{"id":979,"date":"2005-06-07T21:37:15","date_gmt":"2005-06-07T19:37:15","guid":{"rendered":"http:\/\/unknowngenius.com\/blog\/?p=979"},"modified":"2005-07-12T19:54:59","modified_gmt":"2005-07-12T10:54:59","slug":"mathematical-riddle-the-solution","status":"publish","type":"post","link":"https:\/\/unknowngenius.com\/blog\/archives\/2005\/06\/07\/mathematical-riddle-the-solution\/","title":{"rendered":"Mathematical Riddle: The Solution"},"content":{"rendered":"

[Blogger Advisory: unspeakable nerdery, use of foul language.]<\/em><\/strong><\/p>\n

As promised: the answer to yesterday’s riddle<\/a> (I know, suspense is killing you)…<\/p>\n

Upon closer inspection, maybe I went a bit ahead of myself when I postulated the solution was “really easy”. Note that the explanation below, while somewhat tedious and longwinded, should be perfectly intelligible to anybody with very basic notions of arithmetic and a few sober neurons. If you don’t fit either criteria, feel free to skip to the last paragraph.<\/p>\n

The Question<\/h3>\n

Does one stand better chances of: 1) getting at least one<\/strong> ace with 6<\/strong> throws of a die, or 2) getting at least two<\/strong> aces with 12<\/strong> throws?<\/p><\/blockquote>\n

<\/p>\n

The Reasoning<\/h3>\n

First, and without getting into a complete course on stats: whatever probability you find, it better be between 0 and 1. Anything above 1, means in essence that your result would happen more times than the total number of experience (e.g. getting seven ace with six dice), which, most sources agree, would about bring our entire universe down, collapsing under the weight of its own impossibility. <\/p>\n

A probability of 1 means that your event will always<\/em> turn true. In this regard, all those who found p = 1<\/code> to either of the previous case (either throwing 6 or 12 dice) have been wasting their life so far: they should be busy getting rich at the nearest casino, where their own laws of probabilities guarantee<\/strong> them one successful draw for every six rolls (pretty good odds, if you ask me) \ud83d\ude42<\/p>\n

Yet, I have to agree that, at first, it may look like a sensible solution (and this is why stats problems are usually so fun): since your probability of getting a given score (say an ace) for one roll of a die is p(x) = 1\/6<\/code> (with x<\/code> = 1…6), it’s not completely<\/em> farfetched to assume that your total probability to get an ace out of 6 draws would be 6 * p(1) = 1<\/code>. Except that, following the same logic, 7 draws gives you a probability of 7 * p(1) = 7\/6<\/code>, and before you reach the end of this thought, the whole universe has instantaneously disappeared, to be immediately replaced by a more bizarre and inexplicable one, according to Douglas Adams’ 1st Law of Astrophysics<\/a>.<\/p>\n

Unfortunately, the solution is a tad<\/em> more complicated than that:<\/p>\n

Let’s keep things simple and start with two<\/strong> dice. Now, we know the odds of getting any given score for one die are p(x) = 1\/6<\/code>, that is: any of the six faces has an equal probability of coming up. With two dice, the total number of combinations is, quite obviously, 6 * 6 = 36<\/code>. That is, there are exactly 36 different possible results (1 & 1, 1 & 2, 1 & 3 etc.), each with exactly the same odds of happening. Of these, there are 11<\/strong> ways we could get at least one ace (1 & 2, 1 & 3 […] 1 & 6, 2 & 1, 3 & 1 […] 6 & 1 and 1 & 1).
\nOur probability, then, is the ratio of these two numbers: p(at least one ace<\/i>) = [ways to get at least an ace<\/i>] \/ [total number of combinations<\/i>] = 11 \/ 36 = 0.30<\/code>. Roughly one out of three: not bad.<\/p>\n

Now, consider more dice: you would find that 3 dice have 91<\/strong> ways of showing one ace or more, out of 6 * 6 * 6 = 216<\/code> total combinations (p = 0.42<\/code>), 4 dice have 671<\/strong> out of 1296<\/strong> combinations (p = 0.52<\/code>). And so on and so forth. Quite logically, the more dice, the better your chances of getting at least one ace.<\/p>\n

All in all, figuring the odds for “1 ace or more” with n<\/code> dice (basically, situation 1. in the problem above) is rather straightforward for low numbers of dice. It gets tedious for more than 3 or 4 and downright daunting for 12 dice.<\/p>\n

Situation 2. requires a slightly more evolved logic. Let’s reduce the case and assume we want at least two aces out of 3<\/strong> dice. That’s gonna be either A and B, B and C, A and C or all three A, B, C, giving an ace.<\/p>\n

Let’s look at the general case systematically and draw the rule (hang on to your hair, that’s where it becomes a teeny bit technical). <\/p>\n

Say we got n<\/code> dice and want to know the odds of getting exactly<\/strong> r<\/code> aces:<\/p>\n