Coin toss question

Discussion in 'Probability' started by Dave Aronson, Nov 15, 2011.

  1. Dave Aronson

    Dave Aronson Guest

    Maybe this is a stupid question. For a fair coin the probability of 10
    heads in a row is (0.5)^10, right?

    But can we say that as the number of tosses approaches infinity the
    probability of ever getting a 10 head streak approaches 1?
    Dave Aronson, Nov 15, 2011
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  2. Dave Aronson

    danheyman Guest

    yes! Any event with positive probability will occur w.p.1 in
    infinitely many independent trials. Let p>0 be the probability of the
    event. The probability that it doesn't occur in n trials is p^n which -
    danheyman, Nov 15, 2011
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  3. Dave Aronson

    Dave Aronson Guest

    Did you may mean to say that the probability that it doesn't occur in
    n trials is (1-p)^n ?
    Dave Aronson, Nov 25, 2011
  4. Dave Aronson

    danheyman Guest

    danheyman, Nov 26, 2011
  5. Dave Aronson

    Dave Aronson Guest

    Thanks. The reason I asked is because I read a paper on gambler's ruin
    I found in math pages. The link is here:

    It turns out from the first result in the above link that the
    probability of ruin is (1 - prob of reaching N before reaching h ) and
    as N goes to infinity, for r < 1, it is equal to r^h, where h is the
    starting bankroll (r < 1).

    I find this fascinating because if what we discussed holds for
    infinite trials and we define as r^h < 1 the probability of some
    event, in this case the probability of gambler's ruin, as the number
    of trials approaches infinity then ruin is inevitable although the
    bankroll of the gambler is becoming infinite.

    or am I missing something here?
    Dave Aronson, Nov 26, 2011
  6. Dave Aronson

    cjq70 Guest

    Yes, you're getting confused over limits. The link you cited says the
    probability of successfully reaching N is (1-r^h)/(1-r^N), which is 1
    when h = N. So if you set both N and h equal to any arbitrarily large
    number Q, the probability of ruin is zero, and this remains true as Q
    increases. The probability of ruin is (r^h - r^N)/(1 - r^N), and if
    you let both N and h go to the same infinitely large number, the
    numerator is zero. Also, r^h goes to zero as h increases (assuming r
    is less than 1).
    cjq70, Nov 26, 2011
  7. Dave Aronson

    Dave Aronson Guest

    Thanks. That is what I though too until i read the following argument.
    The limit in this case is related to h and N. But we can also have a
    limit on the number of trials this happens, like when flipping coins.

    Let p be equal to the probability of ruin with p < 1, when N is
    larger than a very large number Q. Then as Q increases the probability
    of ruin converges to r^h < 1, if r < 1. Now, given that N is already
    large, the number of trials n increased to infinity, the probability
    of not getting ruined is (1- r^h)^n ---> 0, i.e. ruin is certain.
    As I said, there is the limit w.r.t. N but also a limit w.r.t n when N
    is already large.
    Dave Aronson, Nov 27, 2011
  8. Dave Aronson

    cjq70 Guest

    You're still confused. The gambler begins with h dollars, and wants
    to know the probability of reaching N before going broke, which for
    sufficiently large N (and r<1) is approximately (1-r^h). You're
    describing a completely different game, in which, if the gambler
    reaches N, he gives back all his gains and starts over with h dollars
    for a second "trial". And if he reaches N again, he gives back all
    his gains and starts over again with h dollars for a third trial, and
    so on, up to n "trials". The probability that he successfully reaches
    N in every one of these n trials is (1-r^h)^n, which obviously goes to
    zero as n increases. That's trivial and self-evident. For any game
    in which our probability of winning is less than 1, it's obvious that
    if we keep playing enough times, we will eventually lose a game.
    That's basically the definition of the probability being less than 1.

    This has nothing in particular to do with the gambler's question,
    because our gambler doesn't arbitrarily give all his winnings back and
    start over at h each time he reaches N. More interesting is the
    opposite point, i.e., even if we set N to some arbitrarily huge number
    (essentially infinite), the gambler still has probability of never
    being ruined, assuming r is less than 1. So he has some positive
    probability of playing forever and never going broke. The inevitable
    ruin occurs only with r = 1 or greater.
    cjq70, Nov 27, 2011
  9. Dave Aronson

    Dave Aronson Guest

    Has it crossed your mind that you may not understand what I am getting
    into after all?
    No, that is not what I am describing. This is something you are making
    up and it is not related to my point at all.

    Think of it in another way. The game progresses as the number of
    trials increases. At some point the bankroll is very large but also
    the number of trials is very large. After all, how large is infinity?
    When does the gambler stop and say now I have an infinite bankroll?
    Since inf + inf = inf. the gambler many not stop soon enough and in
    the meantime, his probability of ruin will become 1 and lose it all in
    the first round.
    I answered this above. This is possibly ralated to the paradoxes of

    I think this is not right. Ruin is inevitable in the limit of infinity
    for any r.
    Dave Aronson, Nov 27, 2011
  10. Dave Aronson

    cjq70 Guest

    If we say N is "infinite", then it is basically a random walk,
    beginning at h, and the probability that he will reach zero is r^h.
    This means that he has prior probability 1-r^h that he will NEVER go
    bankrupt. This is only possible if r<1, which implies he has greater
    than 50/50 odds of winning in each round. This shouldn't surprise
    No, you're wrong. Suppose you start a walk at h steps away from zero,
    and on each step you have a 90% chance of stepping forward and a 10%
    chance of stepping backwards. Do you think you will eventually get to
    zero? The answer is obviously no, but you seem to think the answer is

    I suspect you're confused by the fact that if we are, say, 100 steps
    away from zero, there is a finite probability of getting 100 backward
    steps in a row, so if we keep walking, eventually we will get a string
    of 100 backward steps, and return to zero. But that's overlooking the
    fact that soon you will most likely be at 200, so you really need 200
    backward steps in a row, and then a little later you need 2000, and so
    on. As you continue walking, your distance from zero almost certainly
    is increasing, so by the time you eventually get that 100 backward
    steps in a row, you will almost certainly be a trillion steps away
    from zero, so it won't bring you back to zero. Granted, you will
    eventually get a trillion backward steps in a row, but by that time
    you will almost certainly be a gadzillion steps from zero, and so on.

    This is all taken into account by the gambler's ruin formulas with N
    set to "infinity". Starting from h (with r<1), the probability that
    you will NEVER reach zero is 1-r^h. You will just keep getting
    further and further from zero, i.e, the gambler's winnings will keep
    increasing forever.
    Yes, it is exactly what you are describing. You talk about "trials",
    but there are no "trials" in the gambler's game, there is only one
    game that starts with h dollars and ends when he either goes broke or
    reaches N. If you want to talk about a case where there is no N, and
    he is going to play indefinitely, then you would just take the limit
    of the probabilities as N goes to infinity. So the probability of
    eventually going broke is r^h.

    But you are talking about a completely different game, involving n
    "trials". You haven't actually defined a "trial", but from what
    you've said, your "trials" can only be repetitions of the entire
    gambler's game, in which he starts with h dollars and either he
    eventually goes broke or he never goes broke. So you can think of n
    gamblers, each playing this game, and the probability that NONE of the
    n gamblers EVER goes broke is (1-r^h)^n.

    No, the gambler's game consists of him starting with h dollars, and it
    ends when he either goes broke or reaches N. If you want to talk
    about playing with no N, then he either eventually goes broke or else
    he never goes broke. Both of these are possibilitries. The "trials"
    you talk about are for an entirely different game that you invented
    yourself, where there are n gamblers, each starting from h, and you're
    asking for the probability that at least one of them will go broke as
    the number of gamblers goes to infinity. (Yes, this is what you are
    describing, you just don't realize it.)
    No, you're completely confused. Each of your "trials" has to begin
    with h dollars, because otherwise you can't say that the probability
    of not going bankrupt on that trial is 1-r^h. There are no separate
    "trials" in the gambler's game. If you are talking about the bankroll
    continuing to increase, then you are not starting a new trial, you are
    simply continuing the gambler's game, and the probabilities for that
    are already known. There is no "n" in the gambler's game.
    cjq70, Nov 27, 2011
  11. Dave Aronson

    cjq70 Guest

    I meant to say you wouldn't *necessarily* get to zero. You have a
    probability of never reaching zero.
    cjq70, Nov 27, 2011
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