What we need here is the expected value of the high scores. So first we need the probability of each high score.\u00a0\u00a0Then we can multiply each of those\u00a0probabilities by its corresponding score.\u00a0 The sum of those products will be our expected average.<\/p>\n
What\u00a0is the probability of\u00a0a\u00a0high score of 1?<\/strong><\/p>\n Well, you would have to get a\u00a01 every time.\u00a0 That probability is (1\/6)5<\/sup>.\u00a0 So we can write:<\/p>\n P(high score is 1) = (1\/6)5<\/sup>.<\/p>\n What\u00a0is the probability of\u00a0a\u00a0high score of 2?<\/strong><\/p>\n Now you need a 1 or a 2 every time. That probability is (2\/6)5<\/sup>.\u00a0 But wait!\u00a0 That includes the cases that turn out to be only 1\u2019s.\u00a0 But we can subtract the probability we just calculated above to get the probability we want:<\/p>\n P(high score is 2) = (2\/6)5 <\/sup>\u00a0– (1\/6)5<\/sup>.<\/p>\n What\u00a0is the probability of\u00a0a\u00a0high score of 3?<\/strong><\/p>\n We use the same plan: find the probability of getting a high score of as much as 3<\/em>, subtract the probability of getting a high score of as much as 2<\/em> \u00a0and we will be left with the probability of getting a high score of exactly 3:<\/p>\n P(high score is 3) = (3\/6)5 <\/sup>\u00a0– (2\/6)5<\/sup>.<\/p>\n And now we have a pattern to follow:<\/strong><\/strong><\/p>\n P(high score is 4) = (4\/6)5 <\/sup>\u00a0– (3\/6)5<\/sup>.<\/p>\n P(high score is 5) = (5\/6)5 <\/sup>\u00a0– (4\/6)5<\/sup>.<\/p>\n P(high score is 3) = (6\/6)5 <\/sup>\u00a0– (5\/6)5<\/sup>.<\/p>\n OK, then we go ahead and multiply each probability by its score and add them up!<\/p>\n When we do, we get an expected value of:<\/p>\n 6 \u2013 (5\/6)5<\/sup> \u2013 (4\/6)5<\/sup> \u2013 (3\/6)5<\/sup> \u2013 (2\/6)5<\/sup> \u2013 (1\/6)5<\/sup> = 5.431<\/p>\n BUT IS THAT RIGHT? Hmm\u2026<\/p>\n I suppose a short computer program could test this. It\u2019s been years since I have written code.\u00a0 But there\u2019s this program called Excel. <\/em>And it has a random-integer-generating function. [I used Randbetween(1,6) ]<\/p>\n As you can see, I actually simulated 10 years of dice rolls. And then I did it again.\u00a0 And again. And again…The last 5 times I did this, I got 5.4055, 5.4127, 5.4389, 5.4477 and 5.4211.<\/p>\n So I believe I am on the right track. Flush with success, I see on Twitter that Mr. Tanton has posted again:<\/p>\n \u201cIf I roll a die five times, how many distinct values should I expect to see?\u201d<\/strong><\/em><\/strong><\/p>\n Well, how much harder can this be? Ha!\u00a0 Stay tuned\u2026here it is<\/a>!<\/p>\n <\/p>\n","protected":false},"excerpt":{"rendered":" A good puzzle is a torment.\u00a0 And there is something interesting about probability puzzles in particular:\u00a0 you can’t always tell how tricky they are until you dig into them for a while.\u00a0 Sometimes\u00a0a\u00a0question that\u00a0seems quite tractable ends up eating up … Continue reading ![\"tanton](\"http:\/\/advancedmathyoungstudents.com\/blog\/wp-content\/uploads\/2017\/06\/tanton-dice-excel-1.png\")
<\/a><\/p>\n