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In a game where we have a normal 52 card deck. Two cards are delt out at a time, if both are red, then I keep the two cards. If both are black, you keep the two cards. If its one of each, then it gets discarded. On each turn that I have more red cards than you have black cards (26 - 26 is a tie), I get paid 100. If not, then nothing happens so I never lose money. I only win money when I have more red cards then you have black. The fair price should be $\frac{1}{3}100 = 33.33$ or $\frac{1}{4}100=25$? It should be the latter since $rb$ or $br$ have twice the probability of me drawing two red cards?

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  • $\begingroup$ As Ross Millikan noted in his answer, this isn't a very interesting game if you wait until all the cards have been dealt before you get paid. Did you mean that you get paid 100 on each turn that you have more red cards than your opponent has black cards? $\endgroup$ – Skatche Jan 16 '14 at 23:21
  • $\begingroup$ Yes each turn. Sorry for the confusion. $\endgroup$ – adfasdf Jan 16 '14 at 23:24
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The fair price is zero because you can never win. There are an equal number of red and black cards discarded, so there are an equal number of each color not discarded.

Added in light of the comment that you pay or not after each pair of cards: For the first pair of cards, the fair price is $\frac {26 \cdot 25}{52 \cdot 51}\cdot 100=\frac {25}{102}\cdot 100$, a little less than $25$ because the chance of drawing two the same is less than the chance of drawing two different. After the first draw, it depends on what has been drawn.

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  • $\begingroup$ Good point, sir, good point. $\endgroup$ – AlexR Jan 16 '14 at 23:15

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