Can anyone please guide me in solving the following problem:

*Suppose I have a well shuffled stantard deck of 52 cards and I divide the deck into two halves (each half containing 26 cards) and keep the first half aside. Now, from the first half, I pick a card at random. What is the probability that my chosen card is a queen?*

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TopNewestDoes splitting the deck of cards into half add/subtract anything from the information available to you?

When you figure that out, try this – Agnishom Chattopadhyay · 1 month, 2 weeks ago

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– Nilabha Saha · 1 month, 2 weeks ago

I am not picking up a card from a deck of 52 cards. I divide the deck into two. I don't know how many Queens went into my half. So, how do I account for that?Log in to reply

– Jesse Nieminen · 1 month, 2 weeks ago

Dividing in half doesn't do anything since you do not know which cards are in the second half.Log in to reply

giventhe number of Queens in the half deck; that is quite simple. However, the problem I am facing is calculating the probability of the number of Queens entering my half deck. – Nilabha Saha · 1 month, 2 weeks agoLog in to reply

– Jesse Nieminen · 1 month, 2 weeks ago

But the deck is well shuffled so limiting your another random choice doesn't do anything because the deck is already shuffled. Shuffling and selecting randomly is the same as first shuffling and then selecting always e.g. the first card. You can have 0, 1, 2, 3 or 4 queens in the first half, but calculating each case separately doesn't change the probability because randomizing twice is as good as randomizing once.Log in to reply

– Nilabha Saha · 1 month, 2 weeks ago

That would be the case when I draw 26 random cards. However, what I mean by dividing the deck into two is that I only pick the top half. So, that is not randomizing events.Log in to reply

– Nilabha Saha · 1 month, 2 weeks ago

If I an still committing a mistake, can you please explain it. Because, I am getting the feeling that your answer indeed might be correct.Log in to reply

– Jesse Nieminen · 1 month, 2 weeks ago

You have shuffled the deck already so the cards are randomized. Thus, dividing the deck in half and picking a random card from the first half is the same as picking it from anywhere from the deck.Log in to reply

– Nilabha Saha · 1 month, 2 weeks ago

I just thought about this problem for a while and got some deep insights. Actually, I just found the answer. But before that, let me explain the points properly. If I have a well shuffled deck and I pick one card from it the P(Q)=4/52 [P(Q) represents probability of drawing a Queen]. Also, from the deck, if I pick a random set of 26 cards from any position (by position, I mean a specific number of cards down from the top of the deck), then that is nothing but a convoluted way of just picking 1 card from the full deck. So, in this case, P(Q)=4/52 again. However, when I pick the top 26 cards, I am picking cards from specific positions, not random positions. So, in that case we differ our choice. Now, finding the probability goes into the realm of permutations and/or combinations.Log in to reply

– Jesse Nieminen · 1 month, 2 weeks ago

Yes, but one randomization (shuffling) is enough to make P(Q) = 1/13 for any card in the deck even if you limit your choices to a part of the deck.Log in to reply

– Nilabha Saha · 1 month, 1 week ago

It is quite confusing because sense might lead us to state that the probability remains unchanged whether or not I draw a card from q part of my deck. If I talk about a specific position, P(Q) will always equal 4/52 regardless of how many cards are the in the part. However, the deep meaning of draw a card from a full deck is that I have the freedom to draw it from anywhere. When I draw a random number of cards from the deck and draw a random card from it, the probability still remains 4/52 as I just shuffled it even more, which doesn't alter the probability at all, as I drew my cards at random. However, when I limit my card draws to specific positions (in this case, the top half), I don't draw my cards randomly. I am actually forced to draw the cards from specific positions. That hinders my freedom of drawing the next card. I have to draw a card from the specific positions of my deck. My freedom of drawing is reduced. That's the confusing truth about conditional probability, my friend.Log in to reply

– Jesse Nieminen · 1 month, 1 week ago

So do you agree that the splitting in half does nothing because shuffling and random draw are basically the same operation and both alone are enough to make the pick random?Log in to reply

– Nilabha Saha · 1 month, 1 week ago

No, picking up cards from specific positions alters our freedom, thereby altering our probability. So, the probability will change if I draw a card randomly only from the top half.Log in to reply

– Jesse Nieminen · 1 month, 1 week ago

It would change if you knew in which order the cards are in, but since the cards are shuffled, every card has \(\dfrac1{13}\) chance to be a queen, and thus a random limited pick is as good as a random pick.Log in to reply

– Nilabha Saha · 1 month, 1 week ago

I am getting quite perplexed at this question. My mind is fluctuating at considering your answer correct and my answer correct. Please give asome time - maybe a few days - to properly understand this question. I am just getting confused.Log in to reply