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Statistics Of Independant Events and Gambling. watch

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    Hi, I was reading an article that mentioned 'Gambler's Fallacy' and I was a bit curious about something and couldn't figure it since I'm astoundingly terrible at statistics.

    So let's say for example, we have a simple coin toss so obviously the chances of landing a head or tails is equally 0.5, now let's say the coin is tossed 10 times and everytime it has been tossed it's landed on tails, the gambler's fallacy states that the gambler is now more likely to pick heads because they believe, since the last 10 tosses have been tails, it's bound to fall on heads soon.

    I understand that since the events are independant then the fact that the last 10 have been tails has no effect on the next flip, hence the probability of tossing heads on the 11th flip remains at 0.5

    The problem I'm having trouble then understanding is that if we want to work out the probability of tossing the tails 10 times in row you simply work out:

    (0.5)^10 = 0.0009765625

    It's then pretty easy to see that the probability of tossing more and more tails in a row rapidly decreases. Surely this naturally implies though that as we throw more tails in a row, the chances of this streak happening decreases and the chances of the streak ending increase and since the only way to end the streak is to throw a heads, this implies the chances of throwing a heads increases?

    I'm really confused about this, on one hand I can see looking at each throw independantly the chances of the head being thrown doesn't increase but when we look at it as a chain of events, then the chance does increase and yet each event is still independant. I seem to be missing something as these two things (perhaps just on the surface) seem to be in confliction with each other.
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    Anyone? I can give rep if that's what you're into =P
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    Maybe you could consider it this way. If you tossed a coin 5 times, the chance of getting TTTTT is (0.5)^5. However, the chance of getting TTTTH is also (0.5)^5. Both of these results have the same probability of occurring.
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    probability of 10 heads, then 1 tail = 0.5^{10} × 0.5 = 0.5^{11}
    probability of 10 heads, then 1 head = 0.5^{10} × 0.5 = 0.5^{11}

    The probability of getting 10 heads then 1 tail, and the probability of getting 10 heads then another head are both 1 in 2048

    Therefore, it is equally likely to flip 11 heads as it is to flip 10 heads and then 1 tail when flipping a fair coin 11 times

    These two probabilities are as equally likely as any other 11-flip combinations that can be obtained, there are 2048 total, all 11 flip combinations will have probabilities equal to 0.5^{11}, or 1 in 2048

    So you dont say that there is a change of probabilty after a number of flips
    every outcome will always have been equally as likely as the other outcomes that did not happen for that number of trials

    \frac{1}{2} will always be the probabilty of getting head or tail on a fair coin

    read bayes theorum
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    Thanks, this cleared it up nicely.
 
 
 
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