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How do you decide if a game is fair using theoretical or experimental probability?
Suppose that each time the game is played you win an amount X, with probability p or lose an amount Y with probability (1-p). To start with it is assumed that you cannot draw a game. Then each time the game is played you expect to win X with a probability p and lose Y (= win -Y) with probability (1-p). Therefore, the expected gain for you, each time the game is played, is X*p - Y*(1-p). The game is fair if this value is 0. If it is greater than 0 the game is biased in your favour. Otherwise it is biased against you. You can easily incorporate multiple outcomes - such as with a lottery: a bigger win for getting more numbers. If X(1), X(2), ... X(n) are the n pay-outs to you (remember that what you pay will be a negative number), and P(1), p(2), ... p(n) are their respective probabilities where the sum of all the p(r) = 1, then your expected win per game is X(1)*p(1) + X(2)*p(2) + ... + X(n)*p(n) and the game is fair if this is 0.
What is the third form of win?
win
What is the hardest math question ever?
n+n-n-n-n+n-n-n squared to the 934892547857284579275348975297384579th power times 567896578239657824623786587346378 minus 36757544.545278789789375894789572356757583775389=n solve for n! the answer is 42
When does minute to win it start?
MY guess would be that it is not Minute to win it but "in it to win it" this meaning that the reason you are there is to win. Hope this answere ur question :)
Adding n to the second power to n?
n2 + n = n(n + 1)
