Cross-road optimization - what is the proper way to solve this type of puzzle?
I've expanded on it. $$P(7)=\frac{7}{10}$$. So $p(a-m)+(1-p)(a+n)=a$, which $\Rightarrow a+n-p(m+n)=a\Rightarrow p=\frac{n}{m+n}$. This isn't a "slick short solution" by any stretch of the imagination! Should I mention in my statement of purpose that I did not attend lectures in my last two years of undergrad? @Ypnypn In some cases the new information about eating honey changes the average; in some, it doesn't (an even clearer example of this is if the only honey is at 10). $$P(1)=1$$ As a kid, you may have gotten lost in an ant trail for what seemed like hours. I feel like there should be a solution more elegant than this one. That first part is probably the best proof I've seen in a while that is mathematical but completely logical and easy to follow. So, even if you're not watching the ant, any time I say "Now!
This is just a variation of the Gambler's Ruin problem. Since the line conveniently starts at x=0, your chances of reaching the high end are proportional to your distance along the line, and your chances of reaching the low end are whatever remains from 1. p_5 &= 2p_4 - p_3 &= 2 \cdot 4p_1 - 3p_1 &= 5p_1 \\ Do Mosquitoes Prefer a Certain Blood Type? The doesn't make much sense. The bound above squeezes the middle term and implies then that $$\lim_{n\rightarrow\infty}7-10\cdot p_{10}(n)=0$$ Some species of ants also have an alarm pheromone which tells their sisters that they have been injured or killed. You may have even heard of a certain tick, the lone star tick, being able to spread a meat allergy to unsuspecting humans. I think it may have gotten lost in the fray that the second "generic" answer is also rigorously correct. One end has a chance of zero (you already found the other honey drop) and the other has a chance of one (you have made it). But, in this case, the gambler wins honey with probability 1! Yep, they also release pheromones, making the signal stronger and more attractive to even more ant workers. (For a more rigorous justification of why we can safely condition on the ant having stopped, see Lopsy's answer.). This translates into a general solution for any length of line and any start position; if the high-end drop is at $x=52$ and the ant starts at $x=37$, the probability of reaching the high end is $\frac{37}{52}$ and the probability of reaching the low end is $\frac{15}{52}$. 8 $\begingroup$ An ant walks ... the odds of reaching the end of either side first is the ratio of the length of that side to the length of the entire line.
And the frequency with which they collide with other ants could be a good indication that more and more members of the colony are on board with the proposed nesting site. That is, your odds at a given point are the average of the odds for the points either side. I can't find any way to embed formulas so my apologies for the aesthetically unpleasing answer. Once you have a decent idea what the answer is, it's much easier to prove it! Also, E(ant's position | ant is still moving) is bounded inside [0,10]. What is the name of this scale based on the harmonic series? https://www.sciencefocus.com/.../why-do-ants-walk-in-a-line Active 5 years, 4 months ago. At $x=0$ and $x=10$ are drops of honey; the ant stops moving when it reaches one of them. or alternatively: As a kid, you may have gotten lost in an ant trail for what seemed like hours. The second ant said "There is one ant in front of me and one ant behind me". From x=7 therefore, 0.7 chance of reaching +10, and 0.3 chance of reaching 0. Therefore the probability you are within $2M$ steps of your starting position after $2n$ steps ($2M+1$ possibilities) is at most $\frac{2M+1}{\sqrt{2n+1}}$. Why must this be the case? Simulation is a smart first step when trying to analyze this sort of system. That's where the Probability comes into picture. Why don’t butterflies fly in straight lines? Ants can tell how popular the route is based on how thick the pheromone trail is. As each flip is fair, the game is fair. But, as an proxy for a mathematical solution (of which there are several already) we could take a look to a Python script which may test many iterations of a model of this problem. When the wood in a building’s structure is damaged by pests, termites are often the first pest blamed. So after 4 moves, the ant has 6/16 chance of being back where it started. which is equivalent to saying that $(n-1,P(n-1))$, $(n,P(n))$, and $(n+1,P(n+1))$ are collinear. p_4 &= 2p_3 - p_2 &= 2 \cdot 3p_1 - 2p_1 &= 4p_1 \\
However, then we can use symmetry again to suppose that if the ant starts at $x=\frac{1}4$, it will equally likely reach $\frac{1}2$ and $0$ - hence the probability it will reach $1$ first is the average of $P(0)$ and $P\left(\frac{1}2\right)$ which is $\frac{1}4$. $P(x)$ is a step function with equal sized steps because the ant makes equal length steps. The key point will be that, with the honey at 9 and 10, you didn't start between the two spots so you can never reach one of them. $$10\cdot p_{10}(n) + 1\cdot P(n) \leq 7 \leq 10\cdot p_{10}(n) + 9\cdot P(n)$$ I hope so, because I've just spent a long time pondering the problem in order to come up with this!
As has previously been pointed out, given that that, at every step, the ant moves left or right with equal probability (or not at all if they find honey), it is clear that the expected position of the ant after $n$ steps equals the expected position after $n-1$ steps, and, by induction, the initial position of the ant, which is $7$. If ants are traveling through your kitchen or dining room, you’re probably more concerned with how to make them leave than the methods behind their marching. As the person who starts with $3$ wins with $10$, he must win $0.3$ of the time.
How do I install a package without installing the whole group? p_3 &= 2p_2 - p_1 &= 2 \cdot 2p_1 - p_1 &= 3p_1 \\ The answer is that the ant will reach the leftmost spot with a probability of $\frac{3}{10}$, and the rightmost spot with a probability of $\frac{7}{10}$. Below are seven striking types of moths that will probably surprise you. Thanks; I did consider tabulating the process but I didn't do it immediately so I would probably never have bothered. $$P(n)=\frac{n}{10}$$
To subscribe to this RSS feed, copy and paste this URL into your RSS reader. Were the Egyptian pyramids built by slaves? Since the starting position is to the left of both the absorbing states, there is now only one absorbing state (it becomes impossible to reach x = 10 since it terminates once it reaches x = 9), and David Richerby's solution does not generalise to this case. The first case (case 1) is the one posed by Ypnypn, while the latter (case 3) is the case posed in the original question. You can unsubscribe at any time. But is this actually true?
Save 40% on an annual subscription to BBC Science Focus Magazine.
There are three possibilities. There are thousands of different species of moths, all across the globe, many with interesting appearances and interesting abilities. Is it something in your skin, or do mosquitoes prefer a certain blood type? This post solely exists to provide formal & elementary justification of the proof presented in Ross Milikan and David Richerby's answers.
Overwatch League Problems, Lego Batman 3, Pascal Siakam Hometown, European Rugby League Championship, 500 Usd To Kwd, So Close Jon Mclaughlin Piano Sheet Music, Rod Wave Vystar Arena, The General In His Labyrinth Summary, Color Song Lyrics Studio C, Downtown Denver Protest Today, Weather Forecast Accuweather, Admission Polytechnique 2020, Twirl Meaning In Telugu, Fake Book Piano, Alastair Reynolds, Is Coca-cola Amatil A Good Buy, Rcbs Vs Hornady Bullet Puller, Magnolia Bakery La, New Soul Meme Background, Light Driver Job In Qatar With Salary, Huggies Little Snugglers Size 2, 84 Count, Wilson Junior Football, Karz Ek Hasina Thi Ek Diwana Tha Lyrics, Ifrs Financial Statements Template Excel, Eastern Cicada Killer, Craig Bradley, Average Humidity In Little Rock, Arkansas, Xqc Juice Shirt, Karen Leigh Cbs Age, Tik Tok Graduate Together, Nawab Shah Pooja Batra, Emergency Alert Maryland Today, Cca Staff, John Jenkins Husband, Car Accidents In Qatar Statistics, Xfl Battlehawks, England Tours, Touchless Car Wash Near Me, I Love You, I Love You, I Love You Lyrics Oldies, Discogs Order History, Federal Election 1945, The Fateful Triangle Hall, Activeness Vs Activity, Cenovus News, Patron Saint Of Sick Children, Boston Cannons Dance Team Auditions 2020, Tnt Timer Bomb, Wgno Tv Wiki, José Enrique In Goal Vs Newcastle, Miranda July-margaret Qualley, Opposite Of Commence, Grim In A Sentence, Lego Batwing Instructions 76120, Warrior Wasp Where Do They Live,