How to determine to most probable permutations?

Question

I'm a little rusty on my math so bear with me. I was thinking about permutations and probability. A coin toss probability of landing on tails (or heads) is 0.5. While the chances of another toss is still 0.5, the collective chances are halved each time, correct? So the chances of getting 5 of the same in a row is 0.03125, about 3%. Now if I want to calculate the possible permutations of 10 coin tosses, that would be 2^10, which is 1024, correct?

How could I calculate all permutations that exclude any set with more than a predetermined number of repeated tosses? I am basically trying to get the most probable sets of all permutations.

Answer 1

In case the actual question is 'out of 10 tosses of a fair coin, which sequence of outcomes is the most likely', you need to use binomial probability, with $P(H)=\frac{1}{2}, n=10$, and find the mode, which is $\lfloor (n+1)p\rfloor$ in this case.

Answer 2

If the question is about the probability of the length $N$ of longest run of consecutive identical results from the $10$ tosses, then the probabilities are:

 n    prob <= n    prob = n
 1       2/1024      2/1024
 2     178/1024    176/1024
 3     548/1024    370/1024
 4     802/1024    254/1024
 5     928/1024    126/1024
 6     984/1024     56/1024
 7    1008/1024     24/1024
 8    1018/1024     10/1024
 9    1022/1024      4/1024
10    1024/1024      2/1024

This distribution has a most likely value (mode) of $3$, a median of $3$ and an expected value (mean) of about $3.66$