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Probability of a drawn card be a diamond given post conditionsby holemole
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#1
Jul1611, 05:24 AM

P: 10

Setting is like this: I have 52cards of 4 shapes. that's 13 each.
I take one card out and put it in a hat. now with 51card I draw 3 consecutive cards. It happens to be that I picked 3 diamonds. Now I want to find the probability of the card in the hat be a diamond. My friend said it should be 10/49 since 3 diamonds are out, but I disagree. I think those 3 diamonds with only one draw doesn't give enough information to favor other shapes than diamond at all, since the one card draw happened before any additional information, and 3 consecutive draws cannot affect previous draw in any way making the first draw independent of the second draw. 


#2
Jul1611, 05:57 AM

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Try considering another scenario:
Setting is like this: I have 52cards of 4 shapes. that's 13 each. 


#3
Jul1611, 08:04 AM

P: 455

Probability is a function of information. It's about what you know and don't know. When you draw those three cards, your information changes, and therefore the probability changes. Most probability paradoxes are about this. People have a tendency to think that probability exists independent of persons and information. They will naturally think that the probability that the card in the hat is a diamond is a characteristic of that physical setup, the card and the hat, and that if you don't change the card and the hat, the probability can't change. But that's wrong. The probability is about you and what you know. So there is no paradox in the idea that the probability changes when your information changes. So, suppose you drew three diamonds, but you didn't show them to your friend, who is also there with you? What's the probability now? You have new information, so your probability is 10/49. But your friend has learned nothing new, so her probability is still 1/4. Which is right? You're both right. Since probability depends on information, the same event came have different probabilities for different people with different information. The probability of a diamond in the hat is different for you and your friend. One again, the probability is not just a characteristic of the event. 


#4
Jul1611, 08:14 AM

P: 71

Probability of a drawn card be a diamond given post conditions
This is the probability for picking 3 cards of diamonds if the one in the hat is Diamond or not:
[tex]\frac{3*\frac{13}{51}*\frac{12}{50}*\frac{11}{49}+1*\frac{12}{51}*\frac {11}{50}*\frac{10}{49}}{4}[/tex] 


#5
Jul1611, 08:56 AM

P: 71

Sorry, i didnt read your question correctly. I would say that probability is 1/4



#6
Jul1611, 08:59 AM

P: 71

0% after you see the 13 diamonds, 1/4 before you see any diamonds.



#7
Jul1611, 09:14 AM

P: 10

i understand that information has changed. Nevertheless I'm not convinced that the information affects the original draw, since the population of the first draw doesn't necessarily change before I somehow pick 13 diamonds from 51 cards, which I didn't. I'm aware that if any new information that forces the population of the original event then it influences the probability like having 2 yellow balls and 2 red balls, and trying to find the probability that first ball is yellow WHEN second ball is yellow, which is not 1/2 but 1/3



#8
Jul1611, 09:15 AM

P: 10




#9
Jul1611, 09:33 AM

P: 455

[QUOTE=holemole;3406228]i understand that information has changed. Nevertheless I'm not convinced that the information affects the original draw[quote]
The new information DOESN'T affect the original draw. I said so in my answer. The original draw either was or was not a diamond. If it wasn't a diamond, it will never be a diamond, no matter what you draw after. If it was a diamond, it will always be a diamond, no matter what you draw next (but you're guaranteed not to draw 13 more diamonds). And that is why part of Hurkyl's answer was "...1/4 before you see any diamonds". But you are leaping to the conclusion that the information doesn't affect the probability of the first draw. That's completely different. "Doesn't affect the first draw" and "doesn't affect the probability of the first draw" are NOT the same. The first doesn't include the second. New information changes the probability without changing the physical event. 


#10
Jul1611, 10:14 AM

P: 10

[QUOTE=pmsrw3;3406254][QUOTE=holemole;3406228]i understand that information has changed. Nevertheless I'm not convinced that the information affects the original draw
However drawing 3 diamonds from 51 cards doesn't necessarily change those number of events since it is always possible to have first card as diamond AND pick 3 diamonds afterwards OR have first as other shape AND pick 3 diamonds after. Original number of possible events stays the same, since drawing 3 consecutive diamonds is just a random event in any case, whether there were 12 diamonds in 51cards or 13 diamonds. Number of events I'm trying to find didn't change either, since I didn't add any condition to the events I want to find. 


#11
Jul1611, 10:33 AM

P: 455

[QUOTE=holemole;3406306][QUOTE=pmsrw3;3406254]
[tex]\frac{\text{number of events that has the results I'm trying to find}}{\text{total number of possible events}} = \frac{1}{3}[/tex]. You wouldn't agree with that, would you? You can't just treat all possible events as equal. 


#12
Jul1611, 10:58 AM

P: 10

However this is NOT what I'm trying to find. Original question never says to find probability that has 3 draws after the first one to be all diamonds. Thus two cases have different number of possible events. 


#13
Jul1611, 11:12 AM

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After drawing 3 diamonds, asking for the unconditioned probability is a purely hypothetical exercise. (although it is important to be able to comprehend such hypothetical questions when appropriate) Edit: A more blatant example that the posterior probabilities must be conditioned is this: I flip over the top card, see that it's a diamond, and put it back. What are the odds the top card is a diamond?Your line of reasoning would say 1/4. 


#14
Jul1611, 11:20 AM

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P: 1,742

Call the event that the card in the hat is diamond A, and drawing 3 consecutive diamonds while one randomly chosen card is in the hat B. We want P(AB) which is 5/98, or approximately 5.1%. I think.



#15
Jul1611, 11:24 AM

P: 455

Suppose you were walking around and you saw a $100 bill on the sidewalk. Would you say, "Well, the probability of picking up a $100 bill on the sidewalk is practically zero. And, I was 'not trying to find number of events concerning' being able to pick up a $100 bill from the sidewalk. I 'just stumbled upon it randomly'. So the chance that I can now pick up a $100 bill from the sidewalk is still practically zero. Therefore, I'm not going to bend down and pick it up, because I know that the chance that I can pick up a $100 bill from the sidewalk is not worth the effort."? Please, please, PLEASE, I beg you, stop saying, "It's just random". That's meaningless. Everything is "just random", to some degree or other. The idea that you can say "It's just random" and ignore it is WRONG. The probability that the first draw was a diamond is multiplied by [tex] \frac{\text{probability of drawing three diamonds if first was diamond}}{\text{probability of drawing three diamonds if first was nondiamond}} [/tex] That's why it's relevant that the two probabilities aren't equal. 


#16
Jul1611, 11:29 AM

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P: 1,742

[tex] \frac{\text{probability of drawing three diamonds if first was diamond}}{\text{probability of drawing three diamonds if first was nondiamond or drawing three diamonds if first was diamond}} [/tex] ? 


#17
Jul1611, 11:35 AM

P: 455




#18
Jul1611, 11:43 AM

P: 10




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