Undergrad Hypergeometric Distribution Calculation in Libreoffice

Click For Summary
The discussion focuses on using the HYPGEOM.DIST function in LibreOffice to calculate probabilities in a card game scenario. The command provided is evaluated to determine the probability of player two flipping a card greater than six after player one reveals a six. It is clarified that for this specific situation, a hypergeometric distribution is unnecessary; only the total number of cards and the number of cards greater than six are needed. The conversation also addresses whether the inquiry is related to homework, confirming it is not. Overall, the main takeaway is that simpler probability calculations can be applied in this context.
Euler2718
Messages
90
Reaction score
3
Given this libreoffice command:

HYPGEOM.DIST(X; NSample; Successes; NPopulation; Cumulative)

>X is the number of results achieved in the random sample.

>NSample is the size of the random sample.

>Successes is the number of possible results in the total population.

>NPopulation is the size of the total population.

>Cumulative : 0 or False calculates the probability density function. Other values or True calculates the cumulative distribution function.

Consider: Two players have 10 cards each, face down. Player one flips a card over and gets a 6. What's the probability of player two flipping a card greater than six.

Is this the right way to use the command:

=HYPGEOM.DIST(1,10,4,52,0)

It gives ~35.8%. Just wondering if I was interpreting the question and the command right.

Edit: I assumed suits are irrelevant and repetition is allowed.
 
Last edited:
Mathematics news on Phys.org
If player 2 flips over a single card, you don't need a hypergeometric distribution. You just need the number of cards available and the number of cards larger than 6. It does not matter how many cards are face down, or not distributed at all.

Is this a homework question?
 
  • Like
Likes Euler2718
mfb said:
If player 2 flips over a single card, you don't need a hypergeometric distribution. You just need the number of cards available and the number of cards larger than 6. It does not matter how many cards are face down, or not distributed at all.

Is this a homework question?

Thanks for the insight.

No, it's not homework. I've just been curious with the probability of cards as of late and I'm not too good at probability.
 
Here is a little puzzle from the book 100 Geometric Games by Pierre Berloquin. The side of a small square is one meter long and the side of a larger square one and a half meters long. One vertex of the large square is at the center of the small square. The side of the large square cuts two sides of the small square into one- third parts and two-thirds parts. What is the area where the squares overlap?

Similar threads

  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 8 ·
Replies
8
Views
5K
  • · Replies 9 ·
Replies
9
Views
5K
Replies
2
Views
1K
  • · Replies 9 ·
Replies
9
Views
3K
Replies
3
Views
2K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 7 ·
Replies
7
Views
3K
  • · Replies 2 ·
Replies
2
Views
3K
  • · Replies 1 ·
Replies
1
Views
1K