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

In summary, the conversation discusses the differences between the discrete and continuous cases for calculating the expected value of the minimum of three integers uniformly distributed from 1 to 13 versus three real numbers from 1 to 13. The real case can be calculated directly using the CDF and PDF, while the discrete case requires a different approach. The conversation also touches on the difference between drawing cards from a deck and picking integers uniformly at random. Finally, the conversation presents a formula for calculating the expected value of the highest card when drawing from a deck of cards and concludes that the expected value for the discrete case is approximately 10.
  • #1
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?
 
Physics news on Phys.org
  • #2
Why not just ##c(k) = (\frac k {13})^3##? And ##p(k) = (\frac k {13})^3 - (\frac{k-1}{13})^3##?
 
  • Like
Likes member 428835
  • #3
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
 
  • Like
Likes member 428835
  • #4
I got ##E = \frac{133}{13} \approx 10## for the discrete case.
 
  • Like
Likes member 428835
  • #5
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##:
$$E = 13 - \frac{36}{13} = \frac{133}{13}$$
 
  • #6
@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.
 
Last edited:
  • Like
Likes Klystron, member 428835 and berkeman

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

What is the purpose of comparing the maximum of 3 random cards from a deck to the maximum of 3 numbers from 1-13?

The purpose of this comparison is to analyze the likelihood of obtaining a higher maximum value from a deck of cards versus a set of numbers. This can provide insights into the randomness and distribution of values in a deck of cards.

How are the 3 random cards selected from the deck?

The 3 random cards are typically selected using a random number generator or by shuffling the deck and then drawing the top 3 cards. This ensures that the selection is truly random and not biased towards certain cards.

What is the range of possible values for the maximum of 3 random cards from a deck?

The range of possible values for the maximum of 3 random cards from a deck is from 3 to 39. This is because the highest possible value for a card is 13, and 3 cards multiplied together give a maximum value of 39.

How does the maximum of 3 numbers from 1-13 compare to the maximum of 3 random cards from a deck?

The maximum of 3 numbers from 1-13 will always be 39, which is the same as the highest possible value for the maximum of 3 random cards from a deck. However, the likelihood of obtaining a maximum value of 39 is much lower in a deck of cards due to the randomness of the selection process.

What other factors may affect the results of this comparison?

Other factors that may affect the results of this comparison include the type of deck being used (e.g. a standard 52-card deck versus a deck with jokers), the shuffling method, and the number of trials conducted. These factors can impact the distribution of values and the likelihood of obtaining a certain maximum value.

Back
Top