Mathematics Asked by filtertips on November 1, 2020

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.

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 $$mathbb{P}(A) = frac{|A|}{52!}quad text{for $Asubseteq Omega$}.$$ 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 $$|A| = 52cdot 3 cdot 50!$$ 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.

Correct answer by Leander Tilsted Kristensen on November 1, 2020

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