When we toss an unbiased coin, the probability of observing both heads and tails is 1/2. I take that to mean that over a really large number of coin tosses the number of times the coin will turn heads will be almost equal* to the number of times the coin will turn tails.

My question is, if we witness a series of coin tosses that happens to have many more number of, say heads, than tails then will we not expect the upcoming coin tosses to be 'mean-reverting' i.e. more inclined to produce tails than heads - only in order to maintain the definition of probability being 1/2 as per the previous paragraph?

I think what I am really asking is whether an unbiased coin-tossing is a Markov process. I would add that if it is, then my understanding of why the probability of heads/tails is 1/2 is wrong.

[*] - If not, then we need more coin tosses such that the ratio of number of heads (or tails) to the total number of coin tosses approaches 1/2 as the number of coin tosses approaches infinity.

  • $\begingroup$ The answer to the question in your second paragraph is "no", as the coin has no memory of what it has shown before. Your assumption is known as the Gambler's fallacy $\endgroup$ – Henry Nov 24 '16 at 12:32
  • $\begingroup$ For a fair coin, the expected difference (in absolute or root-mean-square senses) between the numbers of heads and tails increases when the number of tosses increases, though the expected difference between the proportions of heads and tails decreases when the number of tosses increases. So you need to consider what "almost equal" really means $\endgroup$ – Henry Nov 24 '16 at 12:36
  • $\begingroup$ @Henry Wouldn't a problem like this pose an "inverse" Gambler's fallacy? Instead we should consider if the coin is biased, i.e. $p \neq 0.5$. $\endgroup$ – Cehhiro Nov 24 '16 at 12:46
  • $\begingroup$ @O.VonSeckendorff: Perhaps. If you tossed a coin several times, and it showed "heads" each time, how many tosses would it take before you checked carefully that it did not have two heads? $\endgroup$ – Henry Nov 24 '16 at 13:11

Intuition can be gained from the fact that these averages are completely determined by the tail values, and not the first few terms. That is, for any fixed $k∈\Bbb N$ and any finite number of terms $b_1,…,b_k$ (even if $b_i ≠ a_j$ for every $i,j$)

$$ \lim_{n→∞} \frac{1}{n}(a_1+…+a_n) = \lim_{n→∞} \frac{1}{n}(b_1+…+b_k+a_{k+1} + … + a_n) $$

Hence, there is no need for the $a_j$s to 'correct the bad behaviour of $b_i$s'; the $b_i$s simply don't matter in the long run.

  • $\begingroup$ Extending the RHS in your equality above (replacing $a_{k+1}...a_{k+m}$ with $c_1...c_m$) $$ \lim_{n\to\infty} \frac{1}{n}(b_1+...+b_k+c_1+...+c_m+a_{k+m+1}+...+a_n) $$ what I am really intrigued about is, would we not have more to say about the expected values of $c_1...c_m$ conditional upon observed sequence $b_1...b_k$ which behaved 'badly'? Agreed, that in the long run they all average out, but please help me grasp why in the immediate vicinity of the badly behaving sub-sequence there is no 'attempt' to correct it. $\endgroup$ – intelinsight Nov 24 '16 at 18:10
  • $\begingroup$ There is no reason to suppose mean reversion; in the standard frequentist model of a coin, the value 1/2 has nothing to do with any finite string precisely because of the above identity and what it means to be a 'long time(limit) average'. You are free to create your own model random variable but it won't be what people refer to as unbiased coin tossing. $\endgroup$ – Calvin Khor Nov 24 '16 at 18:41

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