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maybe you can help me out. The following holds: There is a deck of 52 cards. 13 clubs, 13 spades, 13 hearts, 13 diamonds). Normal deck of cards. Nothing special.

And now the following happens. I draw a card, and put it away, without looking at it. I then draw 2 more cards and look at them. Each of those 2 cards were spades. And now I want to know, given this information, what the probability is, that the first card I didn't look at, was spades as well.

I don't even know, how to start, notation wise. Like P(first drawn card = spades| 2nd and 3rd drawn card = spades)? And normally, should it not just be P(first card = spades) =13/52? But I feel, that would be wrong here.

I would try to calc 13/52 x 12/51 x 11/50 = 0.0129. So that would be 1.3% But that doesn't consider the information properly, I think.

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There are $50$ equiprobable candidates for the first card and $11$ of these candidates are spades.

So the probability that the first card is a spade equals: $$\frac{11}{50}$$

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  • $\begingroup$ Indeed, and that reminded me this riddle $\endgroup$ Jan 16 '20 at 10:35
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    $\begingroup$ @RingØ Yes, I can imagine. My solution on that: someone who sticks to his original choice will win if his original choice is correct. Probability on that is $\frac13$. Some who does not stick to his original choice will win if his original choice is wrong. Probability on that: $\frac23$. $\endgroup$
    – drhab
    Jan 16 '20 at 10:37
  • $\begingroup$ Okay, so here is the catch. I think i am supposed to work with Bayes Theorem. Also this task gives 6 points, which means, i am roughly supposed to take 6 minutes to solve it. $\endgroup$ Jan 27 '20 at 11:23
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    $\begingroup$ @ThePianist Yes, the probability of the first card being a spade is ¼ but the question answered concerns conditional probability. Think of a deck of 3 cards of which 1 shows a car and 2 show a goat. The probability that the first card shows a car is 1/3. Let us denote that as P(c1)=1/3. The probability that the first card shows a car under condition that the second shows a goat is 1/2. Notation P(c1|g2)=1/2. Also you could say that the condition provides new insight in the whereabouts of the car.... $\endgroup$
    – drhab
    Mar 9 '20 at 8:27
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    $\begingroup$ @ThePianist ....This would happen if Monty would choose randomly a second door and behind this door a goat was found. But things are different. In fact Monty rearranges the two cards in such a way that the upper card always shows a goat. Denoting this as M we have P(c1|M)=1/3=P(c1) because this condition does not affect the probability and consequently P(c3|M)=2/3 because evidently P(c2|M)=0$ and P(c1|M)+P(c2|M)+P(c3|M)=1. $\endgroup$
    – drhab
    Mar 9 '20 at 8:28
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How about writing $S_i$ for the event that the $i$th card drawn is a spade? Now you need Bayes' formula: $$\Pr(S_1\mid S_2,S_3)=\frac{\Pr(S_1,S_2,S_3)}{\Pr(S_2,S_3)}.$$

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  • $\begingroup$ Okay, cool. So i will have to look at Bayes again. Hm. How do i calculate Pr(S1,S2,S3) and Pr(S2,S3)? I just dont know, what the "," means, between S2 and S3. I know "|" is "given". But the comma, i dont understand. Neither in the context, nor how to calculate. $\endgroup$ Jan 21 '20 at 12:06
  • $\begingroup$ @user3556093 I just used the comma for "and". So $\Pr(S_1,S_2,S_3)$ is the probability that the first three cards are all spades, which is the value you worked out above. $\endgroup$ Jan 21 '20 at 13:11
  • $\begingroup$ Okay, so how do i calculate "and"? $\endgroup$ Jan 27 '20 at 11:21
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Let us look at the desired conditional probability from the point of view of the classical definition of probability.

We need to find the proportion of triples of cards, in which the first card is a spade, among all the triples of cards, where the second and third are spades. It is exactly the same as the proportion of fully spades triples among triples with a second and third are spades. And there is the same number of triples of cards, where the second and third are spades and where the first and the second are spades.

Therefore this proportion is exactly the same as the proportion of fully spades triples among all triples, where the first and second are spades. And it coincides with the probability of pulling out the third spade if the first two spades are already removed. So it is $\dfrac{13-2}{52-2}$.

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  • $\begingroup$ okay. That sounds nice. But i am not sure if this is the way i am supposed to do this task. I got 6 minutes for it. And i think i am supposed to use Bayes Theorem to solve it. $\endgroup$ Jan 27 '20 at 11:24
  • $\begingroup$ Yes, for sure, the first solution should be using Bayes theorem. This is the explanation afterwards. If you get by Bayes theorem the same result as for third spade after two spades are already removed, you will think whether these probabilities are the same. With a little training, you can learn to think this way from the very beginning. $\endgroup$
    – NCh
    Jan 27 '20 at 11:41
  • $\begingroup$ By the way, I think that time is always given for not the most rational solution, but for the standard one. Using rational solutions, you can significantly save time. $\endgroup$
    – NCh
    Jan 27 '20 at 11:46

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