How many cards must be selected from a standard deck of 52 cards to guarantee that at least three hearts are present among them?
42
Understanding the Problem: Guaranteeing Cards from a Deck
This problem asks us to find the minimum number of cards we must draw from a standard 52-card deck to ensure that we have at least three hearts. This type of question involves the Pigeonhole Principle and the concept of a worst-case scenario.
A standard deck of 52 cards consists of four suits: Hearts, Diamonds, Clubs, and Spades. Each suit has 13 cards.
- Hearts: 13 cards
- Diamonds: 13 cards
- Clubs: 13 cards
- Spades: 13 cards
Total cards = $13 + 13 + 13 + 13 = 52$.
The Worst-Case Scenario Approach
To guarantee a specific outcome (in this case, at least three hearts), we need to consider the scenario where we are least likely to achieve that outcome. This means we should assume we draw as many cards as possible that are not hearts before we are forced to draw a heart.
The cards that are not hearts are Diamonds, Clubs, and Spades.
- Number of non-heart cards = Number of Diamonds + Number of Clubs + Number of Spades
- Number of non-heart cards = $13 + 13 + 13 = 39$ cards
In the worst-case scenario, we would draw all 39 of these non-heart cards first.
| Suit | Number of Cards | Classification |
|---|---|---|
| Hearts | 13 | Target Suit |
| Diamonds | 13 | Non-Heart |
| Clubs | 13 | Non-Heart |
| Spades | 13 | Non-Heart |
Calculating the Minimum Number of Cards
After drawing all 39 non-heart cards, the remaining cards in the deck are all hearts. There are $52 - 39 = 13$ hearts left.
If we have drawn 39 cards and none of them are hearts, the very next card drawn must be a heart. This is the 40th card.
To guarantee at least two hearts, we would need to draw one more heart after the first one. This would be the 41st card.
To guarantee at least three hearts, we would need to draw yet another heart after the first two. This would be the 42nd card.
So, the total number of cards drawn to guarantee at least three hearts is the total number of non-heart cards plus the desired number of hearts:
Minimum cards = (Total non-heart cards) + (Desired minimum hearts)
Minimum cards = $39 + 3$
Minimum cards = $42$
Therefore, you must select 42 cards to guarantee that at least three hearts are present among them.
Summary of the Card Drawing Process
Let's visualize the worst-case drawing process:
- Draw all Diamonds (13 cards).
- Draw all Clubs (13 cards).
- Draw all Spades (13 cards).
- Total drawn so far: $13 + 13 + 13 = 39$ cards (all non-hearts).
- Draw the next card: It must be a heart (40th card drawn). You now have 1 heart.
- Draw the next card: It must be a heart (41st card drawn). You now have 2 hearts.
- Draw the next card: It must be a heart (42nd card drawn). You now have 3 hearts.
At this point, having drawn 42 cards, you are guaranteed to have drawn at least three hearts.
Revision Table: Card Drawing Guarantee
| Goal | Worst-Case Scenario | Cards to Guarantee |
|---|---|---|
| At least 1 heart | Draw all 39 non-hearts | $39 + 1 = 40$ |
| At least 2 hearts | Draw all 39 non-hearts | $39 + 2 = 41$ |
| At least 3 hearts | Draw all 39 non-hearts | $39 + 3 = 42$ |
Additional Information: The Pigeonhole Principle
This problem is a direct application of the Pigeonhole Principle. The principle states that if $n$ items are put into $m$ containers, with $n > m$, then at least one container must contain more than one item. A more general form is used here: to guarantee at least $k$ items in one specific container (or category), we must put items into other containers until they are full (worst case) and then add $k$ items.
In this card problem:
- The 'items' are the cards drawn.
- The 'containers' or categories are the suits (or specifically, 'hearts' and 'non-hearts').
- We want to guarantee at least 3 hearts. The worst case is filling up the 'non-heart' category first (the other 3 suits, which have 39 cards in total) and then drawing the desired number (3) from the 'heart' category.
This principle is useful in many areas of mathematics and computer science to prove the existence of certain conditions.
Important Questions from Introduction
In the following table, the left column contains the names of standard graph algorithms and the right column contains the time complexities of the algorithms. Here, n and m are number of vertices and edges, respectively. Match each algorithm with its time complexity.
List I List II Standard graph algorithms Time complexities A. Bellman‐Ford algorithm I. O(m*log n) B. Kruskal’s algorithm II. O(n 3) C. Floyd‐Warshall algorithm III. O(n*m) D. Topological sorting IV. O(n + m) Choose the correct answer from the options given below :
Match List 1 with List 2 and choose the correct answer from the code given below:
List I
(Graph Algorithm)
List II
(Time Complexity)
a) Dijkstra’s algorithm
i) Θ(E log E)
b) Kruskal’s algorithm
ii) Θ(V 3)
c) Floyd-Warshall algorithm
iii) Θ(V 2)
d) Topological sorting
iv) Θ(V + E)
Where V and E are the number of vertices and edges in graph respectively.
The solution of recurrence relation: T(n)=2T(sqrt(n)) + lg(n) is
Modulus of elasticity of concrete, E is calculated using:
In how many types can R.C.C. be classified into?
