How to mentally flip a coin?

Solution 1:

I have freckles. I just encircle with thumb and forefinger a “random" portion of my arm, count the freckles, and take the parity. It takes about 3-5 seconds. I guess that someone who is fairly hairy could do the same counting hair.
Talking of counting hair, I take a small lock of my hair (I have long hair), and then try to split it as evenly as possible in two “half locks”, each with half the hair. I repeat the same process on one of the half locks, and then on one of the quarter locks, until I’m left with a single hair. The parity of the number of halvings is my random bit. It takes me about 5-10 seconds to produce it. I’ve found this to be quicker than the more obvious method of just counting the hairs and taking their parity.
I play a really fast paced word association game: I think of a word, the very first word associated to it, the second, third and fourth (using my hand’s fingers to count the five words helps). My random bit is the order of the third last letters of the last two words. It’s easier to do than to say, it takes no more than $3$ seconds.
This is somewhat general, though I guess everyone should make it specific to one’s interests in life. I let my mind wander and pick $3$ artists (musicians, bookwriters, painters, etc.) and think of how many of the pieces of each I’ve heard/read/seen. I add the numbers and take the parity. A programmer friend of mine says for him it works really well with examples of some obscure software categories I can’t quite remember. I guess a gourmet could use types of food: how many vegetable soups. omelettes, and bottles of red wine he’s had in the last two weeks.
I pick a number that strikes my fancy, and play the $3n+1$ game (halve it if even or triple it and add $1$ if odd, and then repeat) for $20$ steps, using the (parity of) the second last digit. I've not tried this extensively enough to detect any bias, which instead affects the last digit, but I guess one could correct it with the method the OP suggests for white hair counting. This is slowish (about $20-30$ seconds), but still a bit faster than the congruential generator another poster suggested, at least for me.
If you are good at remembering phone numbers of people, their least significant digits are a good source of randomness against people who know relatively little of your contacts. You can always scramble them a little, like reversing them and skipping every other digit.

Solution 2:

This is one strategy that seems to work for me (36). I talk to myself (11). And for every sentence, I just count the letters in my head (46). Then, I can use the least significant bit, or maybe two, as "random" bits (50). It's a game I used to play as a pasttime for its own sake ... (43) and if I add the numbers wrong, well, it doesn't really seem to affect the randomness (76)! The numbers in parentheses are, if you haven't guessed, the letter counts of each sentence (71).

Solution 3:

You could precompute a long sequence of random numbers, then memorize them. Each time you need a random number use the next one in the sequence.

People are able to memorize tens of thousands of digits of $\pi$ and $e$. You could do it once then be set for life.

(This is a conversion of a comment to an answer)

Solution 4:

Think of a five digit number. Add its digits. If it produces an even number, say it counts for heads. If it produces an odd number, say it counts for tails.

I base this on the fact that while there might be some bias on the numbers you would chose on a smaller scale (of less digits), five-or if you can/or the situation allows, even more digits-provides a sufficient "random hedge". i.e-your brain will not pre-compute the result (even or odd).

It also looks much faster than other methods proposed, particularly if you are involved in some form of fast paced game.

Solution 5:

a pseudorandom generator with a reasonable balance of randomness and simplicity would definitely be a possibility

Well, then let's pick a small multiplicative linear congruential generator.

Let $m=251$ $a=33$

Starting from a initial random (well..) value x[0] in the range $1 \cdots 250$ in each iteration we do mentally the following (pseudocode):

x[n] := x[n-1] * 33 ;
x[n] := x[n] % 251 ;
y[n] := x[n] % 2 ;

Ok, the second statement is not very easy to do mentally, but it can be slightly easied by substracting, for example:

while ( x[n] >= 251) {
      if(x[n] >= 2008) x[n] -= 2008;
      else if(x[n] >= 1004) x[n] -= 1004;
      else if(x[n] >= 502) x[n] -= 502;
      else x[n] -= 251;
}