Using summation with probability question

  • Context: Undergrad 
  • Thread starter Thread starter spikemurphy
  • Start date Start date
  • Tags Tags
    Probability Summation
Click For Summary

Discussion Overview

The discussion revolves around calculating the probability of being dealt a bridge hand with at least 5 hearts. Participants explore different mathematical approaches, including the use of binomial formulas and combinatorial reasoning, while addressing notation and the applicability of certain probability models in the context of card games.

Discussion Character

  • Technical explanation
  • Mathematical reasoning
  • Debate/contested

Main Points Raised

  • One participant presents two potential formulas for calculating the probability, involving summation and combinations.
  • Another participant reformats the original formulas for clarity, questioning the placement of delimiters.
  • There is a discussion about the validity of using binomial probabilities for card draws, with one participant arguing that the events are not independent due to the nature of drawing cards from a finite deck.
  • A later reply suggests a corrected approach to calculating the probability of drawing at least 5 hearts, involving combinations of drawing hearts and non-hearts.
  • One participant acknowledges an error in their original message regarding the total number of cards, indicating a misunderstanding of the problem setup.
  • There is a side conversation about the use of LaTeX for mathematical notation, with a link provided for assistance.

Areas of Agreement / Disagreement

Participants express differing views on the appropriateness of using binomial probabilities in this context, with some supporting the use of combinatorial methods while others challenge the initial formulas presented. The discussion remains unresolved regarding the best approach to calculate the probability.

Contextual Notes

Participants highlight limitations in the initial formulas and assumptions about independence in probability calculations. The discussion reflects a need for clarity in notation and understanding of combinatorial principles.

spikemurphy
Messages
4
Reaction score
0
Help With Probability Question

i have been working on this for a week can anyone help?

NOTE: Look at reply from moodoo for proper matematical symbols!

I need the probability of being dealt a bridge hand with at least 5 hearts.

I have to possible answers but I have never done this before and don't know the proper way to type them but here goes...

1.)
summation sign with 13 on the top and k=5 on bottom x (C 13,k)(C 13,13-K)/C 52,13

2)same summation as above x C 13,k (.25)to the power of k(.75)to the power of 13-k/C 52,13

HELP WITH PROBABILITY QUESTION :confused: :confused: :confused:
 
Last edited:
Physics news on Phys.org
Yikes, that's some messy notation you got there...
Let me clean that up for you...

1. [tex]\sum_{k=5}^{13}{\frac{\left(\begin{array}{cc}13\\k\end{array}\right)\left(\begin{array}{cc}13\\13-k\end{array}\right)}{\left(\begin{array}{cc}52\\13\end{array}\right)}}[/tex]

2. [tex]\sum_{k=5}^{13}{\frac{\left(\begin{array}{cc}13\\k\end{array}\right)*.25^{k}*.75^{13-k}}{\left(\begin{array}{cc}52\\13\end{array}\right)}}[/tex]

Although I am a bit shaky on where your delimeters are. Tell me if any of the above aren't what you had in mind.

EDIT: Maybe that's more like it?
 
Last edited:
summation

Yes, that is exactly what I meant. You are amazing. But is this the right answer for my question? Are both of these binomial formulas and is it ok to use a binomial formula with cards?

Thanks so much for your time>
 
Last edited:
spikemurphy said:
Yes, that is exactly what I meant. You are amazing. But is this the right answer for my question? Are both of these binomial formulas and is it ok to use a binomial formula with cards?

Thanks so much for your time>
Both your solutions are INcorrect.

First, Binomial probabilities can only be used for INDEPENDENT events for which the probability of occurrence (or non-occurrence) of any event does NOT influence that for any other. Unfortunately, for a card deck, when you draw anyone card you automatically change the probability of drawing the next (since you do not replace the drawn card). Thus, your solution #2 does not apply. (Binomial probabilities could be used, however, to determine the probability of at least 5 Tails when tossing 52 Fair Coins since each Coin is independent of the others.)

Your solution #1 is closer, but is still not proper. You are correct that there are [tex]\mathbb{C}_{13}^{52}[/tex] total ways of drawing 13 (combinations of) cards from the deck's 52. You must now determine how many of those contain at least 5 Hearts.

Let's start with exactly 5 Hearts. There are [tex]\mathbb{C}_{5}^{13}[/tex] ways of drawing those 5 from the deck's 13 Hearts. Further, there are [tex]\mathbb{C}_{8}^{39}[/tex] ways of drawing the 8 additional cards required for the hand from the remaining 39 deck cards (which are not hearts). Thus:

{Number Combinations of 5 Hearts in 13 Cards from 52 Card Deck} = [tex]\mathbb{C}_{(5)}^{(13)} \cdot \mathbb{C}_{(8)}^{(39)}[/tex]

Similar reasoning can be applied to {5, 6, 7, ... , 12, 13 Hearts}, so the total number of combinations containing AT LEAST 5 Hearts is given by:

[tex](1) \ \ \ \ \ \sum_{n=5}^{13} \mathbb{C}_{(n)}^{(13)} \cdot \mathbb{C}_{(13 - n)}^{(39)}[/tex]

And the required probability is:

[tex](2) \ \ \ \ \ \color{red} Prob(At \ Least \ 5 \ Hearts) \ = \ \frac { \sum_{n=5}^{13} \mathbb{C}_{(n)}^{(13)} \cdot \mathbb{C}_{(13 - n)}^{(39)} } { \mathbb{C}_{(13)}^{(52)} }[/tex]


~~
 
Last edited:
XANTHYM, Thanks so much for your help. When I saw your answer I checked my original message, the 13 in my jumbled formula was an error. I know that the cards have to add up to 52.

Thanks again, spikemurphy

p.s. is there an easy way to use latex, do I need a special program.
 

Similar threads

  • · Replies 1 ·
Replies
1
Views
3K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 13 ·
Replies
13
Views
3K
  • · Replies 3 ·
Replies
3
Views
3K
  • · Replies 29 ·
Replies
29
Views
8K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 11 ·
Replies
11
Views
4K
  • · Replies 6 ·
Replies
6
Views
2K
  • · Replies 5 ·
Replies
5
Views
2K