Probability of a drawn card be a diamond given post conditions

[holemole]

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.

 

[Hurkyl]

Try considering another scenario:

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 13 consecutive cards.

It happens to be that I picked 13 diamonds.

Now I want to find the probability of the card in the hat be a diamond.​

 

[pmsrw3]

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.

I understand what you're thinking. You're partly right. It is absolutely correct that, before you drew the additional cards, the probability of a diamond in the hat was 1/4. And you're right that the three subsequent draws can't affect the previous draw in any way. But you're wrong to think that that can't affect the probability characteristics of the first draw.

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.

 

[Jaynte]

This is the probability for picking 3 cards of diamonds if the one in the hat is Diamond or not:
\frac{3*\frac{13}{51}*\frac{12}{50}*\frac{11}{49}+1*\frac{12}{51}*\frac{11}{50}*\frac{10}{49}}{4}

 

[Jaynte]

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

 

[Jaynte]

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

 

[holemole]

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

 

[holemole]

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

Could you explain to me why this is the case?

 

[pmsrw3]

i understand that information has changed. Nevertheless I'm not convinced that the information affects the original draw

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.

 

[holemole]

i understand that information has changed. Nevertheless I'm not convinced that the information affects the original draw

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.

A probability, as far as I understand, is calculated by the number of events that has the results I'm trying to find(in this case shape=diamond) divided by total number of possible events. My red-yellow ball example adopts new information(second ball shoud be yellow) forces change in number of events I'm trying to find(from simply first draw being yellow to first AND second draw being yellow) and total number of possible events(from 4 to 3 since one ball is fixed).
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.

 

[pmsrw3]

A probability, as far as I understand, is calculated by the number of events that has the results I'm trying to find(in this case shape=diamond) divided by total number of possible events.

Only if all events are equally probable. For instance, suppose I tried to tell you that the probability that a flipped coin comes up heads is 1/3, because it could come up heads, it could come up tails, or it could land balanced on the edge.

\frac{\text{number of events that has the results I'm trying to find}}{\text{total number of possible events}} = \frac{1}{3}.

You wouldn't agree with that, would you?

You can't just treat all possible events as equal.

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.

It is possible, yes. But not equally probable. Diamond followed by three diamonds is less probable than non-diamond followed by three diamonds.

drawing 3 consecutive diamonds is just a random event in any case, whether there were 12 diamonds in 51cards or 13 diamonds.

Rain on July 17, 3011 is "just a random event in any case". Supernova on July 17, 3011 is also "just a random event in any case". That doesn't mean they're equally probable.

 

[holemole]

Only if all events are equally probable.

first single card draw is consisted of equally probable events, isn't it?

It is possible, yes. But not equally probable. Diamond followed by three diamonds is less probable than non-diamond followed by three diamonds.

This is completely true, but we're not trying to find number of events concerning first draw as diamond amongst total number of events where 3 consecutive draws after single draw are diamonds. 3 diamonds is not our 'condition'. It is just a result we stumbled upon randomly.

Rain on July 17, 3011 is "just a random event in any case". Supernova on July 17, 3011 is also "just a random event in any case". That doesn't mean they're equally probable.

I'm not saying two random events are equally probable. I'm saying that 3 consecutive card draws turning out to be all diamonds is not a preset condition. If that's the case, answer IS 10/49. It can easily be calculated by finding number of events that has first draw as diamond AND has next 3 draws as diamonds as well which is 13*12C3, and divide this by total number of events that has 3 draws as diamonds but no condition on first draw which is 13*12C3+3*13*13C3.
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.

 

[Hurkyl]

This is completely true, but we're not trying to find number of events concerning first draw as diamond amongst total number of events where 3 consecutive draws after single draw are diamonds. 3 diamonds is not our 'condition'. It is just a result we stumbled upon randomly.

Er, yes it is your condition. You saw three diamonds, so now all posterior probabilities ought to be conditioned on that fact. You will find that, out of all the experiments you do, less than one quarter of those where you drew 3 diamonds will have a diamond as the first card.

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.

 

[disregardthat]

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(A|B) which is 5/98, or approximately 5.1%. I think.

 

[pmsrw3]

first single card draw is consisted of equally probable events, isn't it?

Yes. And after the first draw, the probability of a diamond is 13/52 = 1/4.

This is completely true, but we're not trying to find number of events concerning first draw as diamond amongst total number of events where 3 consecutive draws after single draw are diamonds. 3 diamonds is not our 'condition'. It is just a result we stumbled upon randomly.

Yes, it is a result that you stumbled on randomly. You seem to think that whenever you use the word "random", that means you can ignore it. But this is not true. Random events still carry information.

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.

I'm not saying two random events are equally probable. I'm saying that 3 consecutive card draws turning out to be all diamonds is not a preset condition. If that's the case, answer IS 10/49.

As a matter of fact, that's wrong. The probability of drawing three diamonds after one diamond has been drawn is much less than 10/49.

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.

No, you're right, the original question doesn't ask you to find the probability of drawing three diamonds after a first draw of a diamond. It asks you to find the probability that the first draw is a diamond, given that the three subsequent draws are diamonds. These are not the same number. But they ARE related (see http://en.wikipedia.org/wiki/Bayes%27_theorem" ). It doesn't matter that this was not what you call a "preset condition". You've now drawn three diamonds. Whether you set out to do that or not is totally irrelevant. You now have that information.

The probability that the first draw was a diamond is multiplied by

<br /> \frac{\text{probability of drawing three diamonds if first was diamond}}{\text{probability of drawing three diamonds if first was non-diamond}}<br />

That's why it's relevant that the two probabilities aren't equal.

 

[disregardthat]

<br /> \frac{\text{probability of drawing three diamonds if first was diamond}}{\text{probability of drawing three diamonds if first was non-diamond}}<br />

Shouldn't this be

<br /> \frac{\text{probability of drawing three diamonds if first was diamond}}{\text{probability of drawing three diamonds if first was non-diamond or drawing three diamonds if first was diamond}}<br />

?

 

[pmsrw3]

Shouldn't this be

<br /> \frac{\text{probability of drawing three diamonds if first was diamond}}{\text{probability of drawing three diamonds if first was non-diamond or drawing three diamonds if first was diamond}}<br />

?

No -- notice I wrote that this is what the prior probability "is multiplied by". That is, I'm giving the factor by which it changes.

 

[holemole]

As a matter of fact, that's wrong. The probability of drawing three diamonds after one diamond has been drawn is much less than 10/49.

10/49 is the probability of the FIRST draw to be a diamond when three draws after the first are all diamonds.

 

[pmsrw3]

So, here's an experiment. It's a little impractical, but it could be simulated. We agree to play the following game:

1. Shuffle a deck of cards.
2. You take the first card.
3. We turn over the next three.
4. If those three are not all of the same suit, this round is over: back to step 1.
5. If the three cards are all the same suit, we look at your card. If your card is the same suit as the three, I give you $3. If your card is of a different suit you give me $0.95. Now, you should be very happy to do this, because you expect a net profit. But I think I'll profit.

What do you say. Should I run it? I'll show you the code, so you'll know I'm not cheating.

 

[holemole]

So, here's an experiment. It's a little impractical, but it could be simulated. We agree to play the following game:

1. Shuffle a deck of cards.
2. You take the first card.
3. We turn over the next three.
4. If those three are not all of the same suit, this round is over: back to step 1.
5. If the three cards are all the same suit, we look at your card. If your card is the same suit as the three, I give you $3. If your card is of a different suit you give me $0.95. Now, you should be very happy to do this, because you expect a net profit. But I think I'll profit.

What do you say. Should I run it? I'll show you the code, so you'll know I'm not cheating.

This is different from the original question because this process, regarding our net profits, would leave out the cases where first is diamond but next three are of different suit. It would only process our earning only with cases where three diamonds are present.

Three consecutive diamonds being drawn does not represent the population from which we're trying to calculate our probability. It does represent if we change the question into 'probability of the first card being diamond only when next three cards are also diamond'. I still think these two cases are different, and your game suggests that they are the same. Whether I'm right or wrong I need some explanation on this part to be fully pursuaded.

 

[pmsrw3]

Three consecutive diamonds being drawn does not represent the population from which we're trying to calculate our probability. It does represent if we change the question into 'probability of the first card being diamond only when next three cards are also diamond'. I still think these two cases are different, and your game suggests that they are the same. Whether I'm right or wrong I need some explanation on this part to be fully pursuaded.

Well, in that case, we are no longer in substantial disagreement. If you agree that the probability that the first draw is a diamond is different from 1/4 in those cases in which the three subsequent draws are also diamonds, then I think you have essentially agreed to what we've been trying to convince you of. Perhaps there is some metaphysical sense in which a different probability of 1/4 can be said to still exist, which takes into account all those possibilities that we now know to be false, but I think you agree that in practice, you should act as if the probability has changed after drawing three diamonds.

 

[pmsrw3]

In[60]:= game[nt_] :=
 Module[{draw, ws = 0, ls = 0},
  Do[
   draw = Last /@ RandomSample[Tuples[{Range[13], {h, d, s, c}}], 4];
   If[draw[[2]] == draw[[3]] == draw[[4]], , Continue[], Continue[]];
   If[draw[[1]] == draw[[2]], ws++, ls++, ls++],
   {nt}
   ];
  {ws, ls}
  ]

In[69]:= game[100000].{-3., 0.95}

Out[69]= 913.65

You owe me $913.65 :-)

 

[Jaynte]

Lol :)

 

[Jocko Homo]

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

So, you understand that subsequent draws may affect the probability of the first draw but...

i understand that information has changed. Nevertheless I'm not convinced that the information affects the original draw

A probability, as far as I understand, is calculated by the number of events that has the results I'm trying to find(in this case shape=diamond) divided by total number of possible events. My red-yellow ball example adopts new information(second ball shoud be yellow) forces change in number of events I'm trying to find(from simply first draw being yellow to first AND second draw being yellow) and total number of possible events(from 4 to 3 since one ball is fixed).
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.

...suddenly, you don't.

Your objection makes no sense. Your argument is that the first draw can still be anyone of the four suits. However, that was true in your ball example! The ball first drawn may be anyone of the two colors, despite a second draw being yellow.

You recognize what people are saying to you with the ball example but, for some reason, you're obstinately denying it with the deck of cards. What do you think the distinction between the two is?

 

[holemole]

Well I thought the ball example and the card example are separate cases, since the card example shows subsequent draw as just one case whereas the ball example explicitly limits the cases in question to those with second ball as yellow.

The more I think about it I think those two questions COULD be same from the fact that conditional probability is afterall a probability of A given B has happened...and in the original question B has happened...I'm not 100% sure if that's the same situation as a question where 3 card be all diamonds is stated prior to any drawing.

 

[Jocko Homo]

Well I thought the ball example and the card example are separate cases, since the card example shows subsequent draw as just one case whereas the ball example explicitly limits the cases in question to those with second ball as yellow.

The more I think about it I think those two questions COULD be same from the fact that conditional probability is afterall a probability of A given B has happened...and in the original question B has happened...I'm not 100% sure if that's the same situation as a question where 3 card be all diamonds is stated prior to any drawing.

"Could be" the same? "Could!?"

You're still rationalizing hard. Your talk of "seperate cases" and "just one case" aren't making any sense. "Stated prior to any drawing" makes no sense, either...

You have a bag with four balls, half of which are yellow and the rest are red. You pull one ball out without looking at it and stick it in a hat. You pull another ball out and it's yellow. What is the probability that the ball in the hat is yellow?

You have shuffled a standard 52 card deck. You draw the top card and put it in a hat without looking at it. You draw three more cards and flip them over. They all turn out to be diamonds. What are the chances that the card in the hat is a diamond?

What specifically distinguishes one scenario from another?

 

[SpectraCat]

A variant of Hurkyl's first illustration that may help is that following:

1) you pick a card out of a shuffled (i.e. randomly ordered) 52-card deck and place it in a hat

2) you now pick 12 more cards, which all happen to be diamonds

3) what is the probability that the card in the hat is a diamond?

You might be tempted to say it must be 1/52, because there is only card remaining out of the original 52 that could satisfy the condition. However that cannot be correct, as can be appreciated from considering another question:

4) what is the probability that the card in the hat is NOT a diamond?

Since you have already seen 12 cards (all of which happen to be diamonds), and only one card out of the ones you have NOT seen is a diamond, the answer to 4 is clearly 39/40. Therefore, since the answers to 3 & 4 must sum to 1, the answer to 3 must be 1/40.

Now, do you see more clearly how the posterior conditions (i.e. knowing the results of 12 subsequent draws) affect the posterior probability of the initial event?