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Probability of a coin flip

  1. Oct 22, 2011 #1
    1. The problem statement, all variables and given/known data

    Say we do 4 coin flips. If the order matters, of which coins come up, total number of possibilities are 2*2*2*2=24 (i imagine this like getting 16 different binary numbers)

    Did I got this right?

    Now If we want to find the probability to get a combination with 2 heads and 2 tails(not with that order, just a combination that has 2 heads and 2 tails).

    [itex]P(2,2)={\Large \frac{4!}{2!\cdot 2!\cdot 2^{4}}}[/itex]

    1 H H H H
    2 T H H H
    3 H T H H
    4 H H T H
    5 H H H T
    6 T T H H
    7 H T T H
    8 H H T T
    9 T H H T
    10 T H T H
    11 H T H T
    12 T T T H
    13 T T H T
    14 T H T T
    15 H T T T
    16 T T T T


    This is analogues to getting a binary number that has exactly 2 zeroes and 2 ones, right?

    Now for the second part:

    Order DOESN'T matter. Say we toss a coin 4 times again. Total number of possibilities are:

    H H H H
    H T T T
    H H T T
    H H H T
    T T T T

    So getting precisely 2 heads and 2 tails, is 1/5, right?

    But how do i calculate the possibilities when order doesnt matter? Other than writing all the possibilities by hand and counting them?

    2. Relevant equations
    3. The attempt at a solution
     
  2. jcsd
  3. Oct 22, 2011 #2

    LCKurtz

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    No, that isn't right. Those 5 outcomes are not equally probable. This problem is best modeled with the binomial distribution. Have you studied that? Here you have n=4 trials with probability p of success (= H) 1/2.
     
  4. Oct 22, 2011 #3
    I haven't studied binomial distribution yet.

    Can you explain this a bit?

    Did I get the number cases right?

    Talking about the one that order doesn't matter.
     
  5. Oct 22, 2011 #4
    Wait a second. That first part that i presented, that with 16 cases. Can I take all the cases that have exactly 2 head and 2 tails, and calculate that probability. Is that the probability to get 2 head 2 tails, without mattering order?

    [itex]P(2,2)={\Large \frac{4!}{2!\cdot 2!\cdot 2^{4}}}[/itex]

    That is my answer?
     
  6. Oct 22, 2011 #5

    LCKurtz

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    There are 4 slots. There are[itex]\binom 4 2 = 6[/itex] ways to choose the two slots that will be heads. So, as you can see from your first list, 6 out of the 16 equally likely outcomes have two H. So the probability is 6/16 = 3/8.
     
  7. Oct 22, 2011 #6
    So I got that right :D omg. Thank you very much. I did a simulation in matlab and it matches. That was the main point of this exercise, to get better in Matlab (probability and statistics simulations)
     
  8. Oct 22, 2011 #7
    It works! I translated this into matlab, and got 37669/100000=0.37669 !! that is correct even to 2 decimals.

    So you got me interested with this binomial distribution.

    But now I have a more complicated case.

    You are rolling a pair of dice, 3 times. What is the chance of getting a pair of fours(4) at least one time.


    I did the simulation in matlab and I got the chance 0.0826.

    Can you help me, with just this case, so I can get a general rule for all other cases.

    Are these combinations with or without repetition, you gave binomial coefficent ,and I dug up on the internet that those are combinations without repetition. But how come its without repetition, and we are repeating H H T T

    Reason why I thought that those were permutations, is this:

    http://regentsprep.org/Regents/math/ALGEBRA/APR2/LpermRep.htm

    i simply told myself my elements are T T H H. How many permutations can I get out of that, if 2 are repeating.

    But I see now that 2 mistakes cancelled each other and got the result right.

    So in this second problem, I have to get at least one pair of fours, I can get 2 pairs, 3 pairs, just 1 pair is minimum.

    But here order does matter right? Because I can get a pair of fours on the first, second and third roll? So these would be permutations?!

    I am really sorry, but I am struggling here to get the intuition for this :(
     
    Last edited by a moderator: Apr 26, 2017
  9. Oct 22, 2011 #8

    HallsofIvy

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    The probability of getting a pair of fours with two six sided dice is (1/6)(1/6)= 1/36.
    The probability of NOT getting a pair of fours, then, is 1- 1/36= 35/36. The probability of NOT getting a pair of fours when you do that three times is [itex](35/36)^3[/itex].
     
  10. Oct 22, 2011 #9
    Omg, I give up, where did you learn all this?! Can you give me a book recommendation ? Please? :(
     
    Last edited: Oct 22, 2011
  11. Oct 22, 2011 #10

    gb7nash

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    Have you learned the multiplication rule?
     
  12. Oct 22, 2011 #11
    Yes I did!! And bayes' theorem and all that good stuff, independent dependent events etc etc.

    But I cannot get a grip on this field!

    You see this dice roll, I was completely focused to get it solved like the coin flip, with the same rule, etc. I need to exercise a lot of this, from some good book.

    Can we solve the first problem with multiplication rule?

    Theory itself isn't the problem, just the timing when and how and where to apply it, is.

    PS I love your signature :D
     
    Last edited: Oct 22, 2011
  13. Oct 22, 2011 #12

    HallsofIvy

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    Wow. Sounds like you just skipped all of the basic stuff and went straight to the deeper stuff. Are you didn't accidently sign up for the second semester of a two semester course?
     
  14. Oct 22, 2011 #13
    Well, here is the story.

    Name of the course is Information Theory and Source Coding. We had a BRIEF into into probability. No permutations, no combinations, no nothing. We went after 3 lectures straight into PDF, CDF etc.

    But we did some examples on our recitations with coins and probability, but only 2 or so. Way less, than I would like. Because I have almost zero knowledge of probability. Funny thing is, that I found pdf, cdf, pmf way more easy to understand.

    But nevertheless, we are given some problems to solve, test and simulate them in matlab, to see if they are ok. Now matlab part for me is no big deal. But the math behind it is.

    We did some examples on stability of a system, but I didn't realise that was connected with this until now.

    My last exercise is:

    Bingo Plus game is consisted from drawing balls, marked from 1 to 10. 5 balls are drawn and the winner has to have all 5 balls. What is the probability of winning a jackpot if the winning combination is 848235.

    All I see is that I have 6 numbers in the winning combination, and 5 balls to be drawn. All I can say is that, its going to be a long night for me :/
     
    Last edited: Oct 22, 2011
  15. Oct 22, 2011 #14

    Ray Vickson

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    It is worth clearing up your misconceptions right at the beginning. In coin-tossing, dice throwing, etc., we often consider identical coins or dice. That means we do not distinguish between the coins or the dice---BUT WE COULD IF WE WANTED TO. For example, you could use different colored dice, or put distinguishing marks on the coins, etc. In such a case, with 4 coins there are 16 different outcomes (you listed them above), but if we are interested only in how many heads or tails occur (not their order), then you get the probability by adding up all the individual probabilities for that number of heads/tails.

    Interestingly, for "quantum-mechanical" coins or dice, if you have so-called Bose-Einstein objects (i.e., Bose-Einstein coins) your 5-outcome counting method above, each with probability 1/5, would be CORRECT. (In the early days of modern Statistical Mechanics that type of non-classical counting came to be realized, and was important in getting agreement between theory and experiment.) However, we always deal with "classical" coins or dice in probability and statistics, so we always assume they are distinguishable in principle.

    Of course, if you have just one coin tossed several times, we automatically have distinguished tosses (toss 1, toss 2, etc.) and so the same "identical but distinguishable in principle" counting method applies.

    RGV
     
  16. Oct 22, 2011 #15
    I think I understand what you are trying to say. I will try to put that in use just now. Thank you!
     
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