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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.

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    $\begingroup$ It will slightly simplify things if you note that we win if Hit, Hit or Miss, Hit, Hit. And intuition should say it is better to use left, right, left. $\endgroup$ – André Nicolas Dec 12 '13 at 3:10
  • $\begingroup$ @AndréNicolas - It's not quite intuitive, but consider the right hand isn't 80% but 100. Now with the right twice, the middle left 'must' succeed. But with the right in the middle, you have two chances for left to succeed. Was this your thought? $\endgroup$ – JoeTaxpayer Dec 12 '13 at 3:16
  • $\begingroup$ Roughly, the point is that the chance of missing twice with the bad hand is fairly low, getting the middle throw in is critical. $\endgroup$ – André Nicolas Dec 12 '13 at 3:21
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The success patterns are actually, Hit-Hit-Hit, Hit-Hit-Miss and Miss-Hit-Hit, as your problem statement says that he needs two consecutive hits to succeed. Your approach is otherwise correct (and yes, you do need to assume independence in order to be able to get an answer here).

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  • $\begingroup$ Oops, completely missed the "in a row" part. Using those success patterns, I get the same probability for strategy 1 (0.48), but I am getting 0.6 for strategy 2! This seems very counter-intuitive to me. $\endgroup$ – Abe Dec 12 '13 at 3:09
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    $\begingroup$ Think of it this way: he has to have a hit on his second chance in order to succeed, while he only needs one of his first and third chances to hit. So it's much more important for him to have a good second throw. $\endgroup$ – universalset Dec 12 '13 at 3:13
  • $\begingroup$ Good point. Thanks for your help! $\endgroup$ – Abe Dec 12 '13 at 3:13

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