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Coin Tosses

13 March 2009 10 Comments
Coin Tosses

My buddy Dan has a post on probability here [LINK]. I confess I never fully grasped probability in this sense. With a single coin toss heads has 50% chance and tails has 50% of coming up. It follows that with 10 coin tosses you’d expect 50% heads, 50% tails, and as things tend to an infinite number of tosses, actual results get closer to 50/50.

But what are the chances of getting 10 heads in a row? Fairly low, right? (how low I’ve no idea). So if I do 9 consecutive coin tosses and each time I get a head, surely that means the odds of the 10th coin toss being a HEAD are pretty low and of being a TAIL have dramatically increased beyond 50%?

When you look at the toss in isolation it’s 50/50 but when taken as part of a series, do the preceding results have bearing on the outcome of the next throw?

Dan plz explain.

10 Comments »

  • jim shanahan said:

    Francis – I can answer that question. Probability of an event will always remain the same, but the possibility of a certain outcome changes depending on the previous outcomes. This is easy to prove mathematically. See http://saliu.com/Saliu2.htm
    Jim

  • Dan said:

    First things first, assuming you’ve got a fair coin (i.e. the slight differences in texture and weight on either side don’t bias its outcome), you have a 0.098% chance of ten heads in a row. Or it is expected to happen once in every 1,024 trials.

    The previous events have no bearing on the outcome of the next toss. After you’ve tossed nine heads in a row, the chances of your tenth toss being a head is 1 in 2, or 50%. Just as they were for each of your previous tosses.

    Conditional probability is somewhat different from the scenarios you’re painting. The chances of getting seven or more heads in ten tosses are 17.19%. But if you start your run with three heads in a row, then the chances of “victory” go up to 50%, or 1 in 2. Because you only four or more of your remaining seven tosses to come up heads.

    To confuse things a little, covering probability from a more philosophical perspective might do your head in.

    :-)

  • Catherine Murphy said:

    Fran I did probabilities for my LC long long ago don’t remember much but surely commonsense will tell you without any fancy maths what your friend Dan said, previous results will have no bearing on the 10th toss its the same as when you first picked up the coin, 50/50. The chances of ten heads in a row though must be fairly low so I doubt this dilemma will ever be yours to deal with!!

  • jim shanahan said:

    Probability of an event will always remain the same, this is as Dan says.. Previous events have no bearing on next outcome.

    Possibility of a certain outcome (7 heads from 10 tosses)changes depending on previous outcomes. As Dan says, this is true. Possibility of getting 7 heads from 10 tosses is 17.19% but if you have six heads and are on seventh toss chances are 50% you will get at least seven out of 10 heads.It has gone from 17.19 to 50%.

    Use Ian Saliu’s formula FFG to calculate the degree of certainty of an outcome within a certain number of events.

  • Francis (author) said:

    It seems this is known as the “Gamblers’ Fallacy”

    The gambler’s fallacy can be illustrated by considering the repeated toss of a fair coin. With a fair coin, the outcomes in different tosses are statistically independent and the probability of getting heads on a single toss is exactly 1 / 2 (one in two). It follows that the probability of getting two heads in two tosses is 1 / 4 (one in four) and the probability of getting three heads in three tosses is 1 / 8 (one in eight).

    Now suppose that we have just tossed four heads in a row, so that if the next coin toss were also to come up heads, it would complete a run of five successive heads. Since the probability of a run of five successive heads is only 1 / 32 (one in thirty-two), a believer in the gambler’s fallacy might believe that this next flip is less likely to be heads than to be tails. However, this is not correct, and is a manifestation of the gambler’s fallacy.

    While a run of five heads is only 1 in 32 (0.03125), it is 1 in 32 before the coin is first tossed. After the first four tosses the results are no longer unknown, so they do not count. Reasoning that it is more likely that the next toss will be a tail than a head due to the past tosses—that a run of luck in the past somehow influences the odds in the future—is the fallacy.

    http://en.wikipedia.org/wiki/Gambler%27s_fallacy

  • Jim Shanahan said:

    Hi Fran
    When I refer to possibility of a certain outcome I am talking about a series of events results occurring in a row or within a certain number of events, ie. certain number of heads out of 10. If you have the first four in a row tosses resulting in 4 heads, and you’re looking for 5 heads from ten tosses you’re odds are greater after the successful 4 tosses than when you started toss 1, that you will achieve your goal of getting 5 heads within 10 tosses as you still have 6 more chances. However as you state, for any single event(such as the single toss of a coin) probability is always the same and not influenced by previous events. The gamblers fallacy is as Ian Saliu states…in most walks of life persistence pays off but in gambling it can lead to bankruptcy! Don’t worry I am aware of this fallacy. Thats why I rarely do the lotto!

  • Francis (author) said:

    I went back and read some more. Jim, to use your example, if you are looking for exactly (not minimum) 5 heads from ten tosses, the odds are not 50%. Dan keep me honest here but I got 24.6% (my binomial coefficients might be a little rusty). I didn’t use Saliu’s thing whatever that is, just excel.

    My initial reaction was 5 heads out of 10 must be 50% but this it seems is not so, as I calculated 62% of getting 5 heads or more out of ten. Weird.

    In any case, this number is only meaningful at the start, BEFORE you’ve tossed any coins. Once you’ve tossed a coin the outcome is known so any odds you’re going to calculate would need to remove that from the equation.

    It may be semantics but my point is that it’s not correct to say “you’re odds are greater after the successful 4 tosses than when you started toss 1″. It’s more correct to say the new calculation is based on the probability of a SINGLE head being tossed in 6 tries.

    Exactly one single head in 6 tries would be “one over 2 to the power of 6 times (6 binomial 1)” which turns out to be… about 9%

    You’re interested in p(1) + p(2) + p(3) + p(4) + p(5) which looks like about 96% so yeah, the odds are good. However, only because the number of trials has reduced to 6 and the target outcome is now 1 head.

  • dan said:

    Francis,

    Think of two tosses. (As it were.) The odds of getting 1 or more heads is greater than 50%. The odds of getting zero heads (or two tails) are 0.25 (25%). So the odds of 1 or more heads are 1-0.25 or 75%.

    Apply the same logic to ten tosses and it shows why 5 or more heads equates to more than 50%.

    Dan.

  • Mike said:

    People must not forget, probability is an idea, the only thing that ever really matters is outcomes. you can scratch your head running numbers all day long, but probabilities in an unbiased world are always the same. On the contrary, creating a fair situation for anything to happen is also, in our world of understanding, very statistically improbable. Statistics is taking into account every variable and creating the likely hood that something will happen. creating a situation with all variables being known leads to a 100% outcome. the idea of a fair coin toss is the notion that the only variable is chance and nothing more. This can not be simulated, only idealized about. Just as the idea that Infinity is relevant in mathematical equations. Though the existence of this idea has helped prove many equations in math, it being only a notion makes it subject to error within the real world. I guess the point of all of this is to say that in a perfect “fair” world coin tosses, throwing die, playing the roulette wheel will all have the same chance every time. But, it is not unfair to think that seeing a trend may lead to that trend being correct. If Jimmy flips a coin, you have a 50/50 shot of guessing the right outcome. But if he always flips it with the head facing up to start, with the same force, from the same height, at the same angle, a trend will occur in the coin tosses one way or another. if he flips 100 heads in a row, you bet it all on heads, and he flips a tail, you just became the statistical outlier (or in laymen terms, unlucky). In the end, there is no right or wrong decision, just some basic idea of statistics to help guide you. But don’t forget, statistics may guide you away from the trend that was created in an “unfair” sample set.

  • Dave Sutula said:

    I have a probability problem:

    What is the probability of guessing a 1 in 4 chance correctly 10 times in a row. That is, if you have four colors (Red, Orange, Yellow and Pink) and have to guess it ten times in a row, what are the chances you could get it right?

    Then, to complicate matters, what if you had to get those ten correct guesses anywhere along a string of 200 random ‘pulls’?

    So, a guy is pulling colored tiles from a bag. he’s going to pull 200 times. The trick is to have your ten guesses show up in order anywhere within those 200 random samples.

    I think it would be the probability of getting one right (.25) times the same thing ten times (.25 • .25 • .25 etc, etc) and then multiplying those odds by 190 chances (because the last nine ‘pulls’ cannot be the start of a new string…

    Can anyone help me with this?

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