are not the same. Quoted:
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Independent events always have the same probability. I teach this shit for a living. If you teach this shit for a living, did you catch he might need something involving a lambda? Yeah, the number of phone calls per day would be Poisson, I think. I'm taking prob theory right now. Calls to a phone number are not random events. Please don't confuse probability with unpredictability. They are not the same. Whenever there is uncertainty in an event occurring, it has a probability. We don't know how many phone calls we will receive per hour. Thus, it is random. Do you have a background in probability theory, just curious. |
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This isn't for a class. This is the result of observing an argument between a superstitious person and a non superstitious person. i.e., "Don't mention that we haven't had a call in two weeks or we'll have one in the next hour!" So I got to thinking, what is the chance, in real math terms? Quoted:
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What level stats class? As soon as I saw probability and phone calls, something else immediately came to mind... This isn't for a class. This is the result of observing an argument between a superstitious person and a non superstitious person. i.e., "Don't mention that we haven't had a call in two weeks or we'll have one in the next hour!" So I got to thinking, what is the chance, in real math terms? The phone calls are a Poisson process. https://www.probabilitycourse.com/chapter11/11_1_2_basic_concepts_of_the_poisson_process.php |
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Calls to a phone number are not random events. Please don't confuse probability with unpredictability. They are not the same. Quoted:
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Independent events always have the same probability. I teach this shit for a living. If you teach this shit for a living, did you catch he might need something involving a lambda? Yeah, the number of phone calls per day would be Poisson, I think. I'm taking prob theory right now. Calls to a phone number are not random events. Please don't confuse probability with unpredictability. They are not the same. Don't confuse discrete with continuous. |