Probability Expected Values with Cards

  • Thread starter Thread starter Kalinka35
  • Start date Start date
  • Tags Tags
    Cards Probability
Click For Summary
SUMMARY

The discussion focuses on calculating the expected number of matches when drawing cards from a standard deck of 52 playing cards, where a match occurs when the drawn card corresponds to its rank in a sequential manner. The random variable X represents the total number of matches, with each Xi indicating whether the ith card is a match (1) or not (0). The probability of a match at any position is calculated as P(Xi=1), starting with a 4/52 chance for the first card being an ace. The complexity arises from the dependency of probabilities on previously drawn cards, necessitating a careful approach to counting matches.

PREREQUISITES
  • Understanding of probability theory and random variables
  • Familiarity with combinatorial counting methods
  • Basic knowledge of card games and standard deck structure
  • Experience with expected value calculations
NEXT STEPS
  • Study the concept of expected value in probability theory
  • Learn about conditional probability and its applications in card games
  • Explore combinatorial methods for counting outcomes in probability
  • Investigate advanced techniques for calculating expected values in dependent scenarios
USEFUL FOR

Students studying probability, mathematicians interested in combinatorial problems, and educators teaching concepts related to expected values and random variables.

Kalinka35
Messages
48
Reaction score
0

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.
 
Physics news on Phys.org
i think you need to think about how you define your random variable..

If Xi is the probability 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 probability 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 probability of drawing cards will be dependent on what has been dealt previously... anyway just stuff to think about
 

Similar threads

Replies
31
Views
7K
  • · Replies 9 ·
Replies
9
Views
6K
  • · Replies 25 ·
Replies
25
Views
2K
  • · Replies 8 ·
Replies
8
Views
2K
Replies
1
Views
2K
  • · Replies 4 ·
Replies
4
Views
3K
Replies
17
Views
4K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 16 ·
Replies
16
Views
7K
  • · Replies 12 ·
Replies
12
Views
4K