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What are the odd of a single coin toss after many consecutive ones?

If I have a fair dice and I rolled a six $5$ times in a row, it would appear to me that the next row have a lesser chance of being a six.

However the laws of probability state that the next row would still have a $\frac{1}{6}$ chance of being a six (and not being lower, despite the number of times six had just appeared).

This seems very counter-intuitive. I was wondering what's a good analogy / explanation to understand this phenomenon ?

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    See [independence](http://en.wikipedia.org/wiki/Independence_(probability_theory)), and see [here](http://math.stackexchange.com/questions/41794/what-are-the-odd-of-a-single-coin-toss-after-many-consecutive-ones) for the same question only about coins.2011-12-13
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    Successive dice rolls (of a fair die) are independent events.2011-12-13
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    Suppose you walk into a room in which someone had been throwing a die. You walk in just as he makes a roll. What is the probability he throws "6"? Would it matter what he had rolled before?2011-12-13
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    `the laws of probability state that the next row...` The laws of physics, I'd rather say. In our physical model of things (dice), the outcome of a "good" die only depend on physical processes that can be very complex, but which should not depend on past outcomes: the die does not have any property that makes it remember its past, does it?2011-12-13

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The die cannot remember what was rolled in previous trials, so each trial must have an equal chance of rolling a six. The misconception that the probability should be altered based on previous trials is called The Gambler's Fallacy.