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Probability question on flipping a coin

  1. Apr 23, 2005 #1
    I know that when flipping a coin for example with each flip you have a 50/50 chance of getting heads or tails. Each flip has nothing to do with the one before. This goes with any random number picking method. Like they say slot machines cannot be ready to hit (payoff). Isnt there though a probability of it hitting? Or like flipping a coin, even though it is 50/50 chance of heads or tails. It wont always land on say heads. After so many times of it being heads in a row. It seems that it would eventually have to be tails. The more times of it being heads in a row seems would increase the chances of it landing on tails. Like what is the maximum times a coin has been flipped and landed on the same side? Eventually it has to (or will) land on the other side. Like a slot machine can't (wont) not hit forever. After so many times of it not hitting. It will eventually hit. Or if you were rolling a 6 sided die what is the max number of times you would have to roll it to finally hit all 6 numbers? Is there an equation for figuring out probabilities like this?
    Last edited: Apr 23, 2005
  2. jcsd
  3. Apr 23, 2005 #2
    You contradict yourself and answer your question yourself too :

    "Each flip has nothing to do with the one before"

    "The more times of it being heads in a row seems would increase the chances of it landing on tails"

    So you have to choose : either it's a memoryless process or it is not (or you invent a kind of quantum process, up to you). However, you can imagine probabilistic models in which the probability of the next toss depends on the previous outcomes, but this has nothing to do with the basic axiomatic probability theory.
  4. Apr 23, 2005 #3


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    Everyday speech is almost never about 100% absolutes. If I were to say "You will get a heads if you flip a coin 40 times", I don't mean that there's a 100% chance of getting a heads in those 40 flips. I mean that the probability of not getting a heads is so vanishingly small that it's irrelevant. (In this case, it's less than one in a trillion!)

    This goes true for everything. There are some obvious ones, like "When I flip this switch, the light comes on". (Of course, I think it's only obvious because this has happened to most people) There are less obvious ones like "When I ask my computer to compute 1+1, it will give me 2".

    In fact, there is a very small probability that your computer won't give you 2, but it's so small it's irrelevant for most people.

    But, when you're doing a heavy-duty numerical calculuation, you have to have some way of detecting these errors. Similarly, there's no point in having an algorithm that has less chance of giving you a wrong answer than the chance of getting a hardware error!

    This leads to another thing about probabilities that many people get wrong -- when you have the potential for a very large number of trials, things that were vanishingly small can become not so small.

    For example, you should never have to worry about something that has a 1 in a trillion chance of happening to a person each day, right? Well, it happens 6 times every 3 years!
  5. Apr 23, 2005 #4
    You can decide beforehand what outcome you want the probability for, and then calculate that probability. I can give you the probability for getting 50 heads in a row then 1 tails. It's equal to (0.5)^51.

    For a dice of there is no max number because in principle, I could keep rolling forever and never get all six numbers. For n rolls, the probability of not getting a "3", for example, in any of the rolls would be (5/6)^n. This probability becomes tiny when n is large.

    And if you were to win the lottery today (1 in 40 million, say), you would still have just a good a chance of winning the lottery tomorrow as anyone else.
  6. Apr 24, 2005 #5
    Yes, but it's the same as having 51 times head....so the question asked at first has the same answer : even if you get 50 times head, you have equal chance to get head or tail at the next trial...(Since this was your hypothesis of work using probabilities, it will remain at the end)
  7. Apr 24, 2005 #6
    I think Juvenal that you answered it, or partly. I'll a ask different way to make sure I'm right.

    You are flipping a coin. I know with each flip you have an equal chance at getting heads or tail. But does the number of times you get one side in a row in some way increase the chances of getting the other side?

    For example (I'll use made up numbers cause I have no idea what they would be.) Say the chances of flipping 50 heads in a row is 1/100...60 =1/500...
    100 =1/100 etc. You are getting ready to roll your 100th roll. The previous 99 have been tails. Do you have better odds/chance to get heads the next roll? What would you bet on?
  8. Apr 24, 2005 #7
    You answered the question yourself. You just said "each flip you have an equal chance at getting heads or tail". So why would it matter what you bet on?

    The probability of getting 99 tails + 1 head = (0.5)^99 * (0.5).

    The probability of getting 99 tails + 1 tail = (0.5)^99 * (0.5).

    Those values are equal. They are tiny of course, since they each refer to only one possible outcome out of 2^(100) = 1267650600228229401496703205376.

    Another way to think about it is this: Suppose you have a hundred coins. All of them are fair. You flip 99 of the coins and get 99 tails. Now for the last, unflipped coin, I can always relabel the sides of the coin. I can change heads to tails and tails to heads. Or I can call one side "antelope" and the other side "zebra". It doesn't matter. Either side has a 50% shot of coming up. In fact, for the previous 99 rolls, I can relabel each side of every coin to get any combination of heads and tails. The definition of heads and tails from coin to coin is completely arbitrary and meaningless, so fretting over whether the last coin ends up "heads" or "tails" is pointless.

    You understand why it's just as likely to get consecutive numbers in a lottery as non-consecutive numbers, right? You can use the logic I described above to make it clear - just change your numbering scheme so that consecutive numbers are no longer consecutive.

    I would suggest reading a basic textbook on probability. Things will become clearer then if you are still not convinced, I hope.
    Last edited: Apr 24, 2005
  9. Apr 24, 2005 #8
    I understand that. I don't think anything I stated implied otherwise.
  10. Apr 24, 2005 #9
    No, I was just using your calculation to answer the original question, so it's directed to UglyEd.

    If UglyEd wants to follow his intuition, he can write something like : p(next_head)=.5*(1+alpha*n(previous_head)/n_total).

    Where alpha is a free parameter in [-1;1]. This models an orderless memory process...you can then complicate as you want your model to make the process depend on the previous flip (If you believe there is such a behaviour of course, which is a strong hypothesis)
  11. Apr 24, 2005 #10
    Thanks juvenal, and kleinwolf. I had know that it was a basic rule that each flip is independent of any others. I had just thought that maybe in some way although the coin doesnt know the difference. After so many times of the same side coming up you would over all have a better chance of getting the opposite side. Since getting 100 heads in a row would probably be very rare. So does that mean the chances of getting 50 heads in a row is the same as getting 100 in a row? Or even though the chances of getting 100 heads in a row is less than getting 50 heads in a row it doesnt matter? Or could you say. Even though the chances of getting 100 heads in a row is 1/1000 you have a 50/50 chance of getting that 1/1000?
    Last edited: Apr 24, 2005
  12. Apr 24, 2005 #11
    You have asked 2 different question, so there are 2 answer, separated in 4 approaches :

    1) axiomatic school exercise : given p(head)=p(tail)=.5

    a) p(50_head_in_a_row)=.5^50>>p(100_head_in_a_row)=.5^100

    b) however p(51_head¦50_head_in_a_row)=p(51_tail¦50_head_in_row)=.5

    2) experimental approach

    The paper mathemetician said : p(head)=p(tail)=.5

    However, the experimentalist sees that n(head_over_50_trials)>>n(tail_over_50_trials)

    He deduces, as he is free to do that : "Hm, experimentally, my coin is never perfect, so I use numerical probabilities out of some preliminary trials"

    so that the experimentalist is free to deduce p(51_head¦50_head_in_a_row)>p(51_tail¦50_head_in_a_row), because the coin is in fact not fair.

    So for him : getting a head more is more probable than a tail more after 50_head_in_a_row, because he believes the coin is not experimentally fair.

    3) intuitive link to the axioms :

    hypothesis : p(head)=p(tail)=.5

    Theorem : There cannot be only head results.

    Proof : If there are only head results, then [tex]p(head)=lim_{n\rightarrow\infty}\frac{n(head)}{n}=1[/tex]. which contradicts the hypothesis.

    This is the only conlusion you can draw from probability axioms.

    4) You take the probabilistic model :


    alpha=0 : p(next_head)=.5 independent of the previous outcomes

    alpha>0 : if you got more head than tail previously, then the next tail is more probable (reason : the coin is in fact not fair)

    alpha<0 : if you got more head than tail previously, then you have more chance to get tail in order to recover the inependent probability. (reason : there is no reason)

    On average over alpha, p(head)=.5
  13. Apr 24, 2005 #12
    Once again - it would be good for you to learn basic probability. Then you can calculate the probabilities yourself and be convinced. Are you really reading what we are writing? It seems to be going completely past you, since you keep asking the same question over and over.

    Getting 100 heads in a row is harder than getting 50 heads in a row. That's because (0.5)^(100) is much much smaller than (0.5)^(50). But getting 99 heads and 1 tail is also harder than getting 50 heads in a row because that probability is also (0.5)^100, which is much smaller than (0.5)^(50).

    How about 100 alternating heads and tails? Yes, it's (0.5)^100. How about 3 heads, followed by 2 tails, repeated 20 times so you get 100 flips. Yes, it's (0.5)^100. In fact, any one sequence of 100 flips that you can choose arbitrarily has a probability of (0.5)^(100). Any one sequence is extremely rare.

    Compare 1, 2, and 3 coin flips:

    1 coin flip: 50/50 chance of getting H/T
    2 coin flips: 1/4 = 0.25 chance of getting HT, TH, HH, TT.
    3 coin flips: 1/8 = 0.125 chance of getting HHH, HHT, HTT, HTH, THH, THT, TTH, TTT

    So for two coin flips, the probability of getting two heads in a row is 0.25. For three coin flips, the probability of getting 3 heads in a row is 0.125. However, as you see, the probability for getting 2 heads, 1 tail is also 0.125.

    Why don't you take a penny, and try to get 4 heads in a row by flipping? Then see what the fifth flip gives you. Then repeat this exercise a lot of times (flip until you get 4 heads, then flip once more to see the outcome). Try it with more heads in a row if you have the patience, or write a computer program to model this.
    Last edited: Apr 24, 2005
  14. Oct 2, 2010 #13
    Trying to judge the outcome of a coin toss based one previous outcomes is like trying to navigate a plane by it's jet stream.
  15. Oct 2, 2010 #14


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    This thread is 5 years old.
  16. Oct 3, 2010 #15
    So should we sing it happy birthday?
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