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Annual returns

Every day a trader either makes 50% with probability 0.6 or loses 50% with probability 0.4. What is the probability the trader will be ahead at the end of a year, 160 trading days? Over what number of days does the trader have the maximum probability of making money? (Thanks to Aaron.)

Solution

This is a nice one because it is extremely counterintuitive. At first glance it looks like you are going to make money in the long run, but this is not the case.

Let n be the number of days on which you make 50%. After 160 days your initial wealth will be multiplied by 1.5n 0.5160—n.

So the question can be recast in terms of finding n for which this expression is equal to 1:

The first question then becomes: What is the probability of getting 165 or more 'wins' out of 260 when the probability of a 'win' is 0.6? The answer to this standard probability question is just over 14%.

The average return per day is 1 - exp(0.6ln1.5 + 0.4ln 0.5) = -3.34%.

The probability of the trader making money after one day is 60%. After two days the trader has to win on both days to be ahead, and therefore the probability is 36%. After three days the trader has to win at least two out of three, this has a probability of 64.8%. After four days, he has to win at least three out of four, probability 47.52%. And so on. With an horizon of N days he would have to win at least Nln2/ln3 (or rather the integer greater than this) times. The answer to the second part of the question is therefore three days.

As well as being counterintuitive, this question does give a nice insight into money management and is clearly related to the Kelly criterion. If you see a question like this it is meant to trick you if the expected profit, here 0.6 x 0.5 + 0.4 x (—0.5) = 0.1, is positive with the expected return, here - 3. 34%, negative.

Dice game

You start with no money and play a game in which you throw a dice over and over again. For each throw, if 1 appears you win $1, if 2 appears you win $2, etc. but if 6 appears you lose all your money and the game ends. When is the optimal stopping time and what are your expected winnings? (Thanks to ckc226.)

Solution

Suppose you have won an amount S so far and you have to decide whether to continue. If you roll again you have an expected winnings on the next throw of

So as long as you have less than 15 you would continue. The expected winnings is harder.

You will stop at 15, 16, 17, 18 and 19. You can't get to 20 because that would mean playing when you have 15, and throwing a 5. So we must calculate the probabilities of reaching each of these numbers without throwing a 6. At this point we defer to our good friend Excel. A simple simulation of the optimal strategy yields an expected value for this game of $6.18.

kg of berries

You have 100 kg of berries. Ninety-nine percent of the weight of berries is water. Time passes and some amount of water evaporates, so our berries are now 98% water. What is the weight of berries now?

Do this one in your head. (Thanks to NoDoubts.)

Solution

The unexpected, yet correct, answer is 50 kg. It seems like a tiny amount of water has evaporated so how can the weight have changed that much?

There is clearly 1 kg of solid matter in the berries. If that makes up 2% (100 less 98%) then the total weight must be 50 kg.

Urban planning

There are four towns positioned on the corners of a square. The towns are to be joined by a system of roads such that the total road length is minimized. What is the shape of the road?

(Thanks to quantie.)

Solution

One is tempted to join the towns with a simple crossroad shape but this is not optimal. Pythagoras and some basic calculus will show you that the arrangement shown in the figure is better, with the symmetrically placed crosspiece in the middle of the 'H' shape having length 1 — 1/V3 if the square has unit side. Obviously there are two such solutions.

 
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