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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?
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Quantitative Researcher

Interviewed at Jane Street

4.4★
Dec 30, 2016

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?

On the last interview I got this question: Given cards numbered from 1 to 9 all flipped up and two dies. We throw the dies, if the sum of the numbers is greater than 9, we throw again. If the sum is less or equal than 9 we have the following options: -flip down the card with the sum as a number on it or -flip down two cards, showing each thrown number. So if we throw 3 and 4 we can choose to flip down 7 or both 3 and 4. If for example 7 is already flipped then we must flip 3 or 4. Similarly if one of the cards 3 or 4 is flipped then we must choose the sum. If we can't flip down anything the game ends and our score is the number of flipped cards. What is the best strategy?
avatar

Quantitative Researcher Intern

Interviewed at Jane Street

4.4★
Nov 12, 2018

On the last interview I got this question: Given cards numbered from 1 to 9 all flipped up and two dies. We throw the dies, if the sum of the numbers is greater than 9, we throw again. If the sum is less or equal than 9 we have the following options: -flip down the card with the sum as a number on it or -flip down two cards, showing each thrown number. So if we throw 3 and 4 we can choose to flip down 7 or both 3 and 4. If for example 7 is already flipped then we must flip 3 or 4. Similarly if one of the cards 3 or 4 is flipped then we must choose the sum. If we can't flip down anything the game ends and our score is the number of flipped cards. What is the best strategy?

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