how to weight a coin toss

#1
hello community :), english is not my mother language :) so please understand if i have some mistakes in spelling

can you help me answering if there is a solution for this scenario or not? and if there is a solution can you explain why there is one? and what is different to a usual sequence of coin flipps.
OK here we go:

imagine a scenario:

you do Coin flipps:
- outcome is for each 50%; fair coin toss; H=Head; T=Tails
- in this scenario we determinate that formation of "HHT" will appear first (2 to 1 odds) in relation to the formation "HTH" in one single sequence of coin flipps
- you make 100 sequences of coin flipps.
- if one formation appears in 1/100 seqeunces we will restart coin flipping; from 0 on;
- so: "HHT" will appear 66.6 times before "HTH" will appear. and "HTH" will appear 33.3 times before "HHT" will appear in 100 seqeunces.

Now i have all rules written down.



my question:

now that we know that in 66.6 Sequences (the other Formation WON´T APPEAR, 2nd formation)... is it now changing the behavior of 50% winning probability?
and second question, now that i know that in 66.6 sequences "HTH" wont even appear, can i create, (in that given conditions) a environment to bet and increase my odds in that way, that after 100 seqeunces i made a profit?

so basically im trying to figure out, if under those given circumstances, ---> could i create a way of betting of Every Coin flipp that will appear and increase my probability of winning more than 50%?



my idea, how i even get to think about that is: (and also my idea for solution...)

if i know that in 66.6 % of all 100 sequnces the formation "HTH" wont even appear shouldnt it mean that:

HHH (will probably appear in 66.6/100)
HHT (will probably appear in 66.6/100)
HTT (will probaply appear in 66.6/100)
HTH (wont appear in 66.6/100)
TTT (will probably appear in 66.6/100)
TTH (will probably appear in 66.6/100)
THT (will probablyappear in 66.6/100)
THH (will probably appear in 66.6/100)

that i know now, thats not 50/50 anymore because...
if i chose randomly 3 formation + the formation that will appear in in 66.6/100 cases i have higher odds to win then to bet on 3 randomly formation + the formation that will appear in 33.3/100

should that right weighted be more then 50%? odds?

KR