Scenario:
Coin flipping four times calling heads every time.

If heads are hit one time out of four I win $3.

If it lands tails all four times I lose $10.

How do you calculate who has the advantage?

Depending upon your knowledge of probability and statistics, there are several ways to answer this question.

Perhaps the most basic is to simply write down all the possible sequences of H (heads) and T (tails) that can occur when flipping a coin four times. Such as HHHH, HHHT, HHTH, etc.

Tally how many of these sequences have exactly one H. Also tally how many of these sequences have exactly four T.

You win $3 in all of the sequences of the first type and lose $10 in all of the sequences of the second type.

Putting that all together should allow you to see which "side" of the bet has the advantage.

Of course, these calculations can be more directly performed using knowledge of the probability of sequences of coin flips (see the Binomial Distribution), but it is essentially the same derivation.

Thank you!
There are 16 different combinations in this scenario and only 1 does not include a head.

That leaves me at a 1 out 16 chance of losing my $10 bet.

Is that correct or is there more to this?

Correct, but there is more to it (if I am following what you are saying):

How many sequences have exactly one head and three tails?

There are 3 combinations of 1 head and 3 tails.

Would that mean I win 2 out of 3 times or am I on the wrong path?

I would count again the number of 1 head & 3 tails sequences.

Then use the fact, as you pointed out above, that all sequences are equally likely. This allows you to easily calculate how much would you win if all of the above (1H, 3T) sequences occur. This is simply $3 times the number of such sequences, right?

Now compare this to the amount you would lose if a four-tails sequence occurs. This is simply $10 times the number of such sequences, right?

The larger number tells you which side of the bet has the advantage.

I miss counted how many times it repeated itself. There are 4 event's of 1 head and 3 tails.

Then I would multiply 4 times $3 giving me $12
There is only 1 occurrence of 4 tails so then I multiply 1 times $10 giving me $10.

Am I risking $10 to win $12 in the long run?

How do I calculate and write this down in odds form?

Also as I am thinking about this.... if I risked $10 to win $2.50 in the same scenario wouldn't I have a 50% chance of winning or losing?
Would that mean in the long run it's actually zero risk to flip a coin as long as it's even odds or better for me?

Multiplying by the # of permutations only tells you which side has the advantage, not the average profit. For that, you need to multiply by the probabilities.

You have a 4/16 chance of winning and a 1/16 chance of losing. If you don't redo the series in the event of a tie, then your average profit is:

.25(3) - 10/16 = 0.125

You're still risking $10 to win $3, but in the long run each series earns you 12.5 cents.

Your average profit would be 0, but you'd still have a 25% chance of winning a given series. After playing this series many times (say N), you'd have a 50% chance of being above or below the average profit, N * .125.

Absolutely not. Your bankroll is finite and large downswings are inevitable. If the game is 0 EV for you, if you play long enough you'll have a 100% chance of going broke. Even when you have positive expected profit, you have a nonzero (but less than 100%) chance of going broke.