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Showing posts from November, 2013

The Sultan's Wine Bottles

Follow @ProbabilityPuz Q: A Sultan has a 1000 bottles of wine. He needs to use them in 30 days time for a royal banquet. He knows that his enemies have poisoned exactly one bottle with a type of poison that takes effect in 29 days. He decides to use his soldiers to test which bottle is poisoned. Is there a strategy that minimizes the number of soldiers needed for the task? Probability Theory: The Logic of Science A: The naive approach is to have one soldier per bottle. Every soldier gets a drop from each bottle and they wait for 29 days. The number of the soldier who gets affected on the 29th day shows which bottle is poisoned. However, this strategy is quite expensive in terms of the number of soldiers needed for the Sultan. A far more efficient strategy is the following. Label each bottle with a number. Maintain a ledger which maps a number to each of the patterns 0000000000 -> 1, 0000000001 -> 2, 0000000010 -> 3, 0000000100 -> 4 and so on till you rea

Drawing Increasing Numbers from a Deck of Cards

Follow @ProbabilityPuz Q: You have a 5 decks of cards numbered 1-12. You draw 4 cards at random from each deck. What is the probability that the 4 cards come out in an increasing sequence from at least one of the draws? Dremel CKDR-02 Ultimate 3-Tool Combo Kit with 15 Accessories and Storage Bag A: Note a subtle point here. The number of cards in the deck to draw from doesn't really matter. All we are concerned with is if the numbers are in an increasing sequence. If you draw 4 cards from any sized deck, you are left with 4 distinct cards and that's it! Given 4 cards there are \(4! = 24\) ways (permutations) to place them in any given order. Thus, the required scenario happens with a probability of \(\frac{1}{4!} = \frac{1}{24}\). Therefore, The probability that they are not in an increasing sequence is \(1-\frac{1}{24}= \frac{23}{24}\). The probability that all 5 draws from each of the 5 decks is not in an increasing sequence is \(\big(\frac{23}{24})^{5

Winning the Lottery, Twice!

Follow @ProbabilityPuz You might have heard this piece of news often "XYZ wins lottery for a second time". Such news, like  here, leaves the reader wondering how someone can get that lucky. Needless to say, the winner is on cloud nine knowing s/he has won, not once but twice! But are these events truly rare? We are conditioned to believe that winning a lottery in itself is a rare event let alone winning it twice. Lets explore. Probability Theory: The Logic of Science To begin understanding the rarity of the above event, lets revisit the Birthday Problem. The summary of the problem is we need an astonishingly small number of people in a room to have \(>50\%\) probability that two people would have the same birthday, about 23. To generalize, there are 365 days in a year. If there are 2 people in a room, the probability that both have different birthdays are \(\frac{364}{365}\). If there are 3 its \(\frac{364}{365}\times\frac{363}{365}\) and so on. If you

Points on a Circle

Follow @ProbabilityPuz Q: A total of \(n\) random points are selected from a circle. What is the probability that all of them are on a semicircle when evaluated in a clockwise order? Probability Theory: The Logic of Science A: Assume \(n=4\) as shown in the figure below. Without loss of generality, assume the points were picked in the order shown by the numbers in the figure. Start with the point numbered 1 and look in a clockwise direction. Each of the subsequent points can be either in or out of the semicircle starting from 1. Each of the given points will be within the semicircle with probability \(\frac{1}{2}\). Therefore the probability that all the remaining \(n-1\) points will be in the clockwise semicircle is \(\frac{1}{2^{n-1}}\). This estimate is for a given starting point numbered at 1. To compute the probability for all \(N\) points, we apply the same logic to each of the points in sequence. This yields \(\frac{n}{2^{n-1}}\) which is the probability that