halpmaine 0 Posted October 18, 2011 Share Posted October 18, 2011 Hello. Forgive my lay review here but this is the background to my question: My daughter and I were headed to school one morning earlier in the year. She was reading 101 Dalmations. While stopped at an intersection - I looked over to the vehicle next to me and surprise, surprise, you guessed it - there in the back seat was a DALMATION. This morning on the way to school we got to talking about that very event...that coincidence. It then occurred to me while trying to give my daughter a ball-park notion of the (high) odds of having those two events occurring again at random - that I ought to run the question of determining general probability of a seemingly random event by some math gurus. All disclaimers aside and all things equal, is there a means to estimate the probability of such an event occurring without having much hard data? For example, while very loose on the data, I told my daughter perhaps one could consider the odds of her reading that particular book (e.g. say 1/10^{4}), the odds of seeing that particular breed of dog in the car beside us (e.g. estimate that, say, 1/20 cars will have a dog and of those 1/100 would be a dalmation) - and so forth for time of day, route driven and so forth with all the important variables. Thus, if you multiply all relevant terms (i.e. 1/10,000 * 1/20 * x * y * z) you would come up with some kind of guesstimate pertaining to the odds of her reading a book about a particular breed of dog while simultaneously seeing that same dog in a vehicle next to us - at random. I told her it was probably in the millons-to-billion. Is this the general manner in which the probability for otherwise random events are determined? Again, my apologies for the likely enigmatic presentation of this ? but clearly I am not a math whiz! Thanks for your help. -halpmaine Link to post Share on other sites

timo 583 Posted October 18, 2011 Share Posted October 18, 2011 I don't know what "and so forth" is. I would simply multiply by the approximate number of cars you see. And you should probably add up the probabilities for all breeds of dog (or even better: substitute the probability to read a book about Dalmatians with the probability to read a book that prominently features a breed of dog). It is a manner to determine probabilities that works well for low probabilities. A counter example would be the probability to roll a six in six rolls of a dice, which is not 1/6 (for the probability to toll a dice in one roll) times six (for the number of rolls). Link to post Share on other sites

swansont 7597 Posted October 18, 2011 Share Posted October 18, 2011 This would also be a good example of selection bias. If you had not noticed a dalmation in the next car, you would not have noticed and not have much reason to remember the incident. You would really have to keep track of every conversation you've had on topics where you had a chance to notice a coincidence to get a true accounting of the odds of a coincidence happening at some point. If you did, it's very likely they would approach 1. Link to post Share on other sites

DrRocket 337 Posted October 19, 2011 Share Posted October 19, 2011 Hello. Forgive my lay review here but this is the background to my question: My daughter and I were headed to school one morning earlier in the year. She was reading 101 Dalmations. While stopped at an intersection - I looked over to the vehicle next to me and surprise, surprise, you guessed it - there in the back seat was a DALMATION. This morning on the way to school we got to talking about that very event...that coincidence. It then occurred to me while trying to give my daughter a ball-park notion of the (high) odds of having those two events occurring again at random - that I ought to run the question of determining general probability of a seemingly random event by some math gurus. All disclaimers aside and all things equal, is there a means to estimate the probability of such an event occurring without having much hard data? For example, while very loose on the data, I told my daughter perhaps one could consider the odds of her reading that particular book (e.g. say 1/10^{4}), the odds of seeing that particular breed of dog in the car beside us (e.g. estimate that, say, 1/20 cars will have a dog and of those 1/100 would be a dalmation) - and so forth for time of day, route driven and so forth with all the important variables. Thus, if you multiply all relevant terms (i.e. 1/10,000 * 1/20 * x * y * z) you would come up with some kind of guesstimate pertaining to the odds of her reading a book about a particular breed of dog while simultaneously seeing that same dog in a vehicle next to us - at random. I told her it was probably in the millons-to-billion. Is this the general manner in which the probability for otherwise random events are determined? Again, my apologies for the likely enigmatic presentation of this ? but clearly I am not a math whiz! Thanks for your help. -halpmaine A few points. 1. The discipline of probability theory starts with a probability space -- that is with events and their probabilities being known. Probability theory does not involve the determination of the probabilities of elementary events. 2. The estimation of probabilities is the province of statistics. Statistics quite often relies on assumptions that are convenient but not rigorously justifiable. The results are at best approximate, and when the assumptions are grossly violated can be just plain wrong. 3. What can be proved rigorously is that given an event of non-zero probability, in infinitely many independent trials that event will occur infinitely often with probability one (aka almost surely). So if enough vehicles pull up beside you it is highly likely that eventually one of them will have a dalmation dog in it. 4. Probability theory is one of the most misunderstood and misapplied branches of mathematics. All sorts of nonsense is "proved" or justified based on misapplications of probabilitry theory. 5. Just today I saw a car with the license plate VZK 9025. Just imagine the odds against that! Link to post Share on other sites

Phi for All 5945 Posted October 19, 2011 Share Posted October 19, 2011 Is this the general manner in which the probability for otherwise random events are determined? Remember that you're not asking for the odds of a completely random occurrence. Your daughter was already reading 101 Dalmations, so the odds of that are 1:1. You were in public, making the odds of seeing a dog higher. There might have been a Dachshund in the car behind you and a Labrador doing a head bob in the car in front, but you selected the Dalmatian because it confirmed information you already had. The odds of the Lab being in front of you when your daughter is reading that book are actually the same as the Dalmatian, but the Dalmatian had preselected relevance. Link to post Share on other sites

Iggy 344 Posted October 19, 2011 Share Posted October 19, 2011 ...having those two events occurring again at random ...perhaps one could consider the odds of her reading that particular book (e.g. say 1/10^{4}), the odds of seeing that particular breed of dog in the car beside us (e.g. estimate that, say, 1/20 cars will have a dog and of those 1/100 would be a dalmation) - and so forth for time of day, route driven and so forth with all the important variables. Thus, if you multiply all relevant terms (i.e. 1/10,000 * 1/20 * x * y * z) you would come up with some kind of guesstimate pertaining to the odds of her reading a book about a particular breed of dog while simultaneously seeing that same dog in a vehicle next to us - at random... It sounds right to me. You multiply them. The next time you are in the car with your daughter (or any future such time) the probability of the coincidence repeating itself is the probability that you happen to be discussing 101 dalmatians times the probability that an adjacent car carries a dalmatian. To make it easier, like you say, the probability that 1) you are discussing 101 dalmatians 2) there is an adjacent car visible 3) that car carries a dog and 4) the dog is a dalmatian would be the product of the probability of 1,2,3, and 4. Link to post Share on other sites

the tree 222 Posted October 19, 2011 Share Posted October 19, 2011 It's not always entirely a co-incidence, and perhaps probability theory isn't what you need to explain this. As a six year old, there's a high chance that your daughter is already odly aware of the Baader-Meinhof Phenomenon, even more so than you are. When you're learning new words every single day then it is often something that you really become aware of. The thing to consider here is, if she hadn't recently been thinking about dalmations - she may well not have noticed the dalmation in the next car. Seeing a dog in a car is a completely unextraordinary event, but the reason for this story happening is that she was probably subconsiously on the look out for black and white spots. Link to post Share on other sites

halpmaine 0 Posted October 19, 2011 Author Share Posted October 19, 2011 It's not always entirely a co-incidence, and perhaps probability theory isn't what you need to explain this. As a six year old, there's a high chance that your daughter is already odly aware of the Baader-Meinhof Phenomenon, even more so than you are. When you're learning new words every single day then it is often something that you really become aware of. The thing to consider here is, if she hadn't recently been thinking about dalmations - she may well not have noticed the dalmation in the next car. Seeing a dog in a car is a completely unextraordinary event, but the reason for this story happening is that she was probably subconsiously on the look out for black and white spots. ---------------------------------------------------------------- Agreed except I looked to the adjacent vehicle. But I appreciate what you are saying - and the other replies. However, my description and question thereof pertains to the admitted relatively unscientific approach to arriving at a ball-park estimate. Clearly too many variables exist to realistically duplicate this event in vivo. That said, generally speaking - and commensurate with Iggy's post (thx Iggy) - when discussing this with my daughter (and while still considering her genius-level IQ) wouldn't a pretty quick and dirty method indeed be to multiply some general terms (variables) together to arrive at some kind of odds ratio? Those of you with inquisitve first graders will likely follow me on this - the goal is more about the process...the desire to continue a cool dialogue and consider some basic math/science with my daughter on the way to school than to try and arrive at a definitive, provable answer! Thanks All! -halpmaine Link to post Share on other sites

Bignose 946 Posted October 19, 2011 Share Posted October 19, 2011 The odds of the Lab being in front of you when your daughter is reading that book are actually the same as the Dalmatian, but the Dalmatian had preselected relevance. I suspect it that the odds are not the same if only because Labs are one of the most popular breeds and there are simply more Labs than Dalmatians. Link to post Share on other sites

imatfaal 2481 Posted October 20, 2011 Share Posted October 20, 2011 (edited) The experts could confirm this - but my suggestion would be to work out the chances that the same happens tomorrow. Let your daughter choose her book for the journey as usual and be sure to check out every car next to you at any traffic lights; and in advance we can estimate (with very large margins of error) what the chances are. This avoids any confirmation bias or baadarmeinhoff problems. I think it would be a simplish calculation (chance of book being 101d) x (chance of pulling up next to car with dalmation on journey) chance of book being 101d can bes estimate by assuming that there is an even chance of picking any book and there are X books chance of pulling up next to car with dalmation on journey - needs to be estimated by number of car journeys with a dog in car, percentage of dogs that are dalmations and number of cars you pull up next to per journey. It will end up as a sum that looks like 1 - [chance of it not happening]^number of possible sightings edited cos i think i missed out a 1-- but in fact I had bracket wroing Edited October 20, 2011 by imatfaal Link to post Share on other sites

Xittenn 176 Posted October 20, 2011 Share Posted October 20, 2011 (edited) The probability of such an event can be found using the following equation: [math] Probability = \frac{favarouble \; outcomes}{total \; number \; of \; outcomes} [/math] This information is really difficult to determine in your scenario, as I'm sure everyone is already in agreement. What are the total number of outcomes? Everyone has noted the 'fundamental counting principle' where it is noted that odds are found by multiplying chances of so on and so forth. The fundamental counting principle states that given say 5 shirts and 3 pants one can choose from 15 different outfits to wear when each item is 'chosen' in 'combination.' So what you are really trying to find here is the total of all possible outcomes, by way of the fundamental counting principle. You would then define your set of outcomes that you wish to isolate, and using the first equation [math] p= \frac{fav}{tot} [/math] find the percent probability that the event would occur. Now there is more specialized math that makes finding these answers less difficult. These maths are however not easily summarized in a few sentences. But even still, while avoiding these maths, using logic to draw or envision a tree is difficult. The process could still be explained using a more simplified example. 1) Define the total outcomes or the 'sample space' 1 blue shirt, 3 red shirts, 1 green shirt 'and' 2 jeans, 1 dress pant 5 shirts 'and' 3 pants sample space includes 5 x 3 = 15 possible outfits . . . 2) define the favorable outcomes 3 red shirts 'and' 1 dress pant there will be 3 x 1 = 3 outfits consisting of a red shirt and a dress pant 3) find the probability [math] Probability = \frac{favarouble \; outcomes}{total \; number \; of \; outcomes} [/math] P = 3/15 The problem in defining a solution for your question now is defining these. Given the giantness of the question, logically compiling this data is, mentally, very difficult. But even beyond this other mathematical knowledge is required to simplify it and make it manageable. concepts required: i) Factorials ii) Permutations and Combinations iii) Pascal's Triangle and the Binomial Theorem ( can't find simple overview ) It becomes less giant as you reduce your sample space from statistics based on the country you live in, to the number of passenger vehicles on that given road at that time and who might have a dog that is a Dalmatian. This would naturally require a telephone survey . . . . . . . . . . . . . . . . . . . . Edited October 20, 2011 by Wuz Here Link to post Share on other sites

imatfaal 2481 Posted October 20, 2011 Share Posted October 20, 2011 Wuz - that only covers the probability that it will happen at a certain pre-ordained place (ie the traffic lights outside the park) - the prob question was for a journey of multiple possible occurrences which also requires taking powers of the odds of things not happening ie chances of rolling a six in two rolls of a fair die is 1- [(5/6)^2] Link to post Share on other sites

Xittenn 176 Posted October 20, 2011 Share Posted October 20, 2011 Wuz - that only covers the probability that it will happen at a certain pre-ordained place (ie the traffic lights outside the park) - the prob question was for a journey of multiple possible occurrences which also requires taking powers of the odds of things not happening ie chances of rolling a six in two rolls of a fair die is 1- [(5/6)^2] Only the giantness of the problem was 'given'! Xittenn ( name change in t - 9 days ) Link to post Share on other sites

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