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Am I risking $10 to win $12 in the long run?
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.
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if I risked $10 to win $2.50 in the same scenario wouldn't I have a 50% chance of winning or losing?
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.
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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?
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.