Prime

I do not see any way to solve the stick problem without differential equations. So here it goes: So the probability to cut the stick is about 35%. A very economical coin problem solution has been already provided by Yoruichi-san. I would like to stress one point implicit therein. Some may be under impression that with equal number of coin tosses, probability for A to have more heads than B is 1/2 and vice versa. It is not, as evident from Y’s solution.

I deny any responsibility.

First of all, skipping 1/2 cards does not mean 25% success rate as Prof. T suggested. That was an incorrect guess in both respects (number of cards to skip, and the resulting probability.) Second, I posted this problem here for those who want to give a shot solving it. I did few years ago, and I did not use number e, or any logarithmic integrals for that. The web reference you are citing gives an answer to a general problem, but not the solution. It, in turn, cites a theorem equating the series to an integral, but does not show the proof. Original problems, especially math, are very few on this forum, if any. You can always find a precedence in math literature and on the internet. But what's the point of looking up the solution? To me a process of solving problem is rewarding. What you cite inside the spoiler is a correct general answer, but not a solution. So I take your answer as an opinion that this problem is too difficult for this forum and I should not have posted it here. As I already stated, the problem is complex. However, for the actual answer, the number of cards to skip is an integer and the probability is a rational number. And you don't need to use any differential equations, or natural logarithm for the solution. <_< P.S. Armcie actually cited a correct approach earlier in this topic.

I also checked the "braingle" website referred to in your post. It has a similar problem, which was posted there 4 years ago with a wrong answer (50, 25%). So the topic lives on.

The problem at Unreality's post would be the same if the deck there was re-written with a whole new set of numbers after each game, not just reshuffled.

I checked your post again and did not find any correct answers there, or even hint thereof. Not even by Chuck Rampart, who simply made a reference to an incorrect solution he saw at another website. There is much to discuss with respect to this problem for those who are into solving difficult math problems. If there are indeed such members (or will show up in future) who find this problem interesting, it's best to discuss it on my post. My formulation of the problem is more precise and I myself solved it few years back and hope I am still able to repeat the feat. Also, I happen to know the correct answer and can guide someone who shows some progress in solving this problem. I'll repeat, we are looking for the answers in the form of how many cards to skip and what is the exact probability to select the highest number. I can tell you off the bet, 50 cards and 25% are not the answers. I appologize for not finding your post before I started my topic here. I made rather extensive search for different types of arrangements in which this problem appeared in popular literature, but missed your post somehow. The arrangement in which I gave here is my own.

In view of the above, I have to take back what I said in my previous post about "not even starting to move in the right direction". Sorry, I wasn't paying close attention.

The supplement question in your post "what if the deck was shuffled between each game" makes it basically the same problem. Whereas, continuing the game with the same deck after an incorrect guess is a different problem. I browsed quickly through your topic and didn't see anyone even beginning to move in the direction of the solution. If people can solve this problem, they can move on to the one you posted, where you get more than one attempt to select the highest number. This problem is complex. A strategy has been suggested by Prof. T to skip a number of cards (he suggested 50) remembering the highest number among the skipped cards, and then pick the first number that is higher if that happens. I'm going to ease down on a formal proof that such method presents the best strategy. Assume that it is the strategy. The questions become: 1. How many cards do you want to skip up front? (Seeing that 50 is just an unsubstantiated guess and there is no evidence to support it.) 2. Calculate exact probability to select the highest number with the best strategy. I haven't seen any posts here (or in that related topic) even starting to move in the right direction to accomplish that task. This problem has an elegant answer in terms of a limit for such probability in general (for any number of cards.) However, be forwarned: it took me few evenings to solve. And in the end I had to feed my formulas into a spreasheet and arrive to the solution numerically. (Too much work just using pen and paper.) Knowledge of calculus could help in solving this problem. I see it as a dragon of probability problems and I hope I managed to scare many a brave knight venturing to attack it.

And there may be another flaw with that approach. The 50% probability for an event to occur and the average number of trials for an event to occur are not the same thing. For example, it takes on average 6 rolls to roll number "6". However, "6" comes up with a 50% probability in less than 4 rolls. See my preceding discussion with Bonanova. (His winning an actual gambling match notwithstanding.)

Infinite series may add up to a finite number (converge). See my topic Dice Gamblers' course for discussion and relevant math.

That formula is incorrect. Formula for the number of times at least one number is not rolled is more complex and yields smaller number of combinations.

I was typing my posts while watching TV and got a bit slow. I missed few of your posts. Yoruichi-san gets full marks for the solution! That's an honest mathematical proof. I'll re-write with comments and organize it a little here:

So how do you arrive to 1/p? I believe, it is basic math. No books/reference are needed. If you need formula for Geometric series, I can provide. Although, I never remember it myself and have to derive it every time I need it.

Roll the die repeatedly until "6" comes up. Count the number of rolls. I hypothesized, on average, you'd roll the die 6 times. Sometimes, "6" comes up on the first roll, other time it may come up only after 15 rolls, but the average is going to be 6 rolls.

That's basically the proof. However, a bit more work is required to wrap it up. The series is not a geometric progression. (There is no common ratio between the members of the sequence.) How do you calculate the sum E(n*(5/6)(n-1)), where n tends to infinity? With a little effort such math problems may be solved without using any books/reference. That in itself could be rewarding.

That sounds very appetizing. I disagree, however, with posting answers to own problems. Different people come upon them (puzzles) at different times. And what's wrong with leaving it unanswered, anyway?

With recent surge of probability problems, I think this one is due. Prove (mathematically) that it takes on average six rolls of a single die to roll "6". This math may be useful in solving probability problems like the recent one with dice.

Good effort on devising a method! I disagree with your calculations, though. Especially, the 25%.

The king of your country decided to attend to the state’s budget matters. For that purpose, he wants to first conduct some layoffs of his court staff, and second, beef up the military funds by borrowing money from foreign investors. He calls upon you as one of his top mathematical advisors to provide your learned opinion on the borrowing matter in the following manner: In his study hall, there is a box containing 100 cards, each with a number representing the dollar amount to be borrowed from foreign investors. The cards are drawn from the box at random one by one and presented to you for an opinion. You can reject a card with a number on it, then it is tossed away and the next card is drawn. If you accept the card, then the number on it is taken to be your advice and the session stops. You know for a fact that unless you select the card with the largest number on it, your employment will be terminated. What strategy can you adopt to maximize your chance to select the largest number? With the best strategy, what is the probability to retain your position as a top mathematical advisor?

The bet is simply: if you roll dice 4 times in a row and "6" didn't come up -- you win. If "6" came up -- I win. (No need to complete 4 rolls, when "6" comes up). The probability not to roll "6" in 4 consecutive rolls is simply (5/6)4, which is less than 1/2. (Even though it takes 6 rolls on average to roll "6").

I was referring to proving the above statements. Anyway, we could make a separate post of that as a basic (not necessarily simple) probability problem. It could be formulated something like: Prove that it takes on average 6 rolls of a die to roll "6". (I see it as considerably easier problem than the one given here.) I'm about to post, what I think is rather difficult probability problem. If you haven't encountered it before, you'll enjoy it.

Bona, You and CR rely on theorems of probability, which I don't see as a common knowledge on this forum. Case in point: average trials to achieve an event of probability P. If you ask how many rolls of die on average does it take to roll "six" (or any other individual number), most people will answer -- 6. But can you prove it? (That's where geometric series comes in.) With respect to this problem, I stand my ground: the basic formula for the average is infinite series: 6*P6 + 7*P7 + 8*P8 + … Where Pn is the probability to have all six numbers to show up exactly after n rolls. This formula relies only on the basic definition of average. It must add to 14.7 in this case. However, each individual Pn is hard to estimate.

I am going to pick an issue here. I presume, you mean an even money bet. But the average number of trials for an event to occur and 50/50 chance are not the same thing. In a fair coin toss, you need on average two tosses for it landing heads up. And yet a single toss has probability 1/2 for heads. We know that it takes on average 6 rolls of a fair die to roll 6. I'll gladly take an even money bet on 6 showing up within just 5 consecutive rolls, or even just four rolls.

Now, I begin to feel insecure. The OP labels this problem "simple" and Chuck Rampart gives a formula for solving without giving any rational. Is the problem really so trivial and obvious? Is the solution so evident, it requires no proof? Although 14.7 seems to be the solution, I feel an urgency to obfuscate the issue.