Modified Compound Interest Question

Neil3030

Hi guys. I was wondering if somebody could help me in devising a modified compound interest question that factors in losses at a certain rate. So for example, if an investment/gambling strategy was to work out:

80% of the time - return is .4 units
20% of the time - loss is 1 unit

Now I know the formula to use to calculate the long run yield is:

(.8*.4) - (.2*1) = 0.12

and I know that the compound interest formula is:

Future Value = Present Value*(1+rate)to the power of period.

i.e., starting at 1 unit, after 10 successful returns at 12% you'd have:

1*(1+0.12)to the power of 10 = 3.11

But this is obviously not considering the probability of failure.

How do I factor the above failure probabilities, and their consequential impact on the accruing value, in to the above equation(s)?

gerry87

Neil3030 wrote: »

Hi guys. I was wondering if somebody could help me in devising a modified compound interest question that factors in losses at a certain rate. So for example, if an investment/gambling strategy was to work out:

80% of the time - return is .4 units
20% of the time - loss is 1 unit

Now I know the formula to use to calculate the long run yield is:

(.8*.4) - (.2*1) = 0.12

and I know that the compound interest formula is:

Future Value = Present Value*(1+rate)to the power of period.

i.e., starting at 1 unit, after 10 successful returns at 12% you'd have:

1*(1+0.12)to the power of 10 = 3.11

But this is obviously not considering the probability of failure.

How do I factor the above failure probabilities, and their consequential impact on the accruing value, in to the above equation(s)?

By return of .4 and loss of 1 do you mean increase of 40% and loss of 100% or absolute values.

For example say the amount is 100, does an increase bring it up to 140 or 100.4?

If it's 100 => 140, then you can only end up with money at the end if you have a winning streak because with any loss you lose 100%. Then it's the amount you'd get at the end of the winning streak times the probability of getting that far.

1*(1.4^10)*(.8^10) => 1*(1.4*.8)^10 = 3.11

If it's 100 => 100.4, then it's a different problem since the % increase/decrease changes each time.

Then it's just the expected return times the number of trials.

[(.8)*(.4) + (.2)*(-1)]*10= 1.20

Neil3030

Yeah it's the former situation where a 100 stake would return 40 if successful and -100 if unsuccessful.

I see what you're saying about a streak being neccessary.

What if I were to start with a kitty of, say, 200 and always invest 15% of the total kitty, to cover losses?

gerry87

Neil3030 wrote: »

Yeah it's the former situation where a 100 stake would return 40 if successful and -100 if unsuccessful.

I see what you're saying about a streak being neccessary.

What if I were to start with a kitty of, say, 200 and always invest 15% of the total kitty, to cover losses?

Then your expected return would be

(.8*(.85)*(1.4) + .2*(.15)) -1 = -.018%

How much you'd have after x successful tries and your expected value are two different questions.

Neil3030

I'm not sure the return is -.018%, I did a mockup of 100 trials in excel, with 20 randomly designated failures and 80 randomly designated successes.

Using a system that used 15% of the total kitty for all trials and where a success brought about a 40% return and a failure brought about a 100% loss (of the trial amount, i.e., 15% of the total bank used in the relevant trial) I went from 100 to 482.42 in the 100 trials.

In this case, the return was
482.42 = 100*(1+x)^100
4.8242 = (1+x)^100
1.015860919143732 = 1+x
x = .016

I then ran 1000 trials using the same proportional criteria and went from 100 to 158146079.1, suggesting a return of:
158146079.1 = 100*(1+x)^1000
1581460.791 = (1+x)^1000
1.0143762174966315 = (1+x)
x = .014

So it looks like the return is around the 1.4% mark. Is there a formula I can use that would plug in the probability of win (80%), the probability of fail (20%) the return on win (40% of stake) the loss of fail (-100% of stake) and spit out that long run return figure? This would save me having to run simulations all the time.

FoxT

Op, the way this is worded is unclear, so I am going to assume the following:

You take 1 euro & put it into a machine that will gamble n times. Accumulated wins are rolled over into the next gamble. if you lose at any stage, the game will halt & you win nothing.

Probability of winning = (0.8)^n

the value of each win is 1.4.

Since the winnings get 'reinvested' each time, after n iterations your winnings are (1.4)^n.


The gain after n iterations then is


(Probability of winning * winnings) - initial investment

ie (0.8)^n * (1.4)^n - 1.


This looks attractive as an investment but beware! Odds of winning are low in the long run, whilst the gain if you win is high. It is a bit like buying a lotto ticket. Suppose you have a 1 in a million chance of winning, and the prize is 5 million euro. Then theoretically, a ticket is worth 5 euro. If the tickets sell at 1 euro, you could argue that your ROI is 400%, which is rubbish of course.

Have done a quick Monte carlo on Excel here, the above looks to be in the ballpark.

gerry87

Neil3030 wrote: »

I'm not sure the return is -.018%, I did a mockup of 100 trials in excel, with 20 randomly designated failures and 80 randomly designated successes.

Using a system that used 15% of the total kitty for all trials and where a success brought about a 40% return and a failure brought about a 100% loss (of the trial amount, i.e., 15% of the total bank used in the relevant trial) I went from 100 to 482.42 in the 100 trials.

In this case, the return was
482.42 = 100*(1+x)^100
4.8242 = (1+x)^100
1.015860919143732 = 1+x
x = .016

I then ran 1000 trials using the same proportional criteria and went from 100 to 158146079.1, suggesting a return of:
158146079.1 = 100*(1+x)^1000
1581460.791 = (1+x)^1000
1.0143762174966315 = (1+x)
x = .014

So it looks like the return is around the 1.4% mark. Is there a formula I can use that would plug in the probability of win (80%), the probability of fail (20%) the return on win (40% of stake) the loss of fail (-100% of stake) and spit out that long run return figure? This would save me having to run simulations all the time.

I could well be wrong, I attached a worked example. If you want to stick up your spreadsheet.

Just to make sure we're on the same page...

You start with 200, if you win you win 238 (200*.85*1.4) plus the 30 you didn't gamble, if you lose you have 30.

If you lose, then lose again, you have 200*.15*.15, or 6.

If you lose first, you start with 200, then go down to 30, then if you go again and win, you win (30*.85*1.4) plus the 6 you didn't gamble, so 40.2.

Are those right?

Take a look at the spreadsheet and see if it looks right.

With that method I got a return of 10.2%, formula would be

(.8*(1+.85*.4) + .2*.15)-1 = .102

Neil3030

I'm not sure I'm following your spreadsheet, but here's what I did:

In the attached spreadsheet, rows represent iterations.

Column A is your opening bank for each iteration.

Column B is each iteration's stake, i.e., 15% of opening bank.

Column C is the return if successful, i.e., 40% of the value in Column B (stake).

Column D is whether the iteration was successful or a failure, 0 = success, 1 = failure.

The logic is as follows: for all cells A(n) (except A(1) as this is fixed at 100) -

For wins (0) -> if D(n-1) = 0 -> A(n-1)+C(n-1) (i.e., the previous opening bank + the successful return from the previous iteration)

For losses (1) -> if D(n-1) = 1 -> A(n-1)-B(n-1) (i.e., the previous opening bank - the losing stake from the previous iteration)

In the attached spreadsheet there are 1,000 rows, 800 wins and 200 losses, randomised using rand().

The bank grew from 100 to 158146079.1, or at a compound rate of 1.4%

gerry87

Oh ok, i thought you were holding 15% back, not betting the 15%. In that case I'm getting:

(.8*(1+.15*.4) + .2*.85)-1 = .018 or 1.8%

That look right?

Neil3030

gerry87 wrote: »

Oh ok, i thought you were holding 15% back, not betting the 15%. In that case I'm getting:

(.8*(1+.15*.4) + .2*.85)-1 = .018 or 1.8%

That look right?

Yeah it looks pretty good. Is there a reason it seems to predict a higher return than the simulation?

I ran another simulation, using 20% bank each iteration and got a 1.7% return - your formula predicts 2.4%. (0.8*(1+0.2*0.4)+0.2*0.8)-1

I then ran another simulation, using 10% bank stakes, upping the win percentage to 90% and reducing the return per win to 30%. The simulation said 1.6%, while your formula suggests it would be 1.7%.

Have I the formula right in both instances?

I'm assuming it to be:
(a(1+bc)+(1-a)(1-b))-1
where:
a = probability of win
b = stake, in terms of proportion of bank staked per iteration
c = winnings for win, in terms of proportion of stake

FoxT

file enclosed (I hope). Betting 15% of your total capital each time is a winner compared to the 1st model, because you can survive a string of losses, and your probability of winning * benefit of a win is high.

I cant get a return of 1.8% out of it though - but I am looking for the 1000th root toward the end, which is bound to be out a little.

Also, even after 1000 iterations, the gain fluctuates wildly, between say 1.0 * 1.3%

This is a much more complex one to analyse than the problem as first stated

I had a hunch this process would be much more stable, and the gain (ie the equivalent rate of compound interest) would stabilise after a few hundred iterations. So my hunch was wrong or the sheet is wrong!

In any event, it looks like a sure fire winner. DO you know a millionaire who'd like to play it with me for a few hundred rounds?

gerry87

Hmm, tricky one alright.

You're both getting different values to me, so chances are I'm wrong, but maybe someone can figure out what's wrong with this logic.

1. You start with 200, bet 15% (30).
2. If you win, then you win .4 times your bet, or 12. Your total amount is 212.
3. You win 80% of the time.
4. If you lose, then you lose your whole bet, or 30. Your total amount is 170.
5. You lose 20% of the time.
6. Your expected value for the bet is .8*(12) + .2*(-30) = 3.6
7. Your expected return is 3.6/200 = 1.8%

Neil3030

gerry87 wrote: »

Hmm, tricky one alright.

You're both getting different values to me, so chances are I'm wrong, but maybe someone can figure out what's wrong with this logic.

1. You start with 200, bet 15% (30).
2. If you win, then you win .4 times your bet, or 12. Your total amount is 212.
3. You win 80% of the time.
4. If you lose, then you lose your whole bet, or 30. Your total amount is 170.
5. You lose 20% of the time.
6. Your expected value for the bet is .8*(12) + .2*(-30) = 3.6
7. Your expected return is 3.6/200 = 1.8%

Yeah that seems sound, even leaving the 200 bank out of the equation (.8*.4*.15) - (.2*.15) = 1.8%

Maybe it's an upper limit or asymptotic value that it is defining?

I guess chance variation would account for 3-4 simulations all falling below the formula's value?

gerry87

Neil3030 wrote: »

Yeah that seems sound, even leaving the 200 bank out of the equation (.8*.4*.15) - (.2*.15) = 1.8%

Maybe it's an upper limit or asymptotic value that it is defining?

I guess chance variation would account for 3-4 simulations all falling below the formula's value?

Well it's just the expected value, there'll be a distribution of actual values around that. So each simulation of 1000 trials would be unlikely to actually be 1.8%. If you ran a good few simulations and averaged the return you should get close enough to 1.8%.

That said, you guys seem to be systematically lower than 1.8%, so either my way is wrong or there's some subtle problem in the simulations. I'll try to do a monte carlo later and see what happens.

Neil3030

FoxT wrote: »

file enclosed (I hope). Betting 15% of your total capital each time is a winner compared to the 1st model, because you can survive a string of losses, and your probability of winning * benefit of a win is high.

I cant get a return of 1.8% out of it though - but I am looking for the 1000th root toward the end, which is bound to be out a little.

Also, even after 1000 iterations, the gain fluctuates wildly, between say 1.0 * 1.3%

This is a much more complex one to analyse than the problem as first stated

I had a hunch this process would be much more stable, and the gain (ie the equivalent rate of compound interest) would stabilise after a few hundred iterations. So my hunch was wrong or the sheet is wrong!

In any event, it looks like a sure fire winner. DO you know a millionaire who'd like to play it with me for a few hundred rounds?

Haha, not a person as such, but a market of people contributing to a trend

gerry87

Ok, so I ran a monte carlo with 10,000 trials and 10,000 simulations... 1.4%

Distribution attached, 1.8% is far in the upper tail of the distribution, so safe enough to say the formula's wrong.

Neil3030

gerry87 wrote: »

Ok, so I ran a monte carlo with 10,000 trials and 10,000 simulations... 1.4%

Distribution attached, 1.8% is far in the upper tail of the distribution, so safe enough to say the formula's wrong.

Yeah sure looks that way. Can't for the life of me figure out why though...

On a side note, what package do you use to run Monte Carlos? Don't suppose you could do them in SPSS or Minitab?

(Googling "Monte Carlo + Package" would probably put one on a suspect list for Interpol's tax-division.)

gerry87

Neil3030 wrote: »

Yeah sure looks that way. Can't for the life of me figure out why though...

On a side note, what package do you use to run Monte Carlos? Don't suppose you could do them in SPSS or Minitab?

(Googling "Monte Carlo + Package" would probably put one on a suspect list for Interpol's tax-division.)

I did it in matlab. I'm not sure if you can do it in SPSS or Minitab, i don't know how anyway.

You could do it in Excel through VBA, or if you just took your spreadsheet and compressed the formulas into one column, then just dragged it across a few thousand columns. Other than that probably mathematica or just any programming language.

hivizman

This problem can be looked at as an example of the Binomial Pricing Model, which is used for option pricing (it's considered simpler to use for teaching purposes, and it converges to the well-known Black-Scholes Option Pricing Model).

There are two possible states of the world (or payoffs).

1. With probability 0.8, we receive a return equal to 0.4 times the amount staked.
2. With probability 0.2, we lose the amount staked.

Suppose we start at time zero with capital 1, and we always stake 0.15 times the amount of our capital. Then our capital at time 1, after one round of the bet, will be:

In state 1: 1 + 0.15*0.4 = 1.06 (probability 0.8)
In state 2: 1 - 0.15 = 0.85 (probability 0.2)

Our expected capital at time 1 is 0.8*1.06+0.2*0.85 = 1.018.

Hence our period return is (1.018-1)/1 = 0.018 (or 1.8%).

Let's now have another round.

If we were in state 1 at time 1, then we will bet 0.15*1.06 = 0.159. We have two possible outcomes:

State 1,1 (win, win): 1.06 + 0.159*0.4 = 1.1236
State 1,2 (win, lose): 1.06 - 0.159 = 0.901

If we were in state 2 at time 1, then we will bet 0.15*0.85 = 0.1275. Our two possible outcomes are:

State 2,1 (lose, win): 0.85 + 0.1275*0.4 = 0.901
State 2,2 (lose, lose): 0.85 - 0.1275 = 0.7225

Note how the capital in State 1,2 is the same as the capital in State 2,1 (this is why this is called binomial pricing, as the probabilities of the various possible payoffs after various periods follow a binomial distribution).

We have three possible capital amounts at time 2:

1.1236 (Probability 0.8*0.8 = 0.64)
0.901 (Probability 0.8*0.2 + 0.2*0.8 = 0.32)
0.7225 (Probability 0.2*0.2 = 0.04)

The expected capital (weighting by probabilities) at time 2 is 1.036324. This is equal to 1.018^2, so the expected annualised rate of return is 0.018 (or 1.8%).

So far, so what? Well, this is where I think the problem with the distribution coming from the Monte Carlo simulation lies. The simulation actually calculates the value of capital at the end of however many rounds, compares this to the opening capital, and works out an annual rate of return. However, the weighted average rate of return from this process is not the expected annual return as calculated above. To show this, let's work out the annualised rates of return for the three possible capital amounts at time 2.

First, we have the win, win outcome of 1.1236. This is equivalent to an annual return of 0.06 (6% - note that 1.06^2 = 1.1236). The probability of getting this return is 0.64

Second, we have the win, lose outcome, which is the same as the lose, win outcome, of 0.901. The square root of this is 0.94921, implying an annual rate of return of -0.05079. The probability of getting this return is 0.32.

Finally, we have the lose, lose outcome, of 0.7225. The square root of this is 0.85, implying an annual rate of return of -0.15. The probability of getting this return is 0.04.

So the average annual return is:

0.06*0.64 - 0.05079*0.32 - 0.15*0.04 = 0.016472 (or 1.6472%).

If such a bias creeps in after only two rounds, it no doubt persists after many rounds (for three rounds, the weighted average annual return taken across the four possible amounts of capital after three rounds is about 1.55%).

The distribution of annual returns calculated from the Monte Carlo simulation almost certainly involves the calculation of an annual return for each trial rather than calculating an average return based on the average capital accrued after each trial, and is therefore biased downwards. Hence gerry87's distribution, with a mean of around 1.4 to 1.5%, is not inconsistent with the agreed annual return of 1.8%.

FoxT

That is a great explanation, Hivizman. I noticed a couple of things that really fit well with what you say above:

1 - multiple monte carlos to 1000 periods gave widely varying results
2 - the actual number of losses (p(loss)=0.2, using excel Rand() function) varied in each simulation run between about 180 & 220. This of course had a dramatic effect on the outcome.
3 - Another factor what whether the losses were distributed evenly or came all together - a string of say 10 losses in a row is pretty rare but possible & had an effect on the final result.

I did a bit of googling but couldnt find similar math problems that fit this one..just read the wikipedia page on binomial pricing , v. interesting, thanks!

Fremen

Another thing to note: if you bet all your money every time, you will go broke at your first loss. If you play for an arbitrarily long time, you will go broke with probability 1.

If you don't bet any money, or only bet a tiny, constant amount each go, you're not taking full advantage of the fact that the expected value of the game is positive. You can conclude that the best thing to do is something in between the two strategies.

There's a formula that tells you what the optimal strategy is when you have a given amount of wealth, with known payoff and known probabilty of winning. It's called the Kelly criterion. Assuming I haven't messed up the conversion from probabilities to bookie's odds, you should bet 2/3 of your money each time.

Neil3030

Fremen wrote: »

Another thing to note: if you bet all your money every time, you will go broke at your first loss. If you play for an arbitrarily long time, you will go broke with probability 1.

If you don't bet any money, or only bet a tiny, constant amount each go, you're not taking full advantage of the fact that the expected value of the game is positive. You can conclude that the best thing to do is something in between the two strategies.

There's a formula that tells you what the optimal strategy is when you have a given amount of wealth, with known payoff and known probabilty of winning. It's called the Kelly criterion. Assuming I haven't messed up the conversion from probabilities to bookie's odds, you should bet 2/3 of your money each time.

Thanks for the link, but I doubt betting 2/3 of the kitty would work long term. Sure 3 losses in a row and you're pretty much bunched.

EDIT: Did the formula myself there and it suggested using 30% stakes, when the probability of win is .8, and the return on a winning iteration is .4.

This makes far more sense.

gerry87

Neil3030 wrote: »

Thanks for the link, but I doubt betting 2/3 of the kitty would work long term. Sure 3 losses in a row and you're pretty much bunched.

EDIT: Did the formula myself there and it suggested using 30% stakes, when the probability of win is .8, and the return on a winning iteration is .4.

This makes far more sense.

I think the difference between your and freeman's results was the odds, if you do the odds as 1.4 you get 66% of your bankroll whereas if you do the odds as .4 you get 30% of your bankroll. (.4 to 1 would mean you bet 100 and if you win you lose 60... not a the best bet in the world!)

In this case your odds would be 1.4 to 1 so b=1.4 giving f*=66%. The thing about kelly criteron is you'll never go broke, so it doesn't matter if you lose 10 in a row, since the expected value is positive you should make it back as long as you can repeat the bet indefinitely. You can see on the wikipedia page that even followers of the kelly criteron recommend taking a fraction of what the formula says, so it must be seen to suggest higher stakes than people are comfortable taking.

hivizman

Actually, the way the odds are described in the Wikipedia article on the Kelly Criterion, it is correct to define them as 0.4-to-1 (betting 1 unit gives a gain of 0.4 units - in horse-racing terms the odds are 5-to-2 on, equivalent to 2-to-5 against or 0.4-to-1 against).

I did some simulations of betting different fractions of the kitty and found that, once the fraction got above 0.3, the likelihood of losing all the kitty approached 1 (certainty). However, I included in my spreadsheet a condition that, when the kitty fell below a minimal level (1 cent), then the game stopped and the game was considered to have ended in gambler's ruin. If this condition is not included, then it is true that the kitty never actually falls to zero, but it asymptotically approaches zero, and the overall annualised rate of return approaches -1.

gerry87

hivizman wrote: »

Actually, the way the odds are described in the Wikipedia article on the Kelly Criterion, it is correct to define them as 0.4-to-1 (betting 1 unit gives a gain of 0.4 units - in horse-racing terms the odds are 5-to-2 on, equivalent to 2-to-5 against or 0.4-to-1 against).

I did some simulations of betting different fractions of the kitty and found that, once the fraction got above 0.3, the likelihood of losing all the kitty approached 1 (certainty). However, I included in my spreadsheet a condition that, when the kitty fell below a minimal level (1 cent), then the game stopped and the game was considered to have ended in gambler's ruin. If this condition is not included, then it is true that the kitty never actually falls to zero, but it asymptotically approaches zero, and the overall annualised rate of return approaches -1.

Ah ok, I didn't understand what it meant by net odds, that must be it then.

gerry87

Just outta curiosity I tried a few different stakes to see what each tended to. First graph shows the annualised return in each period for each stake. 2nd one is the annualised return in the 4th period, to see what each one tends to. It seems peaks at 40% stake and goes negative between 65% and 70% stakes.

More than 4 periods it might converge to kelly's 30. Kinda interesting to see kelly's actually working.

Edit: yep, it does.

I could be way off the mark here, but if kelly's criteron essentially maxamises the return with respect to the stake taken, then would you be able to get the true expected value formula that the Neil was looking for (accounting for the bias) from integrating kelly's formula?

Fremen

hivizman wrote: »

Actually, the way the odds are described in the Wikipedia article on the Kelly Criterion, it is correct to define them as 0.4-to-1 (betting 1 unit gives a gain of 0.4 units - in horse-racing terms the odds are 5-to-2 on, equivalent to 2-to-5 against or 0.4-to-1 against).

Yeah, I thought I might mess that up.