Solved – You randomly select two distinct integers between 1 and 100. What is the probability that the larger number is exactly twice the smaller number

combinatoricsprobability

I recently took a HackerRank test for a Data Science position and got this question wrong. I came to 1/200. Here's how:

There are 50 combinations that will make this true. (i.e. {1,2},{2,4},{3,6}…{50,100}). Probability of a specific number being chosen is 1/100. Probability that the specific set will be chosen is (1/100 * 1/100).

Since there are 50 sets,

P=50*(1/100)*(1/100)=1/200

I'm of course assuming that 1 and 100 are included. But this was the wrong answer. Can anyone help me understand my mistake?

Best Answer

Your first mistake is that there are 50 outcomes, there's actually 100 (Edit: See comment below for clarification). This is because getting (1,2) and (2,1) are the results of two seperate outcomes, but in each case the larger number is exactly twice the smaller number.

So the total possible ways of getting this is actually given by the set:

{ (1,2), (2,1), (2,4), (4,2), ..., (50,100), (100,50) }

Which is a list of 100 possible outcomes.

The total number of possible outcomes is $100 \times 99$

Since there are 100 possible numbers to choose the first time, and then 99 for the second since they must be distinct.

Hence the answer is given by:

$P = \frac{100}{100 \times 99} = \frac{1}{99}$

Using the same argument, it is straightforward to prove that the probability for the more general case of choosing numbers from $1, 2, ..., n$ where $n$ is some positive even number is given by:

$P = \frac{1}{n-1}$

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Solved – How do we know that the probability of rolling 1 and 2 is 1/18

Imagine that you threw your fair six-sided die and you got ⚀. The result was so fascinating that you called your friend Dave and told him about it. Since he was curious what he'd get when throwing his fair six-sided die, he threw it and got ⚁.

A standard die has six sides. If you are not cheating then it lands on each side with equal probability, i.e. $1$ in $6$ times. The probability that you throw ⚀, the same as with the other sides, is $\tfrac{1}{6}$. The probability that you throw ⚀, and your friend throws ⚁, is $\tfrac{1}{6} \times \tfrac{1}{6} = \tfrac{1}{36}$ since the two events are independent and we multiply independent probabilities. Saying it differently, there are $36$ arrangements of such pairs that can be easily listed (as you already did). The probability of the opposite event (you throw ⚁ and your friend throws ⚀) is also $\tfrac{1}{36}$. The probabilities that you throw ⚀, and your friend throws ⚁, or that you throw ⚁, and your friend throws ⚀, are exclusive, so we add them $\tfrac{1}{36} + \tfrac{1}{36} = \tfrac{2}{36}$. Among all the possible arrangements, there are two meeting this condition.

How do we know all of this? Well, on the grounds of probability, combinatorics and logic, but those three need some factual knowledge to rely on. We know on the basis of the experience of thousands of gamblers and some physics, that there is no reason to believe that a fair six-sided die has other than an equiprobable chance of landing on each side. Similarly, we have no reason to suspect that two independent throws are somehow related and influence each other.

You can imagine a box with tickets labeled using all the $2$-combinations (with repetition) of numbers from $1$ to $6$. That would limit the number of possible outcomes to $21$ and change the probabilities. However if you think of such a definition in term of dice, then you would have to imagine two dice that are somehow glued together. This is something very different than two dice that can function independently and can be thrown alone landing on each side with equal probability without affecting each other.

All that said, one needs to comment that such models are possible, but not for things like dice. For example, in particle physics based on empirical observations it appeared that Bose-Einstein statistic of non-distinguishable particles (see also the stars-and-bars problem) is more appropriate than the distinguishable-particles model. You can find some remarks about those models in Probability or Probability via Expectation by Peter Whittle, or in volume one of An introduction to probability theory and its applications by William Feller.

Solved – the probability that this is a “Wheel of Fortune” machine

I am not convinced you did anything wrong here. Having looked at the work provided I suspected the answer was correct, so I went ahead and made a quick simulation in Python to test the answer you provided (8/13 or ~0.6154). Note that this simulation assumes the probability of picking a machine to play is uniform across all machines.

import random

machine_list = [1,1,0,0,0,0,0,0,0,0]

def pick():
    p = random.randint(0,9)
    return machine_list[p]

def play():
    p = pick()
    if p == 1: #J
       w = random.randint(1, 2)
        if w == 1:  # win
            return 3
        if w == 2:  # lose
            return 4
    if p == 0: #WoF
        w = random.randint(1, 10)
        if w <= 2:  # win
            return 1
        else:  # lose
            return 2


winlist = []
woflist = []

for i in range(10000000):
    p = play()
    if p == 1 or p == 3:
        winlist.append(p)
    if p == 1:
        woflist.append(p)

print len(woflist) / float(len(winlist))

With the following output:

0.615330464637

Best Answer

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Solved – How do we know that the probability of rolling 1 and 2 is 1/18

Solved – the probability that this is a “Wheel of Fortune” machine

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