Sunday, 17 July 2016

statistics - Calculating dice probability and risk-reward value

Having trouble how to model this cost-risk question with Probability.



Assume that you are given 5 standard six-sided dice, what's the probability you get at least three 2s?



I have calculated this:



There are 6^5 = 7776 events possible.




Bad events include rolling exactly 0 2's, 1 2's, or 2 2's.



Good events are rolling exactly 3 2's, 4 2's, or 5 2's.



Thus the probability that I roll at least three 2's is the sum of the probabilities of the good events.



Answer: P = 276/7776.



But now here is a follow-up: A casino will pay 1000 dollars if I roll at least three 2's, but I have to pay the casino 200 dollars if I don't. Does it make sense to take the bet? Why or Why not?




I'm unsure of how to model the reward-risk of cash given the calculated probabilities?

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