- #36
MarcoD
I am not sure about the die game, it depends on how you read the question. If you get a return each round, I think the answer is seven, if you are allowed to 'modify' the outcome by throwing the die again, 3.5 should be correct.
MarcoD said:I am probably the only one who reads #9 as: an algorithm such that you end up either with exactly 200 or 0 in the end. Right?
Jimmy Snyder said:No, just flip twice, 2 heads is A, 1 head, 1 tail is B, 0 heads is ignored.
BobG said:It doesn't say you have to bet the same amount on each game.
This is correct.MarcoD said:The expected value, I guessed, is: chance you get to the 1st throw * expected return on the 1st throw + chance you get to the 2nd throw * expected return on the 2nd throw + ...
This is not correct.MarcoD said:= 1.0 * 3.5 + 0.5 * 3.5 + 0.25 * 3.5 + ...
This is correct.MarcoD said:= 7
MarcoD said:I am probably the only one who reads #9 as: an algorithm such that you end up either with exactly 200 or 0 in the end. Right?
Jimmy Snyder said:This is correct.
The probability that the game will end after a single toss is .5 and the expected value in that case is 2, i.e. the average of 1, 2, and 3. The probability that the game will end after two tosses is .25 and the expected value is 7, i.e. the average of 4, 5, and 6 for the first toss and of 1, 2, and 3 for the second.
MarcoD said:If you write down a tree, you can see you're calculating the limit of:
[tex](\frac{1}{6} + ... + \frac{6}{6}) + \frac{3}{6}((\frac{1}{6} + ... + \frac{6}{6})+ \frac{3}{6}((\frac{1}{6} + ... + \frac{6}{6})+ ... ))[/tex]
AlephZero said:I read it as "you want to make a profit of either 200 or 0", i.e. find a hedging strategy so you end up with either 300 or 100.