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How to roll a Dice by tossing a Coin ? Cheenta Statistics Department

How can you roll a dice by tossing a coin? Can you use your probability knowledge? Use your conditioning skills.

Suppose, you have gone to a picnic with your friends. You have planned to play the physical version of the Snake and Ladder game. You found out that you have lost your dice.

The shit just became real!

Now, you have an unbiased coin in your wallet / purse. You know Probability.

Aapna Time Aayega

starts playing in the background. :p

Can you simulate the dice from the coin?

Ofcourse, you know chances better than others. :3

Take a coin.

Toss it 3 times. Record the outcomes.

HHH = Number 1

HHT = Number 2

HTH = Number 3

HTT = Number 4

THH = Number 5

THT = Number 6

TTH = Reject it, don’t ccount the toss and toss again

TTT = Reject it, don’t ccount the toss and toss again

Voila done!

What is the probability of HHH in this experiment?

Let X be the outcome in the restricted experiment as shown.

How is this experiment is different from the actual experiment?

This experiment is conditioning on the event A = {HHH, HHT, HTH, HTT, THH, THT}.

\(P( X = HHH) = P (X = HHH | X \in A ) = \frac{P (X = HHH)}{P (X \in A)} = \frac{1}{6}\)


Beautiful right?

Can you generalize this idea?

Food for thought

  • Give an algorithm to simulate any conditional probability.
  • Give an algorithm to simulate any event with probability \(\frac{m}{2^k}\), where \( m \leq 2^k \).
  • Give an algorithm to simulate any event with probability \(\frac{m}{2^k}\), where \( n \leq 2^k \).
  • Give an algorithm to simulate any event with probability \(\frac{m}{n}\), where \( m \leq n \leq 2^k \) using conditional probability.

Watch the Video here:

Some Useful Links:

Books for ISI MStat Entrance Exam

How to Prepare for ISI MStat Entrance Exam

ISI MStat and IIT JAM Stat Problems and Solutions

Cheenta Statistics Program for ISI MStat and IIT JAM Stat

Simple Linear Regression – Playlist on YouTube

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AMC-8 USA Math Olympiad

Probability AMC 10B 2019 problem 17

What are we learning ?

Competency in Focus: probability

This problem is from American Mathematics Contest 10B (AMC 10B, 2019). It is Question no. 17 of the AMC 10B 2019 Problem series.

First look at the knowledge graph:-

calculation of  mean and median- AMC 8 2013 Problem

Next understand the problem

A red ball and a green ball are randomly and independently tossed into bins numbered with the positive integers so that for each ball, the probability that it is tossed into bin $k$ is $2^{-k}$ for $k = 1,2,3….$ What is the probability that the red ball is tossed into a higher-numbered bin than the green ball?
$\textbf{(A) } \frac{1}{4} \qquad\textbf{(B) } \frac{2}{7} \qquad\textbf{(C) } \frac{1}{3} \qquad\textbf{(D) } \frac{3}{8} \qquad\textbf{(E) } \frac{3}{7}$
Source of the problem
American Mathematical Contest 2019, AMC 10B Problem 17
Key Competency

Probability

Difficulty Level
4/10
Suggested Book
Challenges and Thrills in Pre College Mathematics Excursion Of Mathematics 

Start with hints 

Do you really need a hint? Try it first!
The probability that the two balls will go into adjacent bins is $\frac{1}{2\times4} + \frac{1}{4\times8} + \frac{1}{8 \times 16} + … = \frac{1}{8} + \frac{1}{32} + \frac{1}{128} + \cdots = \frac{1}{6}$ by the geometric series sum formula.
the probability that the two balls will go into bins that have a distance of $2$ from each other is $\frac{1}{2 \times 8} + \frac{1}{4 \times 16} + \frac{1}{8 \times 32} + \cdots = \frac{1}{16} + \frac{1}{64} + \frac{1}{256} + \cdots = \frac{1}{12}$
We can see that each time we add a bin between the two balls, the probability halves.
Thus, our answer is $\frac{1}{6} + \frac{1}{12} + \frac{1}{24} + \cdots$

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