Imagine you play a dice rolling game with a friend. If you roll a '6' in 8 rolls, you win the entire stake, otherwise you lose. Suppose you have rolled 3 times and haven't won yet. Your friend now suggests to skip the 4th roll, but he will compensate you fairly for it. You will still get to roll your 5th, 6th, 7th, and 8th roll. For the compensation of skipping the 4th roll, he will take out a fair amount from the stake and give it to you, keeping the rest of the stake at play. What would be considered a fair compensation? Discussion I talked to a genius and she gave me the following argument "I think of it as two options, to skip or not to skip. I will only skip if the option gives me an expectation that is higher than, or at least the same as, if I don't skip." Expectation if I don't skip: E1= 1/6 * S + 5/6 * (1 - (5/6)^4) * S where S is the stake Expectation if I skip: E2 = c + (1 - (5/6)^4) * (S - c) where c is the compensation We want E1 <= E2. We
Let's say we are observing an event that to us seems random. E.g. The number of cars passing through a road. Suppose we learn that on average, there are m cars passing the road within 1 hour. Goal: Can we make a model to suggest what is the probability of seeing exactly r cars passing the road in an hour? Let's represent one hour as a line, and on the line there are marks to represent the time on which a car passes. Let's suppose there are m marks, to represent the average case of seeing m cars on the road in an hour. Let's divide the 1 hour line into n uniform time segments, n large enough such that in one time segment there is at most 1 car. Out of the n time segments, m of them contains a car. Then it is reasonable to say that the probability of a time segment to contain a car is m/n. Let's analyse as n goes to infinity. We can say P(seeing exactly r cars in an hour) = P(exactly r cars contained in n segments) = \( \binom{n}{r} \left( \frac{m}{n} \right)^r \left