Prime

The way I see it...

No hints, please. You already know the solution, give us a chance.

You cannot know which segment is slowest until all others burned completely. Not that it would help if you did.

You show the resulting infinite regress in reasoning. However, you cut it off at one point alleging that somehow Logician will outcalculate Merlin. I am going to intercede on Merlin's behalf. Do not underestimate him! Just because he is a wand maker, it does not mean he has no logic. The important point: Merlin can always ensure own survival. Not so for poor Logician, who seems to have only a 50/50 chance of guessing what to do after the duel. But let me retract from some of my previous assumptions and make this problem definitively solvable. I suggested that a simple non-wand stick (a zero-wand) should be prohibited. Now I take it back. They can use plain sticks and no one will know better. Suddenly, Merlin's life is in danger. He does not know if the spell is hanging over him after the duel. Let's make another reasonable assumption (already suggested by Unreality): survival is overriding motive, killing your opponent is the next priority, while keeping your good wand from king's hands, marrying his daughter/son not all that important. Now Merlin must devise the way to ensure own survival. That in itself is a riddle so it goes inside the spoiler. Only in this way both survived, explained to King that their wands were of exact same maximum possible strength, married kings sons and daughters...

I don't think so.

Differential equations and vector analysis again? But if each stands in their corner, until they come up with the solution, this room is going to be very quite for a while. How do you gauge the interest then?

The OP did.

How is it different from 2 unevenly burning ropes problems?

What I don't get is how Logician can even begin to think he made sure that Merlin dies? We agreed, he cannot cast any spells on Merlin without Merlin's knowing.

And to continue stumping dead logician into the ground, here is an infinite regress of reasoning to which I arrive: It is always prudent to precast upon oneself before the contest. Merlin can always ensure his survival, for he knows the relative strength of his wand compared to Logician's. Merlin knows that logician's demise is effected by Logician's misjudgement on whether or not to cast cancellation spell upon himself after the tournament. However, if Logician neglected to perform a pre-spell, then using strongest wand at the tournament ensures Logician's demise. (He would not be able to save himself with any cancellation spells thereafter.) Knowing that, Merlin cannot pass this additional chance and must bring his strongest wand to the tournament, Logician knows to always perform a pre-spell. Merlin knows that Logician expects him to bring strongest wand, and thereafter guess correctly to perform cancellation spell after the tournament and survive. Therefore, to con Logician into performing an extra deadly spell upon himself, Merling brings his weakest wand... Logician knows, that Merlin knows...

One must be a weak logician to fall for such a simple Merlin's trick. Having upper hand in wand-making, Merlin can insure his own survival, but not your demise. But first there are some assumptions to make to disambiguate the statement of the problem. 1. You know when spell is cust upon you, but can you sense the relative strength of the wand with which the spell was cast? 2. You always know of the spells cust upon you (by yourself, or others), but do you know of any spells hanging over your opponent? I am going to go ahead and assume that no one (even Merlin) can tell the relative strength of wand without having it in their possession. And that no one knows of how many spells already hang over another person. Also, it is useful to assume that king knows a wand from plain stick and contestants will have to use some wand. No swithcing of wands in the middle of the contest. Being a top wand maker, Merlin makes the weakest possible non-zero wand and shows up with it at the contest. After the contest and before midnight, he uses his strong wand to negate your death spell still hanging over him. Having figured that out, you cust additional spell upon yourself after the contest and also survive. Merlin can ensure own survival with either weakest, or strongest wand at the contest. In first case, he casts additional spell after the contest, in the second -- he doesn't. Without being able to sense relative wand strength, logician's survival becomes a matter of probability. Logician faces a decision of whether to cast an additional spell on himself after the contest. In either case, it is prudent for logician to precast a spell upon himself, when he is protected against Merlin's stronger wand. The statement of the problem already tells us what happenned. Merlin survived, Logician died. Which means that Logician could not guess that Merlin had the weakest wand at the contest, and then made a misdjudgement with respect to the need of additional spell. Trully mind buggling situation arises, when both survive and neither wants to marry kings daughter...

Perhaps, I misuderstood. The contestants first shoot at each other, then each at himself. Still, some ambiguity remains: 1. Do you have to be within shooting range to cast a spell? That is can you cast at someone without them being aware? 2. Can you sense a not canceled death spell hanging over you? That is, do you know whether or not cancellation spell was effective?

There is an ambiguity in the statement of the riddle.

You preclude some female constituency of the Brain den from participating. Many of them do not want to marry king's beautiful daughter.

4 didgits form 10000 combinations. The way I understand, the conditions boil down to: No more than 3 duplicates of a digit, AND if there are 3 duplicates all 3 must not appear in a row, AND no more than 1 digit duplicated, AND no more than two neighboring digits may appear in ascending/descending sequence. It is not entiely clear whether the "9" from 39 is included in the ascending/descending prohibition, and whether "39" is a part of a double digit duplicate rule (e.g. 390139). You are better off calling directory service.

Your best strategy priorities must be alligned as following: 1. Escape with your life (do not get eaten). 2. Save money. Or rather know in advance how much to bring with you. That is find your average loss if you absolutely have to guess the card. (Alpha does not pay you on odd days, just lets you go, if you win.) - Alphas have strange calendars too. You never know whether it's her even or odd day. Nothing's there to stop Alpha from having few odd, or even days in a row. - That's the strategy, I would adopt. So what's the average loss? This problem is different from picking largest number with unknown distribution. Here you get multiple guesses, until the deck is depleted. Good thoughts like the above need not go inside spoilers. Spoilers are for long solutions with lots of numbers.

Having disposed of this problem so quickly and efficiently, we are now ready to revisit the one posted by Unreality. For as far as I can see, it was left without proper solution it deserves. I have restated and clarified the problem and added some very necessary conditions and justifications.

The problem of finding largest number with unknown distribution is an old widely published classic. But let me restate this problem here, as it may be something more original. So Alpha prepares a deck of 100 cards by writing a different positive number on each. She then presents you with the cards one by one in truly random order. You can reject a card and then it is tossed away, or you can make a guess that the number on the card is the highest. When you have guessed correctly, or when the entire deck is gone, the game is over. If your visit to Alpha falls on an odd day, the rules are as following: For each incorrect guess you pay $1. If you go through the entire deck without guessing the highest number, she will have you for dinner. (Alphas have strange diets, I hear.) On an even day, game has a different character. You pay $20 for each incorrect guess. When you have guessed correctly, Alpha pays $100 and lends you her limo to get home. If you went through the entire deck failing to make the correct guess, she looks the other way and lets you escape unharmed. QUESTION 1: what's the average amount of money you expect to lose on an odd day with the best strategy possible? QUESTION 2: devise the best strategy to maximaize your win on an even day. Calculate the average expected win with the best strategy.

I don't think there is any reason to say one interpretation is more correct than the other. My first instinct was A, but I think B is reasonable as well. Yes, I did mess up the equation for the interpretation B (where you make first random cut on the whole stick, and the second on the remainder). It should be, like you show Int(0,1/2,k/(1-k)dk). Interestingly, for the interpretation A an integral works as well and it evaluates 1/4: If the distance between two cuts is x, then the probability for each individual distance is 2(1-x). Whereas probability for each individual x to built triangle is again x/(1-x). Muliplying the two, we get instanteneous probability 2x. The integral of that with x in the interval from 0 to 1/2 evaluates to 1/4.

Bonanova is a genius! What I thought was a difficult problem and which took me few evenings to solve, he solved so quickly, economically, and with such ease! As I said, knowledge of calculus helps. I derived a different series yielding the same result. It allowed me to find the number of cards to skip without the approximation 1/e. However, my series are not nearly as economical and elegant as Bonanova's equation.

I don't believe so. You can check my example, which shows that 25% estimate is wrong. It also shows that 33% is wrong. Here is my analysis of where your line of reasoning may be wrong.

That's quite a research! How fast do you need to shuffle the cards to conduct 100,000 experiments? So experimental data largely conforms to the formula that I hold secret. The solution to the problem still remains: Show the derivation of a formula that estimates the theoretical probability exactly.

And here is "my opinion of the answer" for the probability to win when skipping first 50 cards out of 100. It's just a bit over 34.90% (which is less than the optimal 1/e). Although, I don't really see the point of that exercise.

Here is a simple proof that 25% is incorrect probability estimate when you skip 1/2 of the deck: I'm not going to show my derivation for the formula to get the correct probability by other than heuristics method. As it may spoil the fun for those who want to solve the problem on their own. I sense, you are under impression that I claimed to have solved the problem without differential equations just to brag. But I had another motive in addition to that. My statement was also meant as hint that the problem may be solved in such a way. You can derive series using combinatorial formulas and/or complex equations with probability fractions.