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Yes it will be approaching 1, the same as for 10^24 coin tosses the number of heads would be approaching 50% (or 5*(10^23)).
by V-2 9y ago
Yes it will be approaching 1, the same as for 10^24 coin tosses the number of heads would be approaching 50% (or 5*(10^23)).
- monktastic1 9y agoGiven the definition of "expected value," it is identically 1 (as opposed to merely "approaching it"). Sorry, I'm not trying to be pedantic. https://en.wikipedia.org/wiki/Expected_value https://en.wikipedia.org/wiki/Expected_value
- V-2 9y agoYou're absolutely correct - not being a native English speaker, I read "expected value" as a common phrase, not a probability theory term. Thanks