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I have what I imagine to be a fairly simple probability question.

I have cards with the numbers one to ten on them. The person who draws the highest card wins.

If I play against a single opponent and I draw a 4, my opponent needs to draw a 5-10 with 9 remaining cards. Therefore they have a 6/9 chance of beating me.

However if I have two opponents what is my probability of being beaten? I've tried a couple of ways and all end in certain defeat (which is clearly wrong as they could both draw any one of the remaining 3, 2 or 1.

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    11/12. $ $ $ $ $ $2011-08-03

3 Answers 3

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Well the first still has 6/9 chance of beating you.

The second (if he's needed) has a 6/8 chance. (The first must have chosen 1, 2, or 3, so there's 2+6 total and 6 still will win.)

The second is needed 3/9 times, so it is

6/9+3/9*6/8
=2/3+1/3*3/4
=2/3+1/4
=(8+3)/12
=11/12

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For you to win, both need to draw from the numbers 1,2,3.

The probability of this is $\frac 3 9 \times \frac 2 8$

The probability you lose is 1-probablility you win.

I just mention this because it is sometimes true that the complement of a set which seems complicated is simple, and this is a trick you should know and look out for.

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There are 72 possibilities for drawing the two cards (9 for the first opponent, 8 for the second). You win only if both cards are among 1, 2, 3, and there are 6 possibilities how this can happen (3 for the first, 2 for the second). Therefore your winning chance is 6/72=1/12.

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    -1 This doesn't answer the question.2011-11-22