2 ms·
You are agreeing with my solution? Though in this problem it seems the "average" final number would be 0.625, not 0.75 - since half the time you have an 0.75 e
by danielvf 10y ago
You are agreeing with my solution?
Though in this problem it seems the "average" final number would be 0.625, not 0.75 - since half the time you have an 0.75 expected outcome and the other half a 0.5 expected.
- thatcat 10y agoNo, having 2 chances obviously increases the probability. I don't see what you're averaging, but the formula for probability of an outcome of 2 unrelated events by the general addition rule is p(1 or 2) = (p1 + p2) - (p1*p2). in this case .5+.5-.25=.75; I was just trying to phrase it in an intuitive way.
- laxatives 10y agoI think the point is that your strategy isn't independent of your opponent. Maybe a similar example is nontransitive dice: https://en.wikipedia.org/wiki/Nontransitive_dice https://en.wikipedia.org/wiki/Nontransitive_dice In other words, it isn't sufficient to shoot for the highest score possible.
- murbard2 10y agoAbsolutely. I was inspired by non transitive dice to create the problem.