Problem 11a.
There are 27 possible combinations (3^3) of 3 block cards. There are 6 ways of having three different block cards, so the probability of blocking a mixed combo of that type is 6/27. Given a particular range of attack coming in, there is only 1 way of having three block cards all of that type, so the probability is 1/27.
Problem 11b.
Out of the 27 possible combinations of block cards, only 8 of them (2^3) do not have any medium block cards (there's 2 possible choices for each of the 3 cards). Thus there is a 19/27 probability of a successful block.
Problem 11c.
There are 7 possible combinations of block cards that have two or more medium block cards, and 19 that have at least one. Therefore, given that the attack was blocked, there is a 7/19 chance that there is another medium block card in his hand. If there isn't, there is a 1/3 chance that he will draw a medium block card on his next turn. Therefore the chance of him having a medium block card is 7/19 + (1/3 * 12/19) = 11/19.
Subscribe to:
Post Comments (Atom)
1 comment:
Well, Alex, this was the first math problem you've posted that I was able to solve completely -- (I might have gotten a couple others but I was too impatient). So either I'm getting smarter, or this was a significantly easier group of questions. I'm guessing the latter.
Post a Comment