18 June, 2008

A Probability Problem

A little problem in probability for all (four) readers of this blog. It's just something I'd wondered about long ago, and I think I have the solution. Probability is often a very confusing subject, as you can see in the Monty Hall Problem.

Well, let's say I have a die one coin, which I decide to toss again and again, 100 times, and count the number of heads each time. I'd say I'll get 50 heads and 50 tails on the whole.

Suppose I have already tossed 99 times, and I got heads EVERY time. (Hey, it's possible.) Now if you were to bet on what I'd get the hundredth time, what'd you say? There are three possible answers:

a) I won't get heads again, because, well, luck has favoured me this far, but it may not favour me again.
b) I will get heads again, because, hey, I'm on a roll.
c) You refuse to bet, because the new toss is independent of the others. Or maybe just because both a) and b) seem equally convincing.

Well, we know c) is right. Problem is, why is it right? I mean, sure, independent toss and all, but isn't it even MORE improbable that I get 100 heads one after the other? Also, please don't accuse my rupee coin of being biased. It's a little sensitive these days.

So, my question to you is: give me a concrete reasoning WHY the probability on the 100th toss too is 50-50. (Maybe you don't agree it's c). Do say so. ) Double points if you can give some calculations. Triple if you can drag in Gauss. (More accurately Poisson. Actually, their curves. I mean 'distributions', dude.)

P.S. Of course you're to answer in Comments. You can also comment in comments. Also, if you have given an answer, I'll only approve the comment after I feel everyone's done the best they can. If you do find something in the comments section, it may be a spoiler.

Hint: Read my first comment below (beneath the ones by Saskia, Anne and Jayanth).

8 comments:

  1. Traditionally, dice have 6 sides, none of which shows either a head or a tail.
    But ok, so you meant coin.
    Now, look carefully at a coin. On most coins, indeed, there is a head on one end. That too, is traditional.
    Now, turn the coin over.
    Do you see a tail?
    Ok, take another coin. Do you see a tail?
    You can do this hundreds of times, and you will find that coins actually seldom have tails, although there is an obscure square African coin that supposedly features the tail of a rhino on the other side of the head of the despot that happens to be head of state. This is, however probably just a rumour.
    So, my dear young friend, here is your answer: you will get head again, as coins just don't have tails.
    [Bows]

    ReplyDelete
  2. I had EXACTLY the same concern! In fact, I formulated the problem in EXACTLY the same way too! The way I see it is, as the number of tosses approaches infinity, the distribution must be 50-50. And, 100 is a long way from infinity!

    Took me a while to reconcile with this!

    ReplyDelete
  3. Saskia, lol.

    Jayant, but how do I reconcile myself to the fact that there is an equal probability of getting 99 heads and then one tail, and getting 100 heads in succession? That is my question.

    You have to use the independence of all 100 of the coin tosses to do this, so it follows from the independence condition as in c). But it's a way to see that the equal probability of getting heads or tails in the 100th toss is not in contradiction with the fact that 100 heads is much more improbable than 99 heads.

    If you look carefully, I have already given a hint in the above paragraph. It's a very subtle change of wording.

    ReplyDelete
  4. The premise is ambiguous.

    The result depends on whether you're looking for the marginal probability or the conditional probability.

    If we're talking about marginal probability, the odds of getting a head on the 100th toss is 50-50

    If we're talking about conditional probability, the odds of getting a head on the 100th toss is, if my analysis is correct- 1 in 2^100, which is approximately 1 in 1.2676506 × 10^30

    I'll try and drag in Poisson now-

    If we use a Poisson distribution to approximate this problem, we find that the conditional probability of said event is ((50^99)*(e^50))/(100!), which is approximately 1 in 8.76497879 × 10^31, which implies a deviation of 1.4462677% from the erstwhile manual analysis.

    Using the Poisson distribution to find the marginal probability isn't a good idea, as it is ideal for large numbers. Nonetheless, the calculated probability is (.5^1)*(e^.5)/(1!) which is approximately 0.824360635, which implies a deviation of our manually calculated value of .5 (50-50)

    I might be wrong for two reasons-
    1. I hate probability.
    2. I suck at numbers.

    ReplyDelete
  5. The premise is ambiguous.

    The result depends on whether you're looking for the marginal probability or the conditional probability.

    If we're talking about marginal probability, the odds of getting a head on the 100th toss is 50-50

    If we're talking about conditional probability, the odds of getting a head on the 100th toss is, if my analysis is correct- 1 in 2^100, which is approximately 1 in 1.2676506 × 10^30

    I'll try and drag in Poisson now-

    If we use a Poisson distribution to approximate this problem, we find that the conditional probability of said event is ((50^99)*(e^50))/(100!), which is approximately 1 in 8.76497879 × 10^31, which implies a deviation of 1.4462677% from the erstwhile manual analysis.

    Using the Poisson distribution to find the marginal probability isn't a good idea, as it is ideal for large numbers. Nonetheless, the calculated probability is (.5^1)*(e^.5)/(1!) which is approximately 0.824360635, which implies a deviation of our manually calculated value of .5 (50-50)

    I might be wrong for two reasons-
    1. I hate probability.
    2. I suck at numbers.

    ReplyDelete
  6. This my friend is called chaos theory.

    U are trying to link up individual events(99 coin tosses) to the overall result of the experiment(100th coin toss) ...

    It cannot be random ... it has to be chaos. But none the less its it not a certain dynamic system.

    Chaos Theory states :
    n mathematics, chaos theory describes the behavior of certain dynamical systems – that is, systems whose state evolves with time – that may exhibit dynamics that are highly sensitive to initial conditions (popularly referred to as the butterfly effect). As a result of this sensitivity, which manifests itself as an exponential growth of perturbations in the initial conditions, the behavior of chaotic systems appears to be random. This happens even though these systems are deterministic, meaning that their future dynamics are fully defined by their initial conditions, with no random elements involved. This behavior is known as deterministic chaos, or simply chaos.

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  7. I'll take option c
    Well the only reason is i see no point in betting, especially after watching '21', since we cannot use the principle of variable change here. In the movie '21', the problem that Prof. Mickey Rosa puts forward is different, in a sense that, he tells us we have a car behind three doors, and you are to choose a right door. After revealing a door behind which thrz no car, we know that the it is in our interest to switch our original choice. But in that case we already know that thrz gonna be a car behind one of the doors. Here, we donot know whether there will be at least one heads or no heads at all. so even if we hit 99 heads, the probability still remains 50-50 that we hit 100

    ReplyDelete