Simple Probability Question:


by Jin314159
Tags: probability, simple
Jin314159
#1
Jul21-04, 02:54 AM
P: n/a
I have in my hand a standard deck of cards. You take the top card off the deck and you hold it in your hand without looking at it. Then I pick a card at random from the remaining deck and I show you that it's a heart.

What is the probability that the card in your hand is a heart?
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relinquished™
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#2
Jul21-04, 05:45 AM
P: 79
*thinks...*

I think this is the answer:

In the first draw, there are 13 cards that have a heart. There are 52 standard cards in a deck. The Probability of getting a heart on the first draw is:

Let H be the even of drawing a heart from a deck of 52 cards

P(H) = n(H) / n(S) = 13/52

You see, the probability of getting a heart on the second draw is a conditional probability because it depends on whether or not you get a heart on the first draw BUT the second event does not affect the first event. Your second draw does not affect the probability of the first draw ^_^

That is of course you shuffle the deck really well to make the first draw be a random event, otherwise that's a whole new story ^_^;;
Hurkyl
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#3
Jul21-04, 06:17 AM
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Because both picks are random, the probabilities will be the same no matter what order you do them...

HallsofIvy
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#4
Jul21-04, 08:03 AM
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Thanks
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Simple Probability Question:


Why do you say "no" and then proceed to prove what Hurkyl said?

"Let A be the event that the first draw is heart and B be the event that the second draw is heart. The calculation of "probability of A given B" (i.e. ) goes like this."
so it doesn't matter whether you (1) draw a heart, hold it unseen, turn over another card which is a heart and ask "what is the probability the first card was a heart?" or (2) draw a card, revealing it to be a heart, and then ask "what is the probability that the next card will be a heart?".
Wong
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#5
Jul21-04, 08:14 AM
P: 80
Oh...sorry, maybe I misunderstood what Hurkyl said...better for me to delete the post
mathman
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#6
Jul21-04, 07:16 PM
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It looks to me that the probability is 12/51. Given the known card is a heart, the unknown is simply one of the remaining 51, of which 12 are hearts.
relinquished™
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#7
Jul22-04, 05:04 AM
P: 79
I believe the probaility is still 13/52 or 1/4 because you are asking "what is the probability that the card in the first draw is a heart" because the card in your hand is the frist card drawn. If the question was, "what was the probability that the 2nd card was a heart?" It would have been EITHER 12/51 or 13/51 depending on the first card.

The Event of getting a heart on the first draw is independent from the event of drawing a heart in the second draw, however, the latter is dependent on the former because your Sample Space is reduced from a standard deck of 52 to 51 since you already took a card.
chronon
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#8
Jul22-04, 08:18 AM
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No, the probabililty that the first card is a heart can depend on the second card, as it gives you extra information. Remember that probabilty refers to your knowledge of what the card is - what it actually is has already been decided.

Looking at it another way, suppose that you put the first card aside and then looked at the remaining 51 cards and found 13 hearts amoung them. Would you still say that the probabilty that the first card is a heart is 1/4
Jin314159
#9
Jul22-04, 09:08 PM
P: n/a
It seems like everyone has agreed on 12/51 and I think that's right.

However, this seems to contradict the Monty Hall problem. (And I apologize for those who aren't familiar with the problem, just google "Monty Hall" and you'll find tons of links). In the Monty Hall problem, the probability that the prize is behind the first door you picked is still 1/3, even after Monty opens one of the remaining doors revealing there is no prize.

So in the Monty Hall problem, the probability remains the same even after new information has been introduced. But in this card problem, the probability changes after new information has been introduced. Why?
mathman
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#10
Jul22-04, 11:13 PM
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In Monty Hall, Monty knows where the prize is. As a result he is picking a door where the prize is not. The card deck problem would be like Monty Hall if the second card was not chosen at random, but was always a heart picked out deliberately. In this case, the first card heart probability would be 13/52.
Jin314159
#11
Jul23-04, 02:22 AM
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Quote Quote by mathman
In Monty Hall, Monty knows where the prize is. As a result he is picking a door where the prize is not. The card deck problem would be like Monty Hall if the second card was not chosen at random, but was always a heart picked out deliberately. In this case, the first card heart probability would be 13/52.
Why yes, of course. Thank you, good sir.
relinquished™
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#12
Jul23-04, 08:11 PM
P: 79
Quote Quote by chronon
No, the probabililty that the first card is a heart can depend on the second card, as it gives you extra information. Remember that probabilty refers to your knowledge of what the card is - what it actually is has already been decided.

Looking at it another way, suppose that you put the first card aside and then looked at the remaining 51 cards and found 13 hearts amoung them. Would you still say that the probabilty that the first card is a heart is 1/4
hmm... that makes sense... so no matter which comes first the first card can still be affected by the second... thanx for the clarification...
irony of truth
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#13
Jul24-04, 02:46 AM
P: 93
"What is the probability that the card in your hand is a heart?"

- what do you mean? the drawn, shown card or the the card that is not shown (the top one)?
Jin314159
#14
Jul24-04, 04:39 PM
P: n/a
Quote Quote by irony of truth
"What is the probability that the card in your hand is a heart?"

- what do you mean? the drawn, shown card or the the card that is not shown (the top one)?
The card that is in your hand, that is not shown. We know what the shown card is already a heart. It's probability of being a heart is %100.


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