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Consider a coin toss of a biased coin, where H comes up with probability p

I bet 1 dollar that the outcome of the toss will be H, win reward R if it is H, and lose the 1 dollar stake otherwise. What is the reward R in a fair game?

I thought it would be R such that $pR + (1-p)(-1)=0$ ie $R=(1-p)/p $ so for example $R=25$ if $p=1/26$

However I understand that actually $R=26$ (ie we must include the 1 dollar stake) ?

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    $\begingroup$ The effective gain if you win is $R-1$. $\endgroup$ – André Nicolas Nov 29 '15 at 12:28
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Let $X$ be the amount of money you gain by playing the game. Then $$X = R\cdot\mathsf 1_H - 1,$$ so $$\mathbb E[X] = \mathbb E[R\cdot\mathsf1_H-1] = R\mathbb P(H) - 1 = Rp - 1. $$ For the game to be fair, we require $\mathbb E[X]=0$, which implies that $R=\frac1p$.

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  • $\begingroup$ So if I look at the wealth process after the first bet, which should be a martingale, then $M_0=1$ (the amount I have to bet ) and $M_1 = M_0 + X $ ? $\endgroup$ – user3203476 Nov 29 '15 at 13:07
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    $\begingroup$ Just to be pedantic I prefer to understand your ( great) answer as $X = R 1_H + 0 1_T -1$ ie -1 is paid (deterministic) , R is given with probability p and 0 is given with probability $1-p $ $\endgroup$ – user3203476 Nov 29 '15 at 14:20
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I wish such problems always used a term such as net earnings or the like. Sometimes they use "winnings", sometimes they specify an amount + stake, here they are using the term "reward".

Here I interpret it that you wouldn't consider it much of a "reward" if you were given $1$ dollar on a bet of $1$ dollar, and so the amount includes the stake, in which case the equation would be

$pR - 1 = 0, R = \dfrac1{p}$

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