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In Richard Feynman's lectures on physics, chapter six, part 3, he explains something called the random walk, in which, in a succession of trials, a system moves forward one step or backward by one step, each with probability one half.

He comes to the conclusion that the most likely result is that there are more forward steps than backward steps or backward steps than forward steps.

For me, this is counterintuitive. If heads on a coin represents a forward step, and tails represents a backward step, surely the most likely result is that there will be the same amount of forward steps and backward steps?

innisfree
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Churent Delising
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    If I flip a coin lots of times, it is very unlikely that I will get exactly the same number of heads and tails. This is because there are many more possible outcomes where the number of heads and tails is different. – Mark Mitchison Jun 13 '16 at 10:34
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    Of all possible numbers of trials, exactly half of the possible numbers (the odd ones) cannot have the same number of forwards and backs. – Todd Wilcox Jun 13 '16 at 18:07
  • Related: http://physics.stackexchange.com/questions/69929/stupid-yet-tricky-question-why-do-we-actually-see-the-sun –  Jun 13 '16 at 21:29

3 Answers3

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You are mixing here two kinds of most likely. First, you are correct, that the same amount of forward and backward steps will be the most probable of all possible outcomes. Second, if you add probabilities of all outcomes with unequal result, you will see that getting some unequal result will be much more likely.

The outcomes of coin tossing are described by Binomial distribution. To give a perspective, the probability of getting exactly $500$ heads out of $1000$ fair coin tosses is $0.025$, and this is the most likely outcome (among a thousand of others!)

gigacyan
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I suggest you read about the Binomial distribution. In general, you are correct that if $p=1/2$, the distribution and its properties are symmetric under exchange of successes and failures (or forward and backwards in your example).

If the number of trials is even, the mean, mode (most probable single outcome) and median coincide at zero in your example. If it is odd, there are modes at $\pm 1$ step.

However, whilst the mode is the single most probable outcome, other outcomes summed together could be more probable. Even in the case where the number of trials is even, for $n>2$ the probability at the mode (at zero) is less than one half. That means that the sum of all the probabilities for distance not equal zero is greater than the probability that the distance is zero.

innisfree
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The problem is identical to tossing a coin four times and looking for 2 heads and 2 tails.
There are 16 possible outcomes.

Four heads has a probability of $\frac {1}{16}$
Four tails has a probability of $\frac {1}{16}$
Three heads and one tail in any order has a probability of $\frac {4}{16}$
Three tails and one head in any order has a probability of $\frac {4}{16}$

And the one which will get you back to where you started from two heads and two tails in any order has a probability of $\frac {6}{16}$.
So this is the mode but the probability of this not occurring, $\frac{10}{16}$, is more likely.

As the number of steps increases so the zero final displacement probability decreases.

Many websites have more about this, here is one.

Farcher
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