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I have been thinking about one problem, which is not understandable to me.

Consider the following:

I have a deck of cards (52 cards). I pick card after card (no replacement), but I stop when the current card belongs to the same family (Aces, Kings, etc.) as the card picked before. The number of observations is an event, so we can say in our sample space there are 50 events (2-52), BUT we should add another, which is that nothing happens at all. This last part gives me a huge headache and I don't know how to consider even start thinking about a problem and create a formula for this problem, such as find probability for the formula for an event.

I would really really appreciate it if someone could help me understand this problem. Thank you.

edit: the current card must match the card picked before, not any card picked before.

filtertips
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    The current card matchies the card picked immediately before, or any of the cards picked before? – Misha Lavrov Oct 10 '20 at 15:41
  • Should add: the indicated duplicate links to this question which contains an argument for the general case (not just the standard $52$ card deck). – lulu Oct 10 '20 at 15:43
  • And here is another duplicate (which I include as a solution posted there has some relevant simulation data). – lulu Oct 10 '20 at 15:44
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    @lulu I am not convinced that the duplicates actually answer the question here, which is about how events and outcomes and sample spaces work. – Misha Lavrov Oct 10 '20 at 15:44
  • @MishaLavrov I took the question to be asking for the probability that a shuffle of an ordinary deck yields no consecutive cards of the same rank. Of course, any solution to that has to address the possible outcomes of a random trial. If the OP intended something else, it should be clarified. – lulu Oct 10 '20 at 15:46
  • @lulu I think you're doing a lot of mindreading here, but sure. This question is in fact not exceptionally clear. – Misha Lavrov Oct 10 '20 at 15:47
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    I disagree with decision to close the question. First of all the questions are not the same. The acclaimed duplicate asks only for the probability of no consecutive cards of same rank, whereas this question asks for probabilities of events of the form ${\text{Card $n$ and $n-1$ have same rank}}$. Furthermore the question seems to be focused more on understanding the problem rather than asking for an explicit solution. – Leander Tilsted Kristensen Oct 10 '20 at 16:02
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    I have reopened the question. – lulu Oct 10 '20 at 16:05
  • @MishaLavrov I have edited the post. it must match exactly the card picked before. – filtertips Oct 10 '20 at 16:14
  • Note: re-opening the question caused the link to this duplicate to disappear. @filtertips I suggest you look at the three duplicates I have linked to to see if they answer your question or not. If they don't, then please edit your question to clarify the gaps. – lulu Oct 10 '20 at 16:26
  • @lulu Thank you for the links. These are similar questions but do not answer a specific thing, which is stated in my post. Let's just leave everything open for now and see what else of the community thinks. They can all access your links to see if these are duplicates. I have read your links and it didn't answer my quesiton. – filtertips Oct 10 '20 at 16:30
  • As I say, I can't see any "specific thing" stated in your post but not covered by the duplicates. Please edit your post to clarify your question. – lulu Oct 10 '20 at 16:32
  • @lulu, i don't know what more to add really. I think the distinction is clear, given two more people said they see it. Let's not turn this topic in hunting shadows, allow it for people to answer. thank you. – filtertips Oct 10 '20 at 16:42
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    In case anyone is interested, the probability (computed using a Maple program) that no two consecutive cards have the same rank is given by $$ p=\frac{a}{b} $$ where $$ \left\lbrace \begin{align} a&=672058204939482014438623912695190927357\ b&=14778213400262135041705388361938994140625\ \end{align} \right. $$ which yields $p\approx .04547628233$. – quasi Oct 10 '20 at 18:40

1 Answers1

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Since your main problem seems to be how to properly define the sample space and event space this is what i will focus on.

First of all it is worth noting that there is no unique way of defining the sample space, but a reasonable sample space to consider would be the set $\Omega$ consisting of all ways to shuffle a set of 52 cards. That is $\Omega$ has $52!$ elements and the elements are on the form $\omega = (\text{Ace of clubs}, \text{4 of spades}, \text{Queen of hearts},\dots)$. Now events are by definition subsets of the sample space, which means that there is a total of $2^{52!}$ events in total.

It is fair to assume that any outcome in $\Omega$ has the same probability, so the probability measure we are considering is the probability measure $$\mathbb{P}(A) = \frac{|A|}{52!}\quad \text{for $A\subseteq \Omega$}.$$ Where $|A|$ is number of elements in $|A|$. So calculating probabilities has been reduced to a counting problem. For instance if we want to compute the probability of the event $$A = \{\omega \in \Omega \: | \: \text{The first two cards of $\omega$ have the same rank}\}$$ we would need to count the number of ways we can shuffle a set of cards such that first two cards have the same rank. Since there are $52$ possibilities for the first card and only $3$ for the second card and then $50!$ for the remaining cards, we get that $$|A| = 52\cdot 3 \cdot 50!$$ and thus we get that $$\mathbb{P}(\text{The first two cards have same rank})=\frac{|A|}{52!} = \frac{3}{51} = \frac{1}{17}$$ I hope this helps in understanding the problem and gives you an idea of how to compute the remaining $50$ cases.

Leander Tilsted Kristensen
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  • Thank you very much for your reply. It is very educational. But I believe I have noticed one problem, you didn't consider it.

    Let's consider we want to find out P( 3rd and 4th cards have the same rank) and let's focus on |A| ... We can say |A| = 52 * 51 * 50 * 3 * 48!

    But what if Ace is was picked as the first card and then again as third. Therefore the coefficient on the 4th card should be 2 and not 3, as in this case, only two aces would be in a game. Am thinking right? And if I am, how to solve this?

    Thank you once again, your explanation opened new way of thinking to me.

     – filtertips Oct 10 '20 at 18:04
  • Well there is a reason that i only considered the simplest case, where we only look at the first two cards. Also if i understood correctly, then you are interested in the events $$A_n = {\text{$n$ is the smallest integer, such card $n$ and $n-1$ have the same rank}}$$ for $n=2,3,\dots,52$ and counting the elements of these events becomes slightly more complicated when $n>2$. And my answer doesn't really tell you how to do this only that the problem is indeed a counting problem/ combinatorical problem. – Leander Tilsted Kristensen Oct 10 '20 at 18:23
  • Yes, but I found your answer very eye-opening still. And as well, we need to consider another event n = 2,3...52, AND 0. Because there is a possibility any two consecutive cards don't have the same rank. But I guess we can just subtract this: 1-P(n = 2,3,...,52) – filtertips Oct 10 '20 at 18:30
  • Yes, that is also a possibility. And that particular case with no consecutive cards of same rank is what all the "duplicates" were about. – Leander Tilsted Kristensen Oct 10 '20 at 21:52