## Maths Challenge 9

Posted by Chris on July 21, 2011 – 3:45 pm

An investor puts $10,000 into a weekly trading scheme.

In each week it is equally likely that he’ll make an 80% gain or a 60% loss (there are no other possibilities). One year later (52 weeks), roughly how much money does the investor expect to have? The answer is one of the following:

a) $1.95, b) $14,000, c) $140,000, d) $1.4 million or e) $131 million.

July 21st, 2011 at 7:55 pm

d

July 21st, 2011 at 11:38 pm

D.

[sum( C(52,n) * 1.8^n * 0.4^(52-n)) from 0 to 52] / 2^52 * 10000 = 1420429.32

July 22nd, 2011 at 1:27 am

I also got 1420429.32 looking at the change of the average each week (+10%)

July 22nd, 2011 at 2:36 am

The answer is (e)

July 22nd, 2011 at 2:44 am

10000*((1.8+0.6)/2)^52=131 million

July 22nd, 2011 at 3:36 am

lol replace 0.6 by 0.4, and the answer is indeed d

July 22nd, 2011 at 3:54 am

That’ll learn me, I should have left the question in it’s vaguer form. The last time this was posted, most people decided that you’r going to gain half the time and lose half the time, so they calculated 10000 * 1.8

^{26}* 0.4^{26}= 1.95(3). I assume that that is the most probable outcome.Bekki solved the problem in the way I expected. I was surprised at Jan’s calculation, it is very slick.

Well done everyone.

July 22nd, 2011 at 4:04 am

It was going so well for me. Week 51 I had over a million dollars in the bank. Then the economic crash….

Final answer – $1.95

July 22nd, 2011 at 4:19 am

LOL, thanks Karl. I was gutted that nobody had given that answer

July 22nd, 2011 at 5:10 am

I can see mathematical ways of getting most answers – by transposing figures as per Jan’s post 5 and just plain bad maths – but can’t get as low as $1.95!

Multiple choice answers can be intriguing if the answers are not blatantly/obviously wrong. In this example, all answers were plausible on cursory examination, except for the equally likely chance of +80% or -60%.

Good question, but shouldn’t have thrown in the multiple choice! Make us work for it!!

July 22nd, 2011 at 5:14 am

Obviously thinking a little bit more about the question – what about 3 or 4 weekly outcomes, or sequential outcomes. Set the question again in a year or two when we have forgotten Jan’s slick formula!

July 22nd, 2011 at 6:29 am

Hi Karl. I showed how to get the $1.95 in post 7. The idea is that on average you will gain half the time and lose half the time. So on average you gain 26 times and lose 26 times. Because the final amount is simply the product of all those gains and losses, it dosn’t matter what order they come in, so we get 10000*(1.8*0.6)

^{26}≈ 1.95Jan’s formula is equivalent to Bekki’s, courtesy of the binomial theorem =>

(a+b)

^{n}= Sum(i=0 to n, C(n,i) a^{i}b^{n-i})July 22nd, 2011 at 9:51 am

10000 * ( (0.4+1.8)/2 )^52 = 1420400

D)

July 22nd, 2011 at 3:49 pm

28

July 24th, 2011 at 12:36 am

I don’t see how the answer can possibly be D.

Assume he gains and loses 50/50, he will have $1.95. We cannot therefore say he should “expect” to have $1.4 million.

The formula used to reach the $1.4 million is simply taking the average of all possible outcomes, which is hardly reasonable in this circumstance. With many losses, his money nears $0. With all wins, it reaches $188005374836229120. With many possibilities approaching zero, and some heading towards infinity, the average will rise greatly. But since the odds for his money to decrease are far greater, he should expect to lose money.

For example, look at the possibilities after two weeks.

He has once chance of having $1,600, one chance of having $32,400, and two chances of having $7,200. Taking the average of those (the same formula used to obtain answer D), we get $12,100. He has only one chance of hitting it big, and three chances of ending with significantly less than the average (and less than what he started with). Nobody in their right mind would say that he could expect to gain $2,100 when he has a 3 in 4 chance of losing money.

Essentially, it is a gambling situation; he should expect to lose money (over 50% of the possible results), but has a chance to end with much more than he started.

July 24th, 2011 at 1:02 am

Erik – not sure I follow all your maths for the 2nd week scenario:

(10k + 80%)= £18k – 60% = 10.8k

(10k – 60%) = £6k + 80% = 10.8k

(10k + 80%)= £18k + 80% = 32.4k

(10k – 60%) = £6k – 60% = 3.6k

3/4 of the time he will be ahead..?

July 24th, 2011 at 4:18 am

I Tested results in a spreadsheet over 10 weeks with 4 different scenarios.

Scenario 1: Gains 80% for first week, thereafter alternates between loss and gain for next 9 weeks.

Result: 100000 investment ends up at 98599.

Scenario 2: Loses 60% for first week, thereafter alternates between gain and loss for next 9 weeks.

Result: 100000 investment ends up at 168958

Scenario 3: Gains 80% for first 5 weeks, and loses 60% for next 5 weeks.

Result: 100000 investment ends up at 45144

Scenario 4: Loses 60% for first 5 weeks, and gains 80% for next 5 weeks.

Result: 100000 investment ends up at 664048

It appears that Scenarios 3 & 4, being the most unlikely, give the extremes of gain and loss over the period, so projecting to 52 weeks, I would expect the 520000 investment to end up at somewhere between 512000 and 878000

I must admit I don’t feel very confident in my reasoning, so please shoot me down!

July 24th, 2011 at 4:41 am

Hi Erik. Expected value is the same as average value, see: http://en.wikipedia.org/wiki/Expected_value for more details. It gives me a small entertainment value by using “expect” (and it’s variants), because I know that many people don’t expect expect to mean what it does.

As you note, after two weeks, on average, he would end up with 10000(0.4*0.4 + 0.4*1.8 + 1.8*04 + 1.8*1.8))/4 = $12 100 (each of the four possibilities are equally likely). That is undeniably an average gain and is quite real. Extending this to 52 weeks gives you the 1.4 million. Although he is more likely to lose money in a given year, on average the [rare] big gains far outweigh the more common not so big losses – you can’t possibly lose more than $10 000, but you can dream of getting up to $188 005 374 836 229 120.89… That’s $188 American quadrillion.

July 24th, 2011 at 5:01 am

Hi Rob. You need to use Excel’s RAND() function to simulate the situation. You are unintentionally biasing the results. You should ideally use the entire 52 weeks and run the simulation many times (perhaps millions of times) to get the results. Despite that, your forecast is pretty good.

I don’t know how to do this in the spreadsheet itself. But I could easily do it using the Visual Basic capabilities of Excel.

July 24th, 2011 at 6:51 am

I’ve now simulated this problem. Here are some results based on runs of one million trials each. I’ve chopped the decimal parts off to save space. Each figure represent the average final amount over the year per fund.

635089, 919899, 918043, 656402, 993140, 675399, 701389, 579761, 737826,

833779, 795262, 637020, 717030, 573761, 746535, 618927, 748814, 573310,

560466, 595430, 577154, 608198, 839945, 802373, 577524

I have seen close to $1 000 000 while testing the code. In practise you need to be investing in about 1000 such funds to be pretty safe.

You can see that the results are pretty erratic. This is quite a common phenomenon for these types of situations. I assume that the standard deviation of the distribution is large.

Here’s the code (plain text): http://trickofmind.com/wp-content/uploads/2011/07/investment.txt

July 24th, 2011 at 9:46 am

I’ve added a max amount catcher to my copy of the code. It nearly always comes up with 55 213 480 872.831. So I expect that the random number generator isn’t good enough to simulate this problem sufficiently well.

I’ve just found that it repeats after 16,777,216 calls – that’s definitely not good enough

July 29th, 2011 at 4:37 pm

think e but im 10 my stepfather helped me so 8*52

July 29th, 2011 at 9:35 pm

i say c.