Question about probabilities with playing cards

  • Context: Undergrad 
  • Thread starter Thread starter jldibble
  • Start date Start date
  • Tags Tags
    Cards Probabilities
Click For Summary
SUMMARY

This discussion focuses on calculating the probabilities of card distributions in a game involving a standard 52-card deck. Specifically, it addresses the probability that each of the three other players has at least one heart and the probability that each has exactly one heart after being dealt five cards. The calculations involve combinatorial mathematics, utilizing combinations such as 13C1 and 39C4 for determining specific outcomes. The final probability for each player having exactly one heart is approximately 4.5e-7, indicating a very low likelihood of this event occurring.

PREREQUISITES
  • Understanding of combinatorial mathematics, specifically combinations (nCr).
  • Familiarity with probability theory as it applies to card games.
  • Knowledge of standard playing card deck composition (52 cards, 4 suits).
  • Basic skills in calculating probabilities from given outcomes.
NEXT STEPS
  • Study combinatorial probability calculations in card games.
  • Learn about advanced probability concepts such as conditional probability.
  • Explore simulations of card games to visualize probability distributions.
  • Investigate the impact of different card distributions on game outcomes.
USEFUL FOR

This discussion is beneficial for mathematicians, statisticians, game theorists, and card game enthusiasts interested in understanding the probabilities involved in card distributions and their implications in gameplay.

jldibble
Messages
50
Reaction score
0
Let's say that I'm playing a card game with 3 friends. The game involves a full deck of regular playing cards (52 cards, 4 suits, etc.) For this game, each player is dealt 5 cards. The cards are dealt out 5 cards at a time and in a clockwise direction starting with the person to the left of the dealer (I don't think the manner in which the cards are dealt will be relevant).

Now my question is this; how can I find the probability that each other player, besides myself, has been dealt at least ONE heart.
And then, how could I find the probability that each player, besides myself, has been dealt ONLY ONE heart.


Thanks for the help.
 
Physics news on Phys.org
This is not my field, so a more elegant solution may exist. By this procedure should give the correct result.

First, the only relevant cards are the 15 dealt to the other players. There are 52C15 = 4 481 381 406 320 possible combinations. The deck itself can be seen as 13 hearts + 39 other cards.

The probability that each other player has been dealt exactly one heart is the easiest to calculate. For the first player, this is
13C1 \times 39C4 = 1 069 263
For the second player:
12C1 \times 35C4 = 628 320
For the third player:
11C1 \times 31C4 = 346 115
So the probability is 2 033 698/4 481 381 406 320 ≈ 4.5e-7, or approximately one chance out of 2 203 563.

To find the possibility that each player has at least one heart is more involved. You can sum up the probability that one player has 2 hearts and the two others 1 heart + the probability that two players have 2 hearts and the other 1 heart + the probability that one player has 3 hearts and the two others 1 heart, and so on.
 
jldibble said:
Let's say that I'm playing a card game with 3 friends. The game involves a full deck of regular playing cards (52 cards, 4 suits, etc.) For this game, each player is dealt 5 cards. The cards are dealt out 5 cards at a time and in a clockwise direction starting with the person to the left of the dealer (I don't think the manner in which the cards are dealt will be relevant).

Now my question is this; how can I find the probability that each other player, besides myself, has been dealt at least ONE heart.
And then, how could I find the probability that each player, besides myself, has been dealt ONLY ONE heart.


Thanks for the help.
Can you look at your hand and see the number of hearts in it? If so, that number will affect the probabilities.

Or are you asking this question without having looked at our hand?
 

Similar threads

  • · Replies 3 ·
Replies
3
Views
4K
  • · Replies 9 ·
Replies
9
Views
3K
  • · Replies 4 ·
Replies
4
Views
3K
  • · Replies 31 ·
2
Replies
31
Views
4K
  • · Replies 9 ·
Replies
9
Views
5K
  • · Replies 17 ·
Replies
17
Views
3K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
Replies
4
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K