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The problem I'm contemplating is the following, perhaps a basic question.

We have $52$ cards. The probability of picking a diamond card is $13/52 = 1/4$.

Now, we are picking one card out of $52$ cards.

If we know the picked card is a diamond card, the next probability of picking a diamond is $12/51$ or $4/17$ or $0.235\ldots$

However, if we don't know anything about the picked card. What will be the probability of picking a diamond card from the $51$ cards?

If we are to calculate this probabilities by summing this way,

$$P(\text{second card is a diamond} \mid \text{first card is a diamond}) \cdot P(\text{first card is a diamond}) + P(\text{second card is a diamond card} \mid \text{first card is not a diamond}) \cdot P(\text{first card is not a diamond})$$

It equals to

$$\frac{1}{4} \cdot \frac{12}{51} + \frac{3}{4} \cdot \frac{13}{51}$$ which equals to $0.25$.

How does not knowing the card increase the probability? Can someone please explanation the intuition for this to happen?

Also, why are we adding two probabilities, only one of them is true, right?

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  • $\begingroup$ In a standard deck, the probability of picking a red card is $1/2$ since $26$ of the $52$ cards are red ($13$ hearts and $13$ diamonds). Are we given that the probability of choosing a red card is $1/4$? $\endgroup$
    – N. F. Taussig
    Commented Oct 3, 2019 at 10:43
  • $\begingroup$ Corrected the question, @N.F.Taussig $\endgroup$
    – Skynet094
    Commented Oct 3, 2019 at 10:52
  • $\begingroup$ This tutorial explains how to typeset mathematics on this site. $\endgroup$
    – N. F. Taussig
    Commented Oct 3, 2019 at 11:06
  • $\begingroup$ You calculated two different probabilities, the probability that the second card is red given that the first card is red and the probability that the second card is red regardless of which card was picked first. Both of your calculations are correct. There is no contradiction. $\endgroup$
    – N. F. Taussig
    Commented Oct 3, 2019 at 11:08
  • $\begingroup$ As a nitpick, you say "the next probability of picking a red card is $12/51$ or..." You meant to say diamond here. Both hearts and diamonds are considered to be "red cards" in these style problems as it is assumed we are using a standard 52 card playing deck and both hearts and diamonds are commonly both colored red. $\endgroup$
    – JMoravitz
    Commented Oct 3, 2019 at 12:16

2 Answers 2

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If we picked a card which appears to be a diamond and go on with picking a card from the remaining cards then we are landed in a new situation which asks for a new mathematical model: we pick from $51$ cards are left and $12$ of them are diamonds. The first card, (e.g. diamond with rank $4$) is no longer a candidate.

If we place one randomly chosen card apart and then from the remaining $51$ cards pick out a card then actually nothing has changed. For the card that is chosen as second there are $52$ equiprobable candidates and $13$ of them are diamonds. This has not changed by the fact that at first one randomly chosen card was placed apart.

To get better hold of this try it out with e.g. $2$ socks: a blue one and a red one. Repeat your thinking about it several times and the intuition will land.

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It's will be summation of two independent cases: 1) if first card is red ,then probability for second card is red:(1/4)(12/51) 2)if first card is not red,then probability for second card is red:(1-(1/4))(13/51) Ans:0.25 *here,Two events are dependent ,means second pick will influenced by first one and for second event it will have two scenarios, summation of two independent cases will give total probability.

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  • $\begingroup$ Hi, I have a question, why does not KNOWING any information about the card increase the probability of a card being red? $\endgroup$
    – Skynet094
    Commented Oct 3, 2019 at 10:41
  • $\begingroup$ Updated the question, I had some mistakes in there. $\endgroup$
    – Skynet094
    Commented Oct 3, 2019 at 10:54
  • $\begingroup$ I have also updated my answer $\endgroup$
    – Sspatel
    Commented Oct 4, 2019 at 3:06

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