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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,

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

  2. 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?

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    $\begingroup$ So, I make or lose $\$10$ with equal probability which makes my expected payout $0$. Or have I misunderstood? $\endgroup$ – lulu Mar 12 '18 at 19:44
  • $\begingroup$ Right. But as the 'bookie' why wouldn't my expected payout on a 10 bet be 15? It doesn't matter what the odds or the probability is, my payout is always going to include the full stake so the probability should only be applied to the 'winnings' since. Say somebody bet 10 on heads and the coin is unfair and there is only a 25% chance of heads. Would my expected payout then be 10 + (0.25*10) = 12.50 or would it be 20 * 0.25 = 5? You would never pay out less than the full stake on a winning bet. $\endgroup$ – forfun32 Mar 12 '18 at 19:55
  • $\begingroup$ I think the issue is just semantics. What are you defining as "payout"? The expected profit/loss on the whole transaction is $0$. Once you have a clear and precise definition of what you mean by "payout" then the answer should be evident. $\endgroup$ – lulu Mar 12 '18 at 19:57
  • $\begingroup$ You're probably right. By 'Payout' I mean the amount I expect, as the 'bookie' to be paying out on a bet. I have the bet coming in and is in my pocket, then how much do I expect to leave my pocket after the coin toss given a probability of 0.5 and also given that the original bet is now in my possession. $\endgroup$ – forfun32 Mar 12 '18 at 20:59
  • $\begingroup$ Ok. So you are separating the "stake" from the "payout". Fine. in that case, my payout is $20$ if you win, and $0$ if you lose, so the expected payout is $10$. Note that, unsurprisingly, this is equal to the stake. $\endgroup$ – lulu Mar 12 '18 at 21:06
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There are 2 approaches here, both yielding the same net:

  1. 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.

  2. 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.

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You need to clarify "payout" to whom: yourself or opponent.They are equivalent.

"Payout to yourself" method: $$\begin{array}{l|c|c|r} & x & P & xP \\ \hline win & 10 & 0.5 & 5 \\ lose & -10 & 0.5 & -5 \\ \hline Total & & & 0 \end{array}$$ "Payout to opponent" method: $$\begin{array}{l|c|c|r} & x & P & xP \\ \hline win & -10 & 0.5 & -5 \\ lose & 10 & 0.5 & 5 \\ \hline Total & & & 0 \end{array}$$ Note that when you win you pay $-10$ to your opponent, which actually implies the opponent pays you $10$.

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