Probability of a deck of cards

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Homework Help Overview

The discussion revolves around a probability problem involving a standard deck of 52 cards. The scenario involves drawing cards at random and without replacement until only cards of one suit remain. Participants are tasked with determining the probability that the remaining cards are hearts.

Discussion Character

  • Exploratory, Assumption checking, Conceptual clarification

Approaches and Questions Raised

  • Participants explore the symmetry of the suits and suggest that the probability for each suit being the last one remaining is equal. Some discuss the implications of the last card drawn determining the last suit left. Others propose mathematical expressions to represent the problem.

Discussion Status

There is ongoing exploration of the problem with various interpretations being discussed. Some participants suggest that the probability of the last suit being hearts is 0.25, while others confirm this reasoning. However, there is no explicit consensus on the final approach or solution.

Contextual Notes

Participants note the constraints of the problem, including the requirement to consider the total number of cards and the equal distribution of suits within the deck. There are also reminders about posting homework-related questions in the appropriate forum.

jinbaw
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Hi ! I need help with this problem.
We have a 52 cards deck, and we're drawing at random and without replacement, until cards of one suit only are left. We need to find the probability that the remaining cards are hearts.

Any suggestions concerning where to start?
 
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jinbaw said:
Hi ! I need help with this problem.
We have a 52 cards deck, and we're drawing at random and without replacement, until cards of one suit only are left. We need to find the probability that the remaining cards are hearts.

Any suggestions concerning where to start?

Symmetry.
 
Umm.. Can you elaborate a bit what you mean ?
Thanks
 
How many suits are there in a standard deck?
 
jinbaw said:
Umm.. Can you elaborate a bit what you mean ?
Thanks

There is no difference between the suits. So whatever probability you have for the last suit being spades will also be the probability for clubs, diamonds and hearts. That is, the suits are symmetrical in this problem.

Also, consider this. The last suit left is bound to be the suit of the last card in the deck. So you really only need to know the last card to know what suit will be left last.
 
Well there are four.
I tried taking an event An = {the nth card is a spade}, and maybe take the union of the intersections [tex]\bigcup^{52}_{i=39}\bigcap^{52}_{j=i} A_{j}[/tex]. But this is of no use I guess...
 
Oh.. So the idea is P(the last suit is hearts) = P(last card is a heart) = 1/4 ?
 
I think so, yes.
 
Agree, yes. The answer is 0.25.
 
  • #10
hi

the sample space is 52,

and the required event is 13 (i.e u r saying that at last there should be all hearts.that means we have 13 hearts in a deck.

so the event is 13,

and the probability is n(E)/n(S),(here E=event, S=sample space)
answer is 13/52,

i.e 0.25
 
  • #11
(Moderator's note: thread moved from "Set Theory, Logic, Probability, Statistics")

Please note for the future, homework assignments or any textbook style exercises are to be posted in the appropriate forum in our Homework & Coursework Questions area. This should be done whether the problem is part of one's assigned coursework or just independent study.
 
  • #12
To jinbaw;

Clearing your thought, we must first understand the range of possibility and the deck.

*[1] possibility = probability = x; {0 < x < 1}

Looking at your problem by default, I knew that total number of cards in the deck = sample space = 52 and within the deck there are four suits. {Hearts, Spades, Diamonds, Clubs)

I will then divide my deck into 4 different group of suits. {52/4 = 13}

*Always remember that you are to determine the possibility, therefore drawing at random will never be a problem (in this case).

Now I have confirm 13 cards in deck at random will confirm hearts.

hence, I can say that the possibility = 13(hearts)/52(sample space) = 0.25

therefore 0.25 = x where 0 < x < 1, it agree with the condition [1] above.

To pre-determine your outcome answer: possibility of suit of hearts {13/52 = 0.25}
possibility of suit of spades {13/52 = 0.25}
possibility of suit of diamonds {13/52 = 0.25}
possibility of suit of clubs {13/52 = 0.25}

The total sum of possibility = probability = 0.25(hearts) + 0.25(spades) + 0.25(diamonds) + 0.25(clubs) = 1 confirm outcome of the deck.
 

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