I'm currently creating a game for class. I know that the probability of one card being higher than another is $8$ out of $17$. What I'm trying to determine is the probability of the following:

A player draws $2$ cards without replacement from a deck. What is the probability that at least one of them will be higher than a third, subsequently drawn, card?

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    $\begingroup$ What have you done so far on your three card problem? This would give us a better idea of your background and situation and help us tailor an answer to your specific problem. $\endgroup$ – Ian Miller Mar 23 '16 at 13:35
  • $\begingroup$ So you draw two cards, and then a third one without replacement, and want the probability that one of the two cards is higher than the third card? $\endgroup$ – астон вілла олоф мэллбэрг Mar 23 '16 at 13:37
  • $\begingroup$ @астонвіллаолофмэллбэрг correct. $\endgroup$ – pianoman Mar 23 '16 at 13:41
  • $\begingroup$ @IanMiller my primary focus to this point has been on the probability of one card being higher than the second, without replacement. That's where I got the 8 out of 17. I've attempted to expand to the 3 card situation, but I've not gotten far. It seems that of the 2 initial cards, the only card of importance would be the higher card, but I'm not quite sure how to incorporate that in. If I have any breakthroughs I will add them to the initial question. $\endgroup$ – pianoman Mar 23 '16 at 13:46
  • $\begingroup$ At least one higher, or exactly one higher? $\endgroup$ – Jimmy R. Mar 23 '16 at 14:08

The probability is $\frac23$ if the values are unique, $\frac12$ if two are the same and $0$ if all three are the same.

The probability for the values to be unique is


The probability for exactly two values to be the same is

$$ \frac{13\cdot12\cdot4^2\cdot3\cdot3}{52\cdot51\cdot50}=\frac{72}{425}\;. $$

Just to check that they add up to $1$, the probability for all three values to be the same is

$$ \frac{13\cdot4\cdot3\cdot2}{52\cdot51\cdot50}=\frac1{425}\;. $$

Thus the desired probability is

$$ \frac23\cdot\frac{352}{425}+\frac12\cdot\frac{72}{425}=\frac{812}{1275}\approx64\%\;. $$

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    $\begingroup$ The problem when conditioning on the value of maximum being $k$ for $1\le k\le 13$ leads to problem. Indeed $P(X_3<\max\{X_1,X_2\})$ depends on whether the $\max$ is attained by both the first two or only by one of the first two cards. So, although it seems straightforward, it is not. Contrary, conditioning on the cards having the same value or not (as you did) is a more fruitful :) approach here. +1 $\endgroup$ – Jimmy R. Mar 23 '16 at 15:11
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    $\begingroup$ An exhaustive search of all 132,600 possible combinations agrees with this answer. $\endgroup$ – Ian Miller Mar 23 '16 at 15:16

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