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  2. Buffon's needle problem - Wikipedia

    en.wikipedia.org/wiki/Buffon's_needle_problem

    We can calculate the probability P as the product of two probabilities: P = P 1 · P 2, where P 1 is the probability that the center of the needle falls close enough to a line for the needle to possibly cross it, and P 2 is the probability that the needle actually crosses the line, given that the center is within reach.

  3. Coupon collector's problem - Wikipedia

    en.wikipedia.org/wiki/Coupon_collector's_problem

    In probability theory, the coupon collector's problem refers to mathematical analysis of "collect all coupons and win" contests. It asks the following question: if each box of a given product (e.g., breakfast cereals) contains a coupon, and there are n different types of coupons, what is the probability that more than t boxes need to be bought ...

  4. Balls into bins problem - Wikipedia

    en.wikipedia.org/wiki/Balls_into_bins_problem

    In the simplest case, if one allocates balls into bins (with =) sequentially one by one, and for each ball one chooses random bins at each step and then allocates the ball into the least loaded of the selected bins (ties broken arbitrarily), then with high probability the maximum load is: [8]

  5. Lottery mathematics - Wikipedia

    en.wikipedia.org/wiki/Lottery_mathematics

    It is a hard (and often open) problem to calculate the minimum number of tickets one needs to purchase to guarantee that at least one of these tickets matches at least 2 numbers. In the 5-from-90 lotto, the minimum number of tickets that can guarantee a ticket with at least 2 matches is 100. [3]

  6. Stars and bars (combinatorics) - Wikipedia

    en.wikipedia.org/wiki/Stars_and_bars_(combinatorics)

    The solution to this particular problem is given by the binomial coefficient (+), which is the number of subsets of size k − 1 that can be formed from a set of size n + k − 1. If, for example, there are two balls and three bins, then the number of ways of placing the balls is ( 2 + 3 − 1 3 − 1 ) = ( 4 2 ) = 6 {\displaystyle {\tbinom {2 ...

  7. Secretary problem - Wikipedia

    en.wikipedia.org/wiki/Secretary_problem

    One important drawback for applications of the solution of the classical secretary problem is that the number of applicants must be known in advance, which is rarely the case. One way to overcome this problem is to suppose that the number of applicants is a random variable N {\displaystyle N} with a known distribution of P ( N = k ) k = 1 , 2 ...

  8. Birthday problem - Wikipedia

    en.wikipedia.org/wiki/Birthday_problem

    From a permutations perspective, let the event A be the probability of finding a group of 23 people without any repeated birthdays. Where the event B is the probability of finding a group of 23 people with at least two people sharing same birthday, P(B) = 1 − P(A).

  9. Newton–Pepys problem - Wikipedia

    en.wikipedia.org/wiki/Newton–Pepys_problem

    The Newton–Pepys problem is a probability problem concerning the probability of throwing sixes from a certain number of dice. [1] In 1693 Samuel Pepys and Isaac Newton corresponded over a problem posed to Pepys by a school teacher named John Smith. [2] The problem was: Which of the following three propositions has the greatest chance of success?