Two Cards (1-100), One Double the Other
You draw two cards from a set of cards numbered 1 to 100. The cards are drawn without replacement. What is the probability that one of the numbers drawn is exactly double the other?
Related Concepts
Hint
- Favorable Pairs:
- Think of pairs (x, y) where y = 2x. What's the smallest possible value for x? What's the largest possible value for x such that 2x is still ≤ 100? This will tell you how many such unique pairs exist.
- Since the cards are drawn sequentially (order matters for the draws), for each pair like (1, 2), it can be drawn as (1 then 2) OR (2 then 1). Does this apply here? Yes, because the question says "one number is double the other," not specifically "the second is double the first."
- Total Outcomes:
- How many choices for the first card?
- After drawing one, how many choices for the second card (without replacement)?
- Multiply these to get the total number of ordered ways to draw two cards.
- Calculate the probability.
Explanation: One Number Double the Other
Imagine you have 100 cards, numbered 1 to 100. You pick two cards, one after the other, and don't put the first one back.
What's the chance that one card's number is exactly double the other's?
- Good Pairs: We need pairs like (1, 2), (2, 4), (3, 6), and so on. The biggest small number can be 50, because its double is 100 (which is on a card). So, there are 50 such basic pairs: (1,2), (2,4), ..., (50,100).
- Order Matters for Drawing: For each pair, say (1,2), you could draw the '1' first then the '2', OR you could draw the '2' first then the '1'. Both situations satisfy the condition. So, each of our 50 basic pairs can happen in 2 different orders. That means 50 × 2 = 100 "good" ordered ways to draw the cards.
- Total Ways to Draw Two Cards: You have 100 choices for the first card. After picking one, you have 99 choices left for the second card. So, there are 100 × 99 = 9,900 total possible ordered ways to draw two cards.
- The Chance: Out of 9,900 total ways, 100 of them are "good" (one number is double the other). So the probability is 100 / 9,900.
We are drawing two cards from 100 numbered cards (1 to 100) without replacement. We want the probability that one number drawn is exactly double the other.
1. Identify Favorable Pairs and Ordered Outcomes
Let the two numbers drawn be x and y. We are looking for pairs where either y = 2x or x = 2y. It's easier to list pairs of the form (x, 2x) where both x and 2x are between 1 and 100 inclusive.
- If x = 1, then 2x = 2. The pair is (1, 2).
- If x = 2, then 2x = 4. The pair is (2, 4).
- ...
- If x = 50, then 2x = 100. The pair is (50, 100).
If x were 51, then 2x would be 102, which is not in our set of cards. So, there are 50 such unique pairs of numbers where one is double the other: (1,2), (2,4), ..., (50,100).
Since the cards are drawn sequentially without replacement, the order matters. For each unique pair (x, 2x), there are two possible orders in which they can be drawn:
- Draw x first, then 2x.
- Draw 2x first, then x.
For example, for the pair (1,2), we can draw (1 then 2) or (2 then 1). Both satisfy the condition.
Therefore, the total number of favorable ordered outcomes is:
Number of unique pairs × Orders per pair = 50 × 2 = 100
2. Determine the Total Number of Possible Ordered Outcomes
We are drawing two cards without replacement from 100 cards. The order of drawing matters for the total sample space if we consider ordered favorable outcomes.
- Number of choices for the first card: 100
- Number of choices for the second card (after one is drawn): 99
The total number of possible ordered outcomes (permutations of 2 cards from 100) is:
Total Ordered Outcomes = 100 × 99 = 9,900
(This is also P(100,2) = 100! / (100-2)! = 100! / 98! = 100 × 99)
3. Calculate the Probability
The probability of the event is the ratio of favorable ordered outcomes to the total ordered outcomes:
P(one number is double the other) = (Favorable Ordered Outcomes) / (Total Ordered Outcomes)
P(one number is double the other) = 100 / 9,900
Simplifying the fraction:
P(one number is double the other) = 1/99
As a decimal, this is approximately:
1/99 ≈ 0.010101... or about 1.01%
Alternative using Combinations (Slightly More Complex to Frame): If we used combinations, we'd count unordered pairs. - Favorable unordered pairs: 50 (as found: (1,2), (2,4), ..., (50,100)). - Total unordered pairs: C(100,2) = (100 × 99) / (2 × 1) = 50 × 99 = 4,950. - Probability = 50 / 4,950 = 1/99. Both approaches lead to the same result, but the ordered approach often aligns more directly with the "drawing one then another" phrasing for many people.
Ponder This: If the cards were drawn *with* replacement, how would the probability change? Would the number of favorable outcomes change? Would the total number of outcomes change?