Am I going about this problem the right way?
In a certain game a participant is allowed three attempts to score a hit. In three attempts he must alternate which hand he uses: thus he has two possible strategies: right hand, left hand, right hand or left hand, right hand, left hand. His chance of scoring a hit with his right hand is 0.8; while it is only 0.5 with his left hand. He is successful at the game provided he scores two hits in a row. What strategy gives the player the better chance of success?
Intuitively, I want to say it is the first strategy - right, left, right, simply because that gives you two attempts with a higher probability of success. But how do I actually demonstrate this mathematically?
Using strategy 1: there are 3 different possible ways to succeed. HIT-HIT-MISS, HIT-MISS-HIT, or MISS-HIT-HIT. The probability of each of these should be (0.8*0.5*0.2), (0.8*0.5*0.8), and (0.2*0.5*0.8), assuming they are independent (not sure if this is a logical assumption to make? but also don't know how I could do this problem otherwise with the information given). Adding these three together gives me a probability of success of 0.48.
Using strategy 2: again, same different 3 success patterns. This time, the probability of each is (0.5*0.8*0.5), (0.5*0.2*0.5), and (0.5*0.8*0.5). This gives me a total success probability of 0.45.
So, this would make strategy 1 more optimal, but only barely.
Is this right? I'm not sure how else I would go about figuring this out.