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Do Past, Unrelated Events Affect Future Probabilities?

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  1. Dec 30, 2013 #1

    Jow

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    For a long time now I've been thinking about something I find odd about probabilities. For example, let's say that I win the lottery, something that is extremely unlikely. What is the probability that I will be struck by lightning? The odds of someone winning the lottery and getting struck by lightning are extremely low because both events are unlikely, so for them to both happen to one person is extremely improbable. However, just because I've won the lottery doesn't mean the chances of me getting struck by lightning are any lower. If the probability of getting hit by lightning is 1 in 1000000 (made up statistic) for everyone, my having won the lottery doesn't change that, therefore the chances of me being hit is still 1 in 1000000. Which is the probability of me getting hit? 1 in 1000000 or the likelihood of me winning the lottery and getting hit by lightning (which would be lower than 1 in 1000000).
    As another example that is related, would my reasoning be correct that if I purposely got myself into a serious car accident, but not fatal, I should be able to go back on the road with less fear as it is unlikely for me to get into two serious car accidents during my life time? This doesn't seem right, but technically aren't the odds of me getting into another serious accident lower because the odds of me getting into two accidents is really small?
    I don't know if that makes much sense as I didn't articulate my confusion very well, but I have been wondering about this for a while, so as a summary question: Does the probability of an event occurring change because of past, unrelated events?
     
    Last edited: Dec 30, 2013
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  3. Dec 30, 2013 #2

    Simon Bridge

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    Your example is an example of the gamblers fallacy ... i.e. if I toss five heads in a row, the odds of getting six in a row is so small that the next toss is more likely to come up heads.

    The event X you are interested in is the event that A and B both happen to you (you are both struck by lightning and win the lottery in your lifetime).

    You are really asking about two different situations ...
    situation 1. A and B have yet to happen P(X)=P(A,B)=P(A)P(B)
    situation 2. A has already happened, in this case: P(X)=P(A,B|A)=P(B) ... because P(A|A)=1

    The probability represents your knowledge of possible future events.
    This probability changes as your knowledge changes - which is what happens when an event goes from being possible in the future to being certainly in your past.
     
  4. Dec 30, 2013 #3

    Jow

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    That makes sense. However, it just seems to me that probabilities have no physical roots in reality (they are just quantities invented by humans to make predictions with imperfect knowledge) and therefore problems like the gambler's fallacy are more philosophical than physical.
     
  5. Dec 30, 2013 #4

    mathman

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    There is a small caveat. If you get into an accident, the probability you will get into another may be slightly reduced. You might drive more carefully after the first accident.
     
  6. Dec 30, 2013 #5

    D H

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    Another small caveat. If you win the lottery big time, you have a vastly increased probability of undertaking expensive hobbies such as golfing and yachting.

    One day you'll be playing golf with your other well-to-do buddies when a sudden storm breaks out while playing the 17th hole. You're not going to call it quits just for a little storm, are you? Of course not. You're going to finish the game. You finish the 17th and go to the 18th to tee off. You raise your spanking new titanium driver to tee off and … strike one!

    You recover and celebrate by taking your 40 foot sailboat out for a spin. When you see that ominous cloud building up, you'll take your boat back to harbor if you did what mathman suggested and learned a lesson about thunderstorms after the golf episode. If you didn't, well, thunderstorms *love* sailboats and their tall masts festooned with all kinds of metallic equipment.
     
  7. Dec 30, 2013 #6
    In your question, you ask if unrelated events affect probabilities. Well by definition, unrelated means that it's not related. So by definition, unrelated events will not affect the probabilities otherwise they would not be "unrelated".
     
  8. Dec 30, 2013 #7

    Simon Bridge

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    That's pretty much the same as with any of our mathematical models ... except that they usually assume perfect knowledge.
    For the models to make sense we also need to model the imperfections in our experiments and ourselves. We do that the same way we develop models for anything: empirically.
    The link between probability and physics is a fascinating one ... look up "inverse problems" for example ... or, perhaps more obviously, "statistical mechanics". The maths of probability is as real as any math we use in physics.
    http://home.comcast.net/~szemengtan/ [Broken]

    Now "what do we mean by real?" is philosophy.

    Giving advise on games of chance is actually what the branch of mathematics, was invented for. It gets it's rules from empirical observations - it is based in the real physical world of real physical cards and dice.

    To use the math, you work out the probabilities - and then you ask a question.
    The math is used to help answer the question. It is important to get the question right.

    The gamblers fallacy arises from the gambler not being careful enough with language when phrasing the questions - you can clear it up by stating the questions in clearer terms: which is what the notation is for and what I did in the reply to you.

    To that extent, the gamblers fallacy arises from philosophy rather than physics. (Notice the care with language here?) However, the embedded proposition (which turns out to be false), involves real testable predictions with real-world consequences that can and have affect real people's lives and the lives of those around them. In that sense it is as real as anything we study in physics.
     
    Last edited by a moderator: May 6, 2017
  9. Dec 30, 2013 #8

    Simon Bridge

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    Obvious isn't it? You may be wondering, therefore, why you are the first to point this out?
    The "by defnition" approach is probably worth pursuing further though...

    A quick reading of the body of post #1 suggests that the title is poorly worded and that Jow is actually commenting on a particular observation which can be illustrated with dice.

    If you have a fair 6-sided die, you are to roll it twice, and X="event you roll two sixes in a row";
    Then, initially: P(X)=1/36
    But, after one of the sixes has been rolled, P(X)=1/6
    Clearly P(X) has changed as a result of the past event.

    But we note that although X is composed of two independent events (that the die rolls a 6), X itself is not independent of the events it is composed of, therefore it is not surprising that the probability P(X) changes.

    If Y is the event that the die rolls a 6, the P(Y) does not change. P(X) changes depending on how many time Y has happened. i.e. P(X|Y,Y)=1 because X is the event that Y happens twice.

    The question in post #1 is answered.
    Independent events do not affect each others probability.
    The situation in the example does not involve independent events.
     
  10. Dec 30, 2013 #9

    Jow

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    Although I admit that my wording in the title was poor (I even said that I "didn't articulate my confusion very well" in my original post), I used the word "unrelated" to show that Event 1 doesn't actually have any bearing on Event 2 (winning the lottery doesn't actually cause someone to be more prone to lightning strikes, unless winning the lottery causes one to frequent golf courses on stormy nights).
     
  11. Dec 31, 2013 #10

    Simon Bridge

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    Correct ... however, you were not asking about how these events affect each other but how they affected the probability of both happening. You just didn't (appear to) realize it.

    [I was responding to johnqwertiful, who seemed to think you were asking about A affecting P(B).]

    In the body of your text you asked about how having won the lottery may make being struck by lightning less likely considering the chance of both happening is extremely unlikely.
    - the probability is unchanged:

    i.e.
    A = event that you are hit by lightning (at least once in your lifetime)
    B = event that you win the lottery (ditto)

    X = event that you win the lottery and also are hit by lightning.

    P(X)=P(A)P(B) since A and B are independent.

    once A has happened, P(X) changes:

    P(X|B)=P(A) > P(A)P(B) - again, because A and B are independent.

    P(A) is unaffected by the occurrence of event B.

    But X is not independent of A and not independent of B.
    Therefore, it is not surprising that P(X) is affected by the events.

    The result is actually to make X more likely, not less.

    But the probability of getting into two accidents if one of them is deliberate is more likely than the odds of getting into two accidents that are both, well, accidental.

    If the event you get into an accident by accident is A, and X is the event you get into two accidents, then:
    P(X)=P(A,A)=P(A)P(A) ... because the probabilities don't (hopefully) affect each other.

    But having already got into one accident ... the situation has changed:
    P(X|A)=P(A,A|A)=P(A) > P(X)

    The next accident is as likely as the last one, but the probability that you will have at least two accidents in your life has increased.

    What you have been doing is conflating X with A.


    Of course, I may have misread you and you had intended to ask a different question...
     
    Last edited: Dec 31, 2013
  12. Dec 31, 2013 #11
    If you already won the lottery, the probability of getting struck by lightning is the same as it always was.

    But, the probability that a person will win the lottery and get struck by lightning is exceptionally small. (The product of the two probabilities.)

    It's not because one event effected the other, it's because in the first case, the probability of winning the lottery does not matter because it already happened.

    The probability that I would be named Luke, have brown eyes, and be left handed, and get struck by lightning is smaller than just the probability of getting struck by lightning, but if the first three things already happened, why do we care about their probabilities when evaluating the current likely hood of all four occurring?

    And yes, this is exactly the gambler's fallacy. You can say the gambler's fallacy is "philosophical" or whatever you want, but it doesn't take a lot of thought to conclude that it is fallacious.
     
    Last edited: Dec 31, 2013
  13. Jan 1, 2014 #12
    There is a kind of fun Bayesian way in which observations of statistically independent events can affect your degree of belief that a similar future event would occur.

    Consider coin flipping. Let us have two hypotheses about the coin; either it is fair (H0), or not fair (H1) with some fixed bias (say towards heads) q. Regardless of which hypothesis is true, the coin flips are statistically independent events.

    Say I mostly believe hypothesis H0 (fair coin) to begin with. My degree of belief that the next flip will yield heads is thus pretty close to 0.5.

    Next, I flip the coin 10 times and it comes up heads every time. If my prior for H0 is not too strong then this evidence makes me now believe in hypothesis H1 more than H0, so my degree of belief that the next flip will yield heads is close to q.

    Thus observations of statistically independent events has affected my assessment of the probability of future events. This is the origin of the gambler's fallacy I think. If you play the pokies or something a lot, then without statistical training it can be easy to convince yourself that you are seeing patterns to how the game works; your degree of belief thus shifts towards the hypothesis that there is some bias you can exploit rather than the "fair" (or more likely house-biased) hypothesis. It can be perfectly rational given the available information, but most of the time there is information you could gain which would tell you that your perceived pattern is far more likely to just be noise.

    Likewise, if your scientific training is weak then your degree of belief in "good luck", "God watching over you" etc. may be influenced by a few random events here and there, and have a big impact on whether you think you'll be hit by lightning or win the lottery or go to heaven amongst a harem of virgins. It might even be rational, it is just that you have been unfortunate enough not to encounter or absorb the information which would apriori bias you against these hypotheses.
     
    Last edited: Jan 1, 2014
  14. Jan 1, 2014 #13

    Simon Bridge

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    The observation of an unrelated event does not affect the probabilities in the future - but does affect the probabilities of combinations of unrelated events. Beyesian statistics exploits this by being specific about how the probabilities are changed.

    Discussion of the coin example and the light it reflects on credulity:
    http://www.scribd.com/doc/48612323/A-False-Coin
    ... you could use figure 1 backwards to deduce you natural level of skepticism from the number of trials before you'd conclude the coin double-headed.
    Pitched at a lay audience with some (secondary school) knowledge of conditional probability.

    It should be stressed that Beyesian stats is about rational, quantifiable, belief - not the regular human kind you find in the street. This is maths not social anthropology.
     
  15. Jan 1, 2014 #14
    I'm not quite sure what you mean. To be technical, I was saying that the Bayesian posterior predictive probability for some event [itex]A[/itex] can change when you observe some other random event [itex]B[/itex], no matter how unrelated they are from the perspective of each model, because the posterior predictive probability is computed as a kind of model averaged quantity, which is what kills the statistical independence.

    e.g. even if
    [tex] P(A=a|B=b,M_k) = P(A=a|M_k) \;\;\; \forall k [/tex]
    We still have
    [tex] P(A=a|B=b,I) = \sum_k P(A=a|M_k,I) P(M_k|B=b,I) \neq P(A=a|I) [/tex]
    so long as
    [tex]P(M_k|B=b,I) \neq P(M_k|I) \;\;\; \forall k [/tex]
    (i.e. so long as the observation of [itex]B[/itex] actually affects the posterior probabilities of the hypotheses [itex]M_k[/itex]).

    (for some prior information [itex]I[/itex])
     
    Last edited: Jan 1, 2014
  16. Jan 2, 2014 #15

    Simon Bridge

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    I mean that the event that a previous coin turned up heads does not affect the probability that the next coin will turn up heads ... but it does affect the probability of turning up N heads by the time the experiment has concluded (a combination of unrelated events) - and it does affect the probability that the coin is fair, since that is also based on a combination of unrelated events.

    The probability the coin is fair is not independent of the results of the tosses even though the tosses themselves are independent of each other.

    No need to get technical ;)
    (I saved that for the link...)

    I think we've beaten the topic to death now...
     
  17. Jan 2, 2014 #16
    Ahh, but this is just the opposite of what I am saying, so long as you interpret "probability that the next coin will turn up heads" in the Bayesian way, that is, as "how confident I am that the next coin will turn up head". Since your beliefs about the properties of the coin change with every flip then so do your beliefs about future flips, since they depend on what properties you think the coin has.
     
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