注册一亩三分地论坛,查看更多干货!
您需要 登录 才可以下载或查看附件。没有帐号?注册账号 
x
这套帖子里,整理了一些从Glassdoor上收集来的题目,并且附上了解答或者讨论区
【求米呜呜】
1
Here are 10 castles, numbered 1, 2, 3, ... , 10, and worth 1, 2, 3, ... , 10 points respectively. You have 100 soldiers, which you can allocate between the castles however you wish. Your opponent also (independently) does the same. The number of soldiers on each castle is then compared, and for each castle, whoever has the most soldiers on that castle wins its points. Additionally, you lose 0.2 points for each "extra" soldier that you have, in excess of your opponent's army size, at each castle that you win. You still need to deploy all 100 soldiers. In the case of a tie, no one gets points for that castle.
Open question, no discuss in Glassdoor
2
Four 50-sided dice are rolled so that the numbers are all different, then assigned randomly to players A,B,C,D. The players with the highest/lowest numbers pair up, as do the two middle ones. Then the team with the higher total pays the team with the lower total the difference.
You are one of the players. Someone scoundrel approaches you with a hack to fix the game so that you start with a given number 1-50.
Question 1: What is (one of?) the best number(s) to pick?
Question 2: Say you answer X to Question 1. The scoundrel hosts an auction to sell the hack that lets you start with X. How do you bid? (As far as I can tell this is just asking for the expected value of the hack? Maybe they wanted a subtlety about bidding strategies?)
Open question, no discuss in Glassdoor
3
a) I have toss 10 fair coins. What is the probability of an even number of heads.
b) Now 100 fair coins.
c) Now X unfair coins and Y fair coins
The fact is that if you have one fair coin, the game will be fair.
We don’t need to calculate combination numbers, only think about we put the fair coin on the last position. Before the last throw, the result of even and odd may be different, but in the last throw, the probability of head and tail is 1/2. So the final probability of getting even is (P(even before last throw)+P(odd before last throw))/2=1/2
4
I have a 3x3 Grid and in each cell I place a lightbulb. Each of the 9 lightbulbs is then turned on with probability 1/2. What is the probability that there are NO adjacent cells both lit up.
The key point is the lightbulb in thre given a choice between two games, each with a $1 payout, which do you prefer?
Game 1: you are given 4 roles of a single die, and you win if you roll at least one six.
Game 2: you are given 24 roles of a pair of dice, and you win if you roll a pair of sixes at least once.
18
200 pieces of 1 dollar coins, for each one it has equal probability to go into the pot or not. You will bid for that pot (get the money in the pot but you don't know how many coins are exactly in the pot). Anyone who offers the highest bid win the auction. What would you bid (with 1 competitor, 10 competitors)? Now if just we two bid and we are trying our best strategy, but I have the advantage of knowing how many of the first 10 coins go into the pot. What will be our strategies? How much will you bid and what is your expected payoff?
Open question, no discuss in Glassdoor
19
You are given a bag with tiles from 0 to 9 and a decimal point tile. The tiles will be pulled from the bag one at a time and placed on the table. The resulting number will be your prize. Compute the value of this prize.
20
a dice roll with a 100-faces dice, labeled from 1 to 100.
1. You get to roll once and receive the amount of dollars labeled on the face, how much would you like to pay for this roll.
2. How much would you pay if you can roll the dice twice if you are unsatisfied with the first outcome?
3. You can roll the dice infinite times, and costs 1 dollar for each roll except the first one. What is your strategy?
|