{"id":294,"date":"2019-10-04T12:04:43","date_gmt":"2019-10-04T17:04:43","guid":{"rendered":"http:\/\/blogs.acu.edu\/mu_sigma\/?p=294"},"modified":"2019-10-11T11:04:45","modified_gmt":"2019-10-11T16:04:45","slug":"problem-5-fall-2019-will-you-win","status":"publish","type":"post","link":"https:\/\/blogs.acu.edu\/mu_sigma\/2019\/10\/04\/problem-5-fall-2019-will-you-win\/","title":{"rendered":"Problem 5 – Fall 2019 – Will You Win?"},"content":{"rendered":"

Will You Win?<\/b><\/span><\/p>\n

This problem was suggested by Brian Jacobs.<\/i><\/span><\/p>\n

Your friend has a game for you to try. He rolls two n<\/i>-sided fair dice and you roll one n<\/i>-sided fair die. You win if the number on your die is between the numbers on his dice, inclusively. For example, if he rolls a double, then you win only if the number on your die matches the number on his dice. On the other hand, if he rolls a 1 and an n<\/i>, then you win no matter what you roll. What is the probability that you will win if you choose to play this game once?<\/p>\n

Please submit all your work for this problem to mathpotw@acu.edu no later than 5:00 PM on Thursday, October 10th to receive credit for your submission.<\/p>\n

Solution to Will You Win?<\/b><\/span><\/p>\n

Let’s focus on the “gap” between the two numbers rolled by our opponent. For example, if he rolls a double, then the gap is 0. And in the worst case (for him) he rolls a 1 and an n<\/i>, making a gap of n<\/i>-1. For a gap of size k<\/i>, we have k<\/i>+1 possible successful rolls. For example, if he rolls a double, then we have to roll that number and so we have only 1 possibility. The answer to the problem can now be obtained by calculating the probability that he rolls a gap of size k<\/i>, then multiplying that probability by the probability that our roll is within that range given he has that gap, and then summing over k<\/i>. The denominator of all the probabilities will be n<\/i>3<\/sup>, so we compute the numerator.<\/p>\n

There are n<\/i> ways for him to roll a gap of 0 (roll a double) and then one way for us to match his number. There are 2(n<\/i>-1) ways for him to roll a gap of 1, and then 2 ways for our roll to be between his numbers. There are 2(n<\/i>-2) ways for him to roll a gap of 2, and then 3 ways for our roll to be between his. Continuing this reasoning leads to a numerator of<\/p>\n

.<\/p>\n

We will make the algebra that follows easier if we include the first term in the summation symbol, which we accomplish by adding an n<\/i> at the beginning and then subtracting it out at the end. The resulting algebra is as follows:<\/p>\n

= = .<\/p>\n

Using well-known formulas for the sum of consecutive integers and the sum of squares of consecutive integers, we arrive at a numerator of<\/p>\n

= (n<\/i>2<\/sup> + 3n<\/i> – 1)\/3. Therefore the probability of our winning if we play this game once is<\/p>\n

(n<\/i>2<\/sup> + 3n<\/i> – 1)\/3n<\/i>2<\/sup>.It may be worth observing that as n<\/i> increases to infinity, the probability decreases to 1\/3. If n<\/i> is 2, the probability is 3\/4. The probability of winning is more than 1\/2 until n<\/i> = 6, in which case the probability is 53\/108.<\/p>\n","protected":false},"excerpt":{"rendered":"

Will You Win? This problem was suggested by Brian Jacobs. Your friend has a game for you to try. He rolls two n-sided fair dice and you roll one n-sided fair die. You win if the number on your die is between the numbers on his dice, inclusively. For example, if he rolls a double, […]<\/p>\n","protected":false},"author":130,"featured_media":0,"comment_status":"open","ping_status":"closed","sticky":false,"template":"","format":"standard","meta":{"_et_pb_use_builder":"","_et_pb_old_content":"","_et_gb_content_width":"","footnotes":""},"categories":[181534],"tags":[],"class_list":["post-294","post","type-post","status-publish","format-standard","hentry","category-problem-of-the-week"],"_links":{"self":[{"href":"https:\/\/blogs.acu.edu\/mu_sigma\/wp-json\/wp\/v2\/posts\/294","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/blogs.acu.edu\/mu_sigma\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/blogs.acu.edu\/mu_sigma\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/blogs.acu.edu\/mu_sigma\/wp-json\/wp\/v2\/users\/130"}],"replies":[{"embeddable":true,"href":"https:\/\/blogs.acu.edu\/mu_sigma\/wp-json\/wp\/v2\/comments?post=294"}],"version-history":[{"count":2,"href":"https:\/\/blogs.acu.edu\/mu_sigma\/wp-json\/wp\/v2\/posts\/294\/revisions"}],"predecessor-version":[{"id":305,"href":"https:\/\/blogs.acu.edu\/mu_sigma\/wp-json\/wp\/v2\/posts\/294\/revisions\/305"}],"wp:attachment":[{"href":"https:\/\/blogs.acu.edu\/mu_sigma\/wp-json\/wp\/v2\/media?parent=294"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/blogs.acu.edu\/mu_sigma\/wp-json\/wp\/v2\/categories?post=294"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/blogs.acu.edu\/mu_sigma\/wp-json\/wp\/v2\/tags?post=294"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}