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Suppose you have a 52 card deck. We attribute $10$ points to the Ace, $5$ points to the King, $3$ points to the Queen, $1$ point to the Jack, and $0$ to the rest. If we simultaneously pick two cards from the deck, what is the probability of having a sum of points odd?

My first approach was first to calculate the number of issues possible: $\binom{52}{2}$.

Then, to see what issues would give an odd sum:

Let $A$ denote Ace, $K$ King, $Q$ queen, $J$ Jack, and $O$ others. The issues that satisfy our need are: $$B =\left \{ (A,K);(A,Q);(A,J);(K,O);(Q,O);(J;O) \right \} $$

Now all is left to calculate is the cardinal of $B$. As there are $4$ Aces and $4$ Kings, we have 16 of $(A,K)$ pairs. And so on for the $(A,Q);(A,J)$.
Now for each $(\cdot; O)$ we have 4 cards times 9, thus giving in total:

$$|B|= 4*(4 * 4) + 3*(4*9)= 156 $$

So the probability would be $$\frac{|B|}{|\Omega|}=\frac{156}{\binom{52}{2}}=\frac{156}{1326} \approx0.12 $$

Yet the correction show that the probability is equal to $\frac{12}{25}$. Where is my mistake?

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    $\begingroup$ @ThomasAndrews I expect that "impair" here is just French for "odd". $\endgroup$
    – lulu
    Commented Mar 4, 2017 at 13:48
  • $\begingroup$ @ThomasAndrews My bad, I ment odd, as lulu stated. $\endgroup$
    – John Mayne
    Commented Mar 4, 2017 at 13:49
  • $\begingroup$ I don't get the same numbers you are getting. I have posted a quick calculation below...perhaps I am not understanding the question properly. $\endgroup$
    – lulu
    Commented Mar 4, 2017 at 13:49
  • $\begingroup$ First, it should be $3\cdot (4\cdot 4)$ for the cases with an ace. For the $O$ cases, there are $3\cdot (4\cdot 4\cdot 9)$ - there are $4\cdot 9$ non-face cards to choose from. $\endgroup$
    – Thomas Andrews
    Commented Mar 4, 2017 at 13:57

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The exact points per card seem like a distraction. You have $40$ even cards, and $12$ odd....that's all that matters. To get an odd sum, one draw must be odd and the other even.

the probability of drawing first an odd and then an even is $\frac {12}{52}\times \frac {40}{51}=\frac {40}{221}$. Of course the probability of drawing them in the opposite order is the same. Thus your answer is $$\boxed {\frac {80}{221}\approx .362}$$

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  • $\begingroup$ I don't see why are the points a distraction. It indicates specifically which pairs we are interested in. But your generalisation "You have 40 even ranks, and 12 odd...." does simplify the problem greatly. $\endgroup$
    – John Mayne
    Commented Mar 4, 2017 at 13:52
  • $\begingroup$ It makes no difference if the odd number you draw is $1,3$ or $5$. It only matters that it is odd. $\endgroup$
    – lulu
    Commented Mar 4, 2017 at 13:53
  • $\begingroup$ I agree with you. Your reasoning is very clear. Thank you for answering with such speed. $\endgroup$
    – John Mayne
    Commented Mar 4, 2017 at 13:56

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