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Dilbert 
(Maybe I should count only six people, because one of the seven had never realized there's a difference between a cartoon speech bubble and a cartoon thought bubble. Sheesh.) As I take it, Dogbert owes Dilbert $2 million. As a belated birthday gift to me, if you get this, or if you even think you get this, you must tell me. Plus which, if you do and if you like, I'll cite your explanation in this space and give you credit.
Update of May 10, 2001 Quite a few people have clicked the let me know link above offering an explanation why this strip does make sense. So far no one's explanation makes sense to me. Here's a sample from today's mail:
Let me try harder to explain why this strip makes no sense and include some additional information. First, let's look at the first panel, which I've chosen not to discuss so far and which, to my surprise, all of my correspondents have too. The first panel, all by itself, makes no sense, regardless of whatever comes after. In the first panel it is clear that Dogbert had already promised to pay Dilbert a million dollars if Dilbert "went to work like this." "Like this" must mean dressed just as we see Dilbert dressed in the first panel. But Dilbert is in fact clothed, which makes no sense. Why would Dogbert offer to pay Dilbert a million dollars to go to work wearing clothes? (For what it's worth, I notice that Dilbert isn't wearing his characteristic tie that's turned up at the bottom. This might be a mere artist's typo, or it might have some meaning that helps this panel or the entire strip make sense, but if that's the case then I still don't get it.) So, moving on, in the second panel Dogbert says: I'll bet you double or nothing that you won't go to work totally naked. Almost everyone who's emailed me about this crucial panel has made a comment similar to Ann's above, although usually with more words (sometimes a lot more). For some reason, all those correspondents seem to have misconstrued what Dogbert's bet means, somehow coming to believe it means the opposite of what it does. If you too think it means the opposite of what I think it means then the strip does indeed make sense. It's not particularly funny, but at least it makes more sense than otherwise. However, Dogbert's doubleornothing bet does not the mean the opposite of what I think it means. My guess is that it's the word not in his offer that is confusing people. Here's what Dogbert's bet really does mean, and here's how you can prove it to yourself. First, let's remove all the unnecessary information and simplify the offer. Dogbert's offer can be restated without harm as: "I bet you will not go naked." Now, let's consider the reverse of that offer by removing the word not, which results in this: "I bet you will go naked." According to this simplified and reversed version it is clear that Dilbert wins by not going naked, i.e., Dilbert wins if he wears clothes. Presumably Dogbert would not make such a bet, but remember that this is the reverse of the wording of the actual bet. The wording of the actual bet is, "I bet you will not go naked." By adding the word not Dogbert has reversed the meaning of the bet discussed in the paragraph above. And if in the bet above Dilbert wins by wearing clothes, which he clearly does, then just as clearly he wins the actual bet by going naked. Dilbert did go naked, so Dogbert owe him two million dollars. If that's the case, which I claim it is, then why is it funny? All we have is that Dogbert paid Dilbert two million dollars to go to work naked. Furthermore, if that's all there is to it, which I still claim it is, then Dilbert's thought bubble in the last panel makes no sense. Further yet, none of this explains the first panel. The only way the middle panel makes complete sense to me, given that we know Dilbert did go to work naked, is if Dogbert somehow both won the bet and also caused Dilbert to embarrass himself by going to work naked. But that's simply not the case, because, as I think I've proven, Dilbert won the bet. Herringcolored glasses One of my correspondents has suggested that the fact Dilbert is wearing his spectacles in the third panel means he lost the bet, which would produce that desired result of Dogbert's having won the bet and also having caused Dilbert embarrassment. But if that's all there is to the last two panels then, as that correspondent said, it's lame. Furthermore, if that's all there is to it it still does not explain the first panel, which presumably was intended to have some bearing on the other two. I still want help with this, so if you think you've got it figured out, or if you think I've got it figured out wrong, please . Update of November 13, 2002: Although I still don't have the answer to the primary question above, I do now have additional information, brought to my attention by a correspondent with a good memory and a collection of Dilbert compilation books he was willing to leaf through till he found the series of five daily strips (Random Acts of Management by Scott Adams, 2000) that ends with the one above. Thanks go to (no relation) for pushing this debate along a bit. The "Bathrobe Bet" series started on Monday, May 24, 1999, and ended on Friday, May 28th. The three middle strips
have absolutely nothing to do with anything in question and so are not included. Here is the May 24th strip that started the series followed by,
again, the last strip, the one I still can't figure out. 
So, anyway, I still want an answer to my primary question: What is funny about the third panel of the last strip? If you think you know, . Update of April 19, 2007: A correspondent, Adam Perry (no relation), offers an explanation that is both new and reasonable. I'll quote him directly. The Dilbert "I'll bet you double or nothing" strip seems to be a play on the conjunction, that is, Dogbert is ambiguously asserting that he will either bet Dilbert double or he will bet him nothing. When Dilbert tries to claim his reward, Dogbert can escape on the assertion that he bet Dilbert nothing. If I had to bet I'd say this is not what Adams intended, but it does make sense. Update of August 29, 2007: New information, from reader Justin SaberScorpion, whom I quote below, lends weight to the theory above. I can assure you this is probably what Scott Adams intended, because this is actually a trick that Dogbert pulls a lot in the comic  using words with meanings that aren't taken at face value in regular conversation, such as "double or nothing." For example, in one strip he tells the PointyHaired Boss that his magazine has "between one and two billion readers," but at the end of the strip he says, "I figured out how to make three readers sound like a lot." Mr. SaberScorpion also kindly supplied the strip, dated November 18, 2005.
Update of April 22, 2005 Today's strip makes sense, especially to anyone who's ever had to make more than three calls to try to get an FOT computer fixed. But my question has to do with the math. Dogbert says there are 25 possible fixes that must be tried in every possible combination, and obviously he agrees that that would result in 625 "things" the customer has to try. I think the real number is a bit higher than 625. There's only one way to interpret the phrase "25 fixes in every possible combination": You start by trying the first fix, then you try the seond fix, then you try the third fix and so on, up to the 25th fix. We could write this particular combination thusly: 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 But let's simplify that to this notation: 1 2 3 4 . . . 23 24 25 Here's a second combination: 2 1 3 4 . . . 23 24 25 All we've done above is swap the order of the first two fixes, and we have a new combination Dogbert's customer has to try. Here's a third combination, in which we swap the original second and third fixes. 1 3 2 4 . . . 23 24 25 And now we swap the original third and fourth fixes for a fourth combination. 1 2 4 3 . . . 23 24 25 Eventually we'll get through all the combinations in which we swap two consecutive fixes and leave all the others alone, and there will be 24 of them in addition to the original order. The last one will be 1 2 3 . . . 23 25 24 But then there's a whole series in which you swap every other fix, starting this way: 3 2 1 4 . . . 23 24 25 1 4 3 2 . . . 23 24 25 and ending with 1 2 3 4 . . . 25 24 23 Then you need the series in which you swap every third fix, and the series in which you swap every fourth fix, and so on. And it gets more complicated than that, because you can also start with 1 2 3 4 . . . 23 24 25 and then start swapping both the 13th and 14th fixes and the 19th and 20th fixes but no others. And it gets more complicated than that, but I'll stop, because I'm sure you get the idea that there are probably more than 625 such "things to try" if the customer must try "25 possible fixes . . . in every possible combination." Combination versus permutation. And now I must confess to perpetuating, however briefly, an error in Adams's diction. Strictly speaking, in math the term combination would be incorrect here. It should be permutation. Where the customer says there are 625 "things" he'd have to try, he means permutations, and in every case above where I used the term combination, I should have said permutation, so please scroll up, dab on some mental WiteOut, and let it dry in your brain. The difference is whether the order makes a difference. If it does not, such as in a hand of 5 cards you're dealt from a deck of 52, that's a combination. If it does, such as the order in which 5 lottery balls are chosen from a set of 52, that's a permutation. There's a way to calculate the number of possibilities for both combinations and permutations, and I will not bore you with those formulas, but I will point out that many calculators nowadays offer these functions. Look for keys labelled nCr and nPr, which are for n things chosen r at a time, where the middle letter tells you whether it's for Combinations or Permutations. (Of course, all spreadsheets offer these functions; in Excel they are COMBIN and PERMUT, and r is replaced with k. In the unlikely event you're interested, here's the Excel 2000 spreadsheet I used.) In case you're still interested, the number of combinations of 5 cards dealt from a deck of 52 (which would be 52C5) is less than 3 million, whereas the number of permutations, in which the order of the 5 cards does make a difference (52P5), is over 300 million. Another complication. Now that we know Dogbert is really talking about permutations and not combinations, before we calculate the correct number of "things" his customer must try, we have to consider that not all of the possible "things" are exactly 25 fixes long. We also have to count the permutations of only 24 fixes. And we have to count the permutations of only 23 and of only 22 and so on down to the simplest possible "combination," which is the very first fix, all by itself. So, to calculate the correct number, which I reassure you turns out to larger than 625, we need to add 25P25 + 25P24 + 25P23 + . . . + 25P3 + 25P2 + 25P1. Just to get you started, 25P1 equals 25 (duh), 25P2 equals 600, and 25P3 equals 13,800. The correct number. Instead of 625, which is merely 25 squared, the number of things Dogbert's customer must try is actually 42,163,840,398,198,058,854,693,625, or 42 septillion and change. (Thanks to CCH for calculating the exact number using software that wood have cost me $2000.) If we assume that the customer could try each of those 42 septillion "things" in an average of ten seconds (which is wildly optimistic, but never mind), to go through all the permutations would take over 133 quintillion millennia, or exactly 1 billion times the age of the universe to date, . I'm not sure the customer's unwillingness to take on that project could be called laziness. If you disagree with any of these assertions or calculations, and many have, please . If you find large numbers as interesting as I do, go here or, better yet, here. 
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