8/28/2023 0 Comments Monopoly chance cards back pythonI think you will agree that all of these outcomes are equally likely. I have denoted 'heads' by an 'h', and tails by a 't.' I have starred the outcomes that indicate a win for myself. Thus, I believe the following list of possible endings to the game is exhaustive. In a similar manner, if I had not achieved the necessary 2 points for my victory, this would imply that you had in fact achieved at least 3 points and had therefore won the game. For, in those four tosses, if you did not get the necessary 3 points for your victory, this would imply that I had in fact gained the necessary 2 points for my victory. Seeing as I needed only two points to win the game, and you needed 3, I think we can establish that after four more tosses of the coin, the game would have been over. In 1654, Blaise Pascal and Pierre de Fermat corresponded on this problem, with Fermat writing:Īs to the problem of how to divide the 100 Francs, I think I have found a solution that you will find to be fair. If the game is interrupted when H has 8 heads and T has 7 tails, how should the pot of money (which happens to be 100 Francs) be split? Player H wins the game if 10 heads come up, and T wins if 10 tails come up. show ()įermat and Pascal: Gambling, Triangles, and the Birth of Probability ¶ Pierre de FermatĬonsider a gambling game consisting of tossing a coin. We'll start by defining the contents of the urn: But there are only 56 / 2 = 28 combinations, because (W1, W2) is the same combination as (W2, W1). For example, if I want to choose 2 white balls from the 8 available, there are 8 ways to choose a first white ball and 7 ways to choose a second, and therefore 8 × 7 = 56 permutations of two white balls. The second issue is handled automatically by the P function, but if I want to do calculations by hand, I will sometimes first count the number of permutations of balls, then get the number of combinations by dividing the number of permutations by c!, where c is the number of balls in a combination. That makes it clear that selecting 'W1' is different from selecting 'W2'. To account for the first issue, I'll have 8 different white balls labelled 'W1' through 'W8', rather than having eight balls all labelled 'W'. An outcome is a set of balls, where order doesn't matter, not a sequence, where order matters.We have multiple balls of the same color.We'll solve each of the 3 parts using our P function, and also using basic arithmetic that is, counting. So, an outcome is a set of 6 balls, and the sample space is the set of all possible 6 ball combinations. What is the probability of each of these possible outcomes: We select six balls at random (each possible selection is equally likely). (You'd think the urns would be empty by now.)įor example, here is a three-part problem adapted from :Īn urn contains 23 balls: 8 white, 6 blue, and 9 red. This notebook will develop all these concepts I also have a second part that covers paradoxes in Probability Theory.Īround 1700, Jacob Bernoulli wrote about removing colored balls from an urn in his landmark treatise Ars Conjectandi, and ever since then, explanations of probability have relied on urn problems. (This assumes that all outcomes in the sample space are equally likely.) Since it is a ratio, probability will always be a number between 0 (representing an impossible event) and 1 (representing a certain event).įor example, the probability of an even die roll is 3/6 = 1/2. Īs Laplace said, the probability of an event with respect to a sample space is the number of favorable cases (outcomes from the sample space that are in the event) divided by the total number of cases in the sample space. The set of all possible outcomes for the experiment.Ī subset of possible outcomes that together have some property we are interested in.įor example, the event "even die roll" is the set of outcomes. The result of an experiment one particular state of the world. We'll start being methodical by defining some vocabulary:Īn occurrence with an uncertain outcome that we can observe. Laplace really nailed it, way back then! If you want to untangle a probability problem, all you have to do is be methodical about defining exactly what the cases are, and then careful in counting the number of favorable and total cases. when nothing leads us to expect that any one of these cases should occur more than any other. is thus simply a fraction whose numerator is the number of favorable cases and whose denominator is the number of all the cases possible. (You should have some background in probability and Python.) This notebook covers the basics of probability theory, with Python 3 implementations. Peter Norvig, A Concrete Introduction to Probability (using Python) ¶
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