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Showing posts from May, 2013

A Winning Series

Q: You are required to play a game of skill against opponents A & B. You win if you can win two games in a row. You are presented with two options
Play A first, then B followed by A (option A-B-A)Play B first, then A followed by B (option B-A-B)The probability that you will win against A is \(\frac{1}{3}\) and that against B is \(\frac{1}{2}\). Which option maximizes your chances of winning?

The Probability Tutoring Book: An Intuitive Course for Engineers and Scientists (And Everone Else!)

A: At first brush it appears that option BAB is your best option because you play B more often and your chances of winning against B are higher than against A. However, there is a surprising and counter intuitive twist to the tale. You win only if you win two games in a row. To understand it better, let us enumerate out the cases for the A-B-A option. The options are shown in the table below

Each row enumerates out a possible scenario. The final column is the probability of that scenario playing o…

The Glass Coin Toss

Q: A & B agree to stake $100 each for a coin tossing game involving 10 tosses of a coin. Whoever gets more heads wins. However, the coin is made of glass and at the 7th toss, when A has won 4 heads and B has won 3 heads,  it breaks. How should the total of $200 be divided in a fair manner between the two?

The Probability Tutoring Book: An Intuitive Course for Engineers and Scientists (And Everone Else!)

A: A fair way to proceed is to give A all the credit aggregated so far and do the same for B with the obvious intuition being that A deserves more as he was closer to win than B. But how does one compute that? On any given toss, the probability that either would win is \(\frac{1}{2}\). Without loss of generality, it would be fair to say that had the tossing not been stopped, A would have won if he won two or more times. Being disjoint events, this probability works out to
$$
\big(\frac{1}{2})^{3} + \big(\frac{1}{2})^{3} = \frac{1}{4}
$$
Likewise, B would have won if he won all thr…

Identifying the Lazy Student

Q: You ask two students A & B to do a statistical task. The task is to roll two dies, sum the numbers and to repeat it a 100 times. You get back the set of numbers from both the students. However you know that one of them is a lazy student and has rolled just one die and doubled its value and reported it. How do you identify which one of the students is the lazy student?

The Probability Tutoring Book: An Intuitive Course for Engineers and Scientists (And Everone Else!)

A: Let \(X_{1}\) and \(X_{2}\) represent the random variables that shows up when a die is rolled. This can be any of 1 to 6. The sum of the two random variables would have a variance given as
$$
Var(X_{1} + X_{2}) = Var(X_{1}) + Var(X_{2})
$$
If we let \(Var(X_{1}) = Var(X_{2}) = \alpha\) then \(Var(X_{1}) + Var(X_{2}) = 2\alpha\). The lazy student doubles the die score. The variance for a multiplier to the random variable works out as
$$
Var(\beta X) = \beta^{2}Var{X}
$$
So the lazy student's variance would wor…

A Matter of Confidence

Q: An urn contains 3 red and 3 blue balls. A ball is drawn from and it socked away. Two people now draw balls from it. They record the color and put the ball back into the urn. The first person A, does this 7 times and draws a red ball all 7 times. The second person B, does this 20 times and draws a red ball 14 times. Both conclude the urn has a majority of red balls but who among them have more confidence in their prediction?

The Probability Tutoring Book: An Intuitive Course for Engineers and Scientists (And Everone Else!)

A: One is tempted to assume that A, who has done more draws which strongly indicate it to be a red-ball majority urn is likely to have a higher confidence. But this is not the case when seen in the Bayesian perspective.

Let \(H\) denote the hypothesis that the majority are red balls and \(E\) denote the evidence each user collects. As always, Bayes theorem states
$$
P(H|E) = \frac{P(E|H)P(H)}{P(E|H)P(H) + P(E|\neg H)P(\neg H)}
$$
For user A, the probability that s…

Angelina Jolie and Bayesian Statistics

This is less of a puzzle and more a write up based on recent news surrounding Angelina Jolie and breast cancer. This write up focusses on the statistical angle and is not meant to prove or disprove conspiracy theories that float around :). An interesting take is mentioned here and all statistics described here borrow from that article.

The Probability Tutoring Book: An Intuitive Course for Engineers and Scientists (And Everone Else!)

A key point floated around by popular media was that she had an 87% risk of breast cancer because she had the BRCA mutation in her genes. But note, just \(\frac{1}{600}\) actually have that mutation! Even if you did have that mutation, the chances of you eventually getting breast cancer is said to be around \(56\%\). Also note, it is important to understand whether it is exactly that mutation which is causing breast cancer, you have a \(13\%\) chance of getting it anyway (source: Breast Cancer.org). So even if you do test positive for the BRCA mutation, t…