Sunday, 23 September 2012

Probability for Wargamers 2 - The Gambler's Fallacy

I wasn't expecting to have to write this one, but it appears I do :D

The most common misconception about probability is SO common it has a name - The Gambler's Fallacy. It's the mistaken belief that past independent events affect future ones.

To explain it with an example: suppose you toss a coin five times and it comes up heads five times. Assuming its an unbiased coin, what is the probability of the next coin being a head?

The amusing thing here is the gambler's fallacy manifests itself in two ways here, depending on the person concerned: some people say 'it's bound to be another head', and some people say 'tails are due'. Both are wrong - this is, in fact, entirely irrational behaviour.

What if I'd tossed the coin in private, and it had come up heads five times in a row, and I hadn't told you, and asked you to say whether heads or tails was more likely? The odds are still evens, 50%, call it what you will. The coin doesn't have any memory of its previous tosses.

It does rear its head in more subtle ways: let's go back to the previous post. The chance of not getting a 3+ when rolling 3 dice (note, it doesn't matter if you roll three dice together or one after the other, or the same dice three times) is 3.7%. The chance of succeeding is therefore 96.3%.

What if you throw three dice one after the other, and fail on the first dice?

I'm pretty sure some people's (conscious or not) thought process would be "that's alright, Mike said I have a 96% chance of success on three dice, I'm still good!".

Wrong!

You're not. You now have to make 3+ on one of two dice. So the probability of failing to do so is the chance of rolling a 1 or 2 twice, which is 1/3 * 1/3, or 11%. Your chance of success has dropped to 89%.

10 comments:

  1. The probability of getting 3+ in three rolls is precisely the same whether they are rolled simultaneously or one after the other, stopping when you achieve the result.
    P(success) = 1 - P(failure) = 26/27

    But the rolls in sequence can be expressed as a SUM of probabilities, taking each in turn:
    P(success) = P(hit on the first roll) + P(hit on the second roll GIVEN THAT YOU MISSED ON THE FIRST) + P(hit on the third roll GIVEN THE PREVIOUS TWO ROLLS WERE MISSES). Now the second and third terms of the right side of this equation are in fact products of probabilities, as appears in the following equation.
    P(success)
    = P(hit on the 1st) + P(hit on the 2nd) x P(miss on the first) + P(hit on the 3rd) x P(miss on the 2nd) x P(miss on the 1st)

    =2/3 + 2/3.1/3 + 2/3.1/3.1/3
    =18/27 + 6/27 + 2/27
    =26/27 ... as above.

    If you miss on the first roll, that probability collapses: you now in effect have zero chance of rolling 3+ on the first roll, since we have established that you didn't (does that make sense?). But you still have 2/3 chance of success on the 2nd roll, and if that carks out, 1/3.2/3 chance of success on the third.

    Adding gives you
    P(success after missing on 1st roll) = 2/3 + 1/3.2/3 = 2/3[1+1/3] = 2/3.4/3 = 8/9 ~ 89%

    And of course, if you miss on the first two, the probability of hitting on either of those rolls collapses to zero, and you are left with 2/3 of a chance on you final roll - 67%.
    Cheers,
    Ion




    ReplyDelete
    Replies
    1. It is, as I think I noted earlier, easier to note that P(succeeding) == 1 - P(failing all three).

      There's a reason they work out to the same :D

      Delete
  2. I once correctly called the coin toss seventeen times in succession. What are the odds of that? (It totally freaked-out the guy tossing the coin! =)

    ReplyDelete
    Replies
    1. The odds are the same as getting it wrong 17 times but I'm pretty sure the tosser (oh, come on, I couldn't leave that pun just lying there) wouldn't have raised an eyebrow over that :-)

      Delete
  3. That was one of the things I was trying to demonstrate.

    ReplyDelete
  4. AJ The probability that the coin toss can be predicted 17 times in succession might be regarded as 100% - certain - as it's bound to happen sooner or later.

    But seriously, on that occasion, the probability would by 1 chance in 2 raised to the power of 17. That is 1 in 131,072 or 131071:1 against.

    But remember Granny Weatherwax's Law of Probability: Million-to-one shots crop up nine times out of ten. :-)

    ReplyDelete
  5. Heads


    Heads


    Heads


    Heads


    Heads


    Heads


    Quote - Tom Stoppard - Rosencrantz and Gildernstren are Dead (scene 1)

    ReplyDelete
  6. I thought I'd leave off replying to this one for a while.

    After all...

    ...there is an art to the building up of suspense.

    ReplyDelete
  7. I've enjoyed these posts and comments. I have a least one friend who moans about dice luck and the dice gods being against him who should visit here and read these posts.
    Cheers,
    Mike

    ReplyDelete
  8. As a corollary, a gaming buddy of mene hates opposed die rollsfor combats. "Even with great odds, it's no good if you roll a 1 and your opponent rolls a 6" he says. I say it's the same as if you roll 2xD6 and get a total of 2. I even re-wrote a CRT to show it. He still wouldn't have it

    ReplyDelete

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