Probably Cards, I See

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? Note by Nilabha Saha
3 years, 2 months ago

This discussion board is a place to discuss our Daily Challenges and the math and science related to those challenges. Explanations are more than just a solution — they should explain the steps and thinking strategies that you used to obtain the solution. Comments should further the discussion of math and science.

When posting on Brilliant:

• Use the emojis to react to an explanation, whether you're congratulating a job well done , or just really confused .
• Ask specific questions about the challenge or the steps in somebody's explanation. Well-posed questions can add a lot to the discussion, but posting "I don't understand!" doesn't help anyone.
• Try to contribute something new to the discussion, whether it is an extension, generalization or other idea related to the challenge.
• Stay on topic — we're all here to learn more about math and science, not to hear about your favorite get-rich-quick scheme or current world events.

MarkdownAppears as
*italics* or _italics_ italics
**bold** or __bold__ bold

- bulleted
- list

• bulleted
• list

1. numbered
2. list

1. numbered
2. list
Note: you must add a full line of space before and after lists for them to show up correctly
paragraph 1

paragraph 2

paragraph 1

paragraph 2

> This is a quote
This is a quote
# I indented these lines
# 4 spaces, and now they show
# up as a code block.

print "hello world"
# I indented these lines
# 4 spaces, and now they show
# up as a code block.

print "hello world"
MathAppears as
Remember to wrap math in $$...$$ or $...$ to ensure proper formatting.
2 \times 3 $2 \times 3$
2^{34} $2^{34}$
a_{i-1} $a_{i-1}$
\frac{2}{3} $\frac{2}{3}$
\sqrt{2} $\sqrt{2}$
\sum_{i=1}^3 $\sum_{i=1}^3$
\sin \theta $\sin \theta$
\boxed{123} $\boxed{123}$

Sort by:

1. How many queens are there in the deck?
2. What is the probability that a specific card is in the spot picked at random?
3. What is the probability that the card picked at random is any of the queens?

- 3 years, 2 months 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?

- 3 years, 2 months ago

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

- 3 years, 2 months ago

Whenever I divide a deck into two, my sample space decreases from 52 to 26. However, the number of Queens in the smaller sample place is dependent on my random choice. So, we can calculate the probability directly given the 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.

- 3 years, 2 months 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.

- 3 years, 2 months 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.

- 3 years, 2 months 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.

- 3 years, 2 months 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.

- 3 years, 2 months 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.

- 3 years, 2 months 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.

- 3 years, 2 months 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.

- 3 years, 1 month 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?

- 3 years, 1 month 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.

- 3 years, 1 month 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.

- 3 years, 1 month 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.

- 3 years, 1 month ago

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

When you figure that out, try this

Staff - 3 years, 2 months ago