Expected Payout Fair Coin Toss For a bet on a coin toss with a fair coin where P(Heads) = P(Tails) = 0.5. In this example, I am the one accepting the bets. I.e. I"m the 'house' so to speak. I pay out even money on a winning bet. So if you bet \$10, you make \$10 profit. 
If you place a \$10 bet and you lose, I keep the \$10. If you place a \$10 bet and you win, I give you \$20 - that is, $10 winnings plus your \$10 stake. 
What is my expected payout on any given \$10 bet that I accept? Is it a) \$20 * 0.5 = \$10 or is it b) 10 + (10*5) = \$15? 
The idea with b) being that no matter what the odds are, you will always get the full stake back on a winning bet so you don't apply the probability to the stake - it is a constant. As odds change, the only thing that changes is your winnings. 
For example, 


*

*Prob winning = 0.5. Odds = 2. Bet amount, \$10. Actual winnings = \$10. Total payout = \$20. Expected winnings = 10*.5 = \$5. Expected Payout= \$15 

*Prob winning = 0.526315789. Odds = 1.9. Bet \$10. Actual winnings = \$9. Total payout = \$19. Expected winnings = 9 * 0.526315789 = 4.736842105. Expected Payout= \$14.736842105 
Which approach is correct and why (or why is one wrong)? As the 'house' how much do I expect to pay out on a $10 bet with a 0.5 chance of winning?

There are 2 approaches here, both yielding the same net:


*

*Do not include the stake. Then you pay 10 or lose 10 with equal probability, so if chance to win is $p$, expectation is $$10p - 10(1-p) = 20p-10,$$ which equals $0$ if $p=1/2$, as expected.

*Include the stake. Then you unconditionally get 10, and have to pay 0 when you win and pay 20 when you lose. So the expectation is $$10 + 0p - 20(1-p) = 20p-10$$ exactly as above.
