Probability Expected Values with Cards

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The discussion focuses on calculating the expected number of matches when drawing cards from a standard 52-card deck, where a match occurs when the card drawn corresponds to its rank in sequence. The random variable X represents the total number of matches, with individual matches denoted as Xi. The probability of a match at any position is initially considered as 4/52, but this approach is deemed oversimplified due to dependencies on previous draws. The conversation highlights the complexity of counting matches, suggesting that a brute force method may be necessary to accurately account for the probabilities of matches occurring throughout the draw. Overall, the challenge lies in defining the random variable and accurately calculating the probabilities based on the sequence of cards drawn.
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Homework Statement


Cards from a standard deck of 52 playing cards are laid face up one at a time. If the first card is an ace, the second card is an 2, the third card is a three, ..., the 13th card is a king, the 14th is an ace, etc. we call it a "match." The 13n+1th card does not need to be any particular ace, it just needs to be an ace. Find the expected number of matches that occur.


Homework Equations





The Attempt at a Solution


I let the random variable X represent the total number of matches.
So Xi=1 if the ith card is a match, and 0 if it is not a match.
Then I summed over the all the cards from 1 to 52.
The thing that I was having trouble calculating was P(Xi=1). It seems like on any given trial there is a 4/52 chance of getting a match, but this seems oversimplified.
 
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i think you need to think about how you define your random variable..

If Xi is the probabilty of a match at the ith position, then for i=1, the probability of first drawing an ace will be 4/52

if X is the total number of matches, then the probabilty of X depends on how you count matches

do you stop counting matches after your first mismatch...? in that case the probability will be 4/52 multiplied by the probability of not getting a two on the 2nd card (51-4)/51
if 1st card ace...

sounds more likely you want to find the total number of matches, then you must count all the occurences of exactly one match anywhere in the draw, ie:
1-draw an ace, then no other matches...
1-no ace, 2-draw a 2, then no other matches...
... etc.

unless you can find a smart way to do this, the brute force method could get tricky as the probabilty of drawing cards will be dependent on what has been dealt previously... anyway just stuff to think about
 
Question: A clock's minute hand has length 4 and its hour hand has length 3. What is the distance between the tips at the moment when it is increasing most rapidly?(Putnam Exam Question) Answer: Making assumption that both the hands moves at constant angular velocities, the answer is ## \sqrt{7} .## But don't you think this assumption is somewhat doubtful and wrong?

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