Meena and Seema are playing cards. Probability and counting principle. Meena and Seema are playing cards (ace has 1 point, jack has 11 points, queen has 12 points, & king has 13 points). If Meena gets a jack, what is the probability that she would win?
In my book the answer given is $\frac{10}{13}$
But I strongly doubt to it...
My answer:

First approach:
  Case 1 : if meena gets first chance and she picks up jack, and Seema gets a second chance.
  When meena picks , she has to pick one among all the 4 types (club,spade,heart,diamond) of Jacks.(so probability is $\frac{4}{52}$)
  And then after one is gone then there are 51 cards left (..here I am assuming without replacement...is my assumption right...??)
  Now when Seema has to pick she has to pick either of the Ace to 10 (i.e. 10 cards out of every type...so 10*4=40 choices as favourable out of 51 left..)
  (Thus Seema probabilty will be $\frac{40}{51}$)
  So Probability will be $$\frac{4}{52}.\frac{40}{51}$$
  Case 2 : if Seema picks first and then meena picks...
  Whenever Seema pick she has to again the one with value less than jack (i.e. 11), for meena to win ..
  So probabilty for Seema is $\frac{40}{52}$
  And now for meena is $\frac{4}{51}$
  ..(bcoz I am assuming without replacement...)
  So the total probabilty is $$\frac{40}{52}.\frac{4}{51}$$
  So thus the total probabilty for both cases is sum of the two probabilities..
  $$\frac{4}{52}.\frac{40}{51}+\frac{40}{52}.\frac{4}{51}$$
Second Approach:
  Here I am goofing up with sure event definition...I think sure means that has to come so it's probabilty is 1
  Thus as meena gets Jack so her probabilty is 1
  Now either seema could get chance after meena (in that case her probabilty would be $\frac{40}{51}$) or before meena (in that case her probabilty would be $\frac{40}{52}$)
  So thus the total probabilty is $$(1).\frac{40}{51}+\frac{40}{52}.(1)$$
  (I think I am goofing up with sure event...plz help whether this approach is right...)
Plz tell which of my approach is right and why...(or am I totally mistaking....some where..)

 A: On the assumption that if Seema draws a jack they tie so Meena does not win and Seema draws from what is left in the deck after Meena's draw, if Meena draws a jack there are $40$ cards it beats and $11$ cards it does not beat.  Her chance of winning is then $\frac {40}{51} \gt \frac {10}{13}$.  To get the book answer we have to assume Seema draws with replacement, so Meena's card goes back in the deck, the deck is shuffled, and Seema draws.  Then there are $40$ cards Meena beats and $12$ she does not, giving a chance to win of $\frac {10}{13}$.  When asking problems like this it is important to specify the rules carefully.
A: probabilitu biggest card is jack and meena gets it=$\frac{{4 \choose 1}{40 \choose 1}}{{52 \choose 2}}\times \frac{1}{2}$ = a
Probability meena gets a jack = $\frac{1}{13}$=b
Answer is $\frac{a}{b}=\frac{40}{51}$ ,so yeah i think your book is wrong (and i am assuming jack vs jack is a tie).
Your first method is correct but you should also consider it's probability $\frac{1}{2}$ on who starts. Then also divide your answer by $\frac{1}{13}$ because it's a conditional probability.
