# Probability game with three players winning plus losing doesn't equal 1

I have a game of cards numbered 1-10. Three players draw cards, the highest card wins. Cards are not replaced.

The first player draws a 9. I want to work out the probability of that player winning and losing.

The probability of winning is the product of the two other players picking up lower cards in this case

$$\frac{8}{9} \cdot \frac{7}{8} = \frac{56}{72}$$

The probability of losing is the probability that either player draws a higher card (the ten)

$$\frac{1}{9} + \frac{1}{8} = \frac{17}{72}$$

Given that player 1 can only win or lose and there is no way that there can be a draw why do these probabilities add to make more than one?

I've tried a couple of different values and they always come out more than one.

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+1 for showing your working. The error is in your second equation. The correct sum is $\frac{1}{9} + \frac{8}{9} \cdot \frac{1}{8} = \frac{16}{72}$. This is because the last player can only draw a ten if the second player didn't draw a ten. – Zhen Lin Aug 3 '11 at 13:29

The probability of losing is not correct. In such case there are two possible cases about the remaining players:

1) First player draws a lower card (this is the part you forgot in the question) and the second one a higher card;

2) First player draws a higher card and the second one a lower card.

Thereby,

$\Pr[lose] = \Pr[2^{nd} \; player\; wins] + \Pr[1^{st}\; player\; wins] = \frac{8}{9} \times \frac{1}{8} + \frac{1}{9} \times \frac{8}{8} = \frac{16}{72}$

You add all up and it sums to $1$.

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The probability of the latter is $$\frac19 + \frac89 \times \frac18 = \frac29$$ because winning of the third player indicates that the second player didn't draw a 10, so we should multiply this probability. Finally, $\frac{56}{72} + \frac29 = 1$.

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