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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 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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marked as duplicate by Austin Mohr, Zev Chonoles Dec 13 '11 at 17:05

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

See independence, and see here for the same question only about coins. – Zev Chonoles Dec 13 '11 at 16:57
Successive dice rolls (of a fair die) are independent events. – user5137 Dec 13 '11 at 16:58
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? – David Mitra Dec 13 '11 at 16:59
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? – leonbloy Dec 13 '11 at 17:00
up vote 1 down vote accepted

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.

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