I Max of 3 random cards from deck vs max of 3 numbers from 1-13

AI Thread Summary
The discussion focuses on the expected value of the maximum of three integers uniformly distributed from 1 to 13 compared to three real numbers in the same range. The continuous case yields an expected value of 10 using a derived PDF and CDF. In contrast, the discrete case involves calculating probabilities and leads to an expected value of approximately 10 as well, with a formula that incorporates the sum of cubes. Participants clarify that drawing cards from a standard deck differs from selecting integers uniformly, emphasizing the need for precision in defining the sets involved. The conversation highlights the nuances between discrete and continuous distributions in probability theory.
member 428835
Hi PF!

I am wondering the differences between the discrete and continuous case for expected value of minimum of 3 integers uniformly distributed from 1 to 13 vs 3 reals from 1 to 13.

The real case is direct: ##F = ((x-1)/12)^3 \implies f = 3(x-1)/12)^2## for CDF ##F## and PDF ##f##. Thus the expected value for the max of 3 reals in this range is ##\int_1^{13} x f \, dx = 10##. But now for the discrete case: the probability a random variable ##X_i## is less than some integer ##k## I think should be ##P(X_i \geq k) = (13-k+1)/13 \implies P(X \geq k) = ((13-k+1)/13)^3## but I really don't know how to proceed. Is there a direct way to arriving at the CDF that I'm missing?
 
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Why not just ##c(k) = (\frac k {13})^3##? And ##p(k) = (\frac k {13})^3 - (\frac{k-1}{13})^3##?
 
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Just to point out, drawing three cards from a deck is not the same as picking three integers uniformly at random, unless you replace the cards each time you draw.

Also. ##P(X\geq k)## is the cdf, well one minus that is. Were you not sure how to get the pdf (which as Perok points out is just the difference of consecutive pdfs).Note ##k^3-(k-1)^3## is actually a quadratic polynomial, so the answers are more similar than they might initially appear
 
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I got ##E = \frac{133}{13} \approx 10## for the discrete case.
 
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In general, for a uniform choice from ##1-n## (with replacement) of ##m## cards, then I get the expected value of the highest card to be:
$$E = n - \frac{1}{n^m}\sum_{k =1}^{n-1}k^m$$With ##m = 3##, we have the sum of cubes:
$$\sum_{k =1}^{n-1}k^3 = \frac{n^2(n-1)^2}{4}$$And$$E = n - \frac{(n-1)^2}{4n}$$And with ##n = 13##:
$$E = 13 - \frac{36}{13} = \frac{133}{13}$$
 
@joshmccraney, your thread title is misleading: "max of 3 random cards from deck vs max of 3 numbers from 1-13"

A deck of cards has 52 cards in it, in four suits. Unless your deck has just 13 cards -- A, 2, 3, ..., J, Q, K -- in one suit, it's different from the set of integers 1 through 13.

The first post seems to be asking a different question -- the difference between a discrete set and a continuous set.
 
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