Why Proportional Bets Skew Outcomes in a Fair Coin-Toss Game

# Random Brownian Simulation Startling Results


# Random Brownian Simulation Startling Results












I was playing around in Excel the other day, simulating possible equity curve/P&L paths for a simple game I designed. The game is really trying to find an optimal risk managment strategy.

I start of with an initial Capital: 100. In each time increment you have the opportunity to risk a % of your capital on a coin flip, at each time period, you can either gain/lose the amount you bet on each flip. E.g. in the initial flip, if you gambled 25% of your capital, you can gain 25 or lose 25. The possibility of winning on each coin flip is 50% and the possibility of losing is 50%.

I simulated over 4000 possible P/L paths over 10 coin flips and I kept coming out with a win ratio less than 50. The more I increased the amount invested on each flip, the more I lost. The more coin flips I played the more my win% dropped. (Note, my definition of a win ratio is by the end of the 10th toss, how many of the simulations ended with a final capital greater than my initial capital of 100)

So I decided later to alter the % I invest on each coin flip based based on whether I won/lost the prior coin flip. In other words, if the last coin flip was win, the % amount I invest would be $x$ and if I lost the prior coin flip the % amount I invest would be $y%$. I came out with pretty startling results. I found that if I lost the prior flip, I should increase the amount I invest on the next flip. For the 4000 simulations among 10 flips, I came out with a 53% win ratio for % amount invested $x=3$ and $y=16$, ie, your increasing your bet size as you lose more. Is there any justification for this? Has anyone come to a similar conclusion from a similar simulation. Would love to post the excel spread sheet if anyone would like to take a look it.

## Answer by Peter (score 6)

https://quant.stackexchange.com/a/9832

When I run this simulation I see the same results, and it makes sense.

For the straight 50%/50%, I found that my win ration was about 38% and my loss ratio 61%. The reason it wasn't 50/50 was that if I had consecutive up flips my value could keep going up, but if I had consecutive down flips I would 0 out and the sequence would have to end as I had lost all my money... With it being random the chance of zeroing out isn't that unlikely.

When I increased the number of flips from 10 to 100 my win ratio dropped to 9% and my loss ratio is about 90%.

When I tried using your alternative method of modifying the %'s based on the previous result, the results were even worse. At 10 flips only 7 out of 10000 had made any money, at 100 flips 0 survived...

Slight Correction It's not that you are zeroing out, it's that if you subtract 25% and then add 25% you don't end up with the same number. So if you have 3 consecutive subtractions followed by three consecutive additions starting from 100 you end up with ~82.

When I change the code to use absolute values for +/- rather then percents I get a 30% win and 60% loss. When I remove the stop at 0 so that it can go negative it comes out to 50/50. This is regardless of whether or not I use a fixed amount on win/lose, or if I used the alternate x/y strategy.

Interesting side note I'm not sure what your game is actually going to be, but perhaps this page on Random Walks might be of interest: http://en.wikipedia.org/wiki/Random_walk. They have a nice graphs on the right hand side showing multiple series starting at 0 and moving in a random up/down pattern.

## Answer by Svisstack (score 0)

https://quant.stackexchange.com/a/9734

Computers can't generate true random numbers without special devices, please read http://en.wikipedia.org/wiki/Hardware_random_number_generator for more information about that.

In normal computer or excel pseudo random number generation often looks like walking over and over again over same hard coded table with seeding based on time with some modulo operations between.

I think your conclusions with better results using some parameters in break-even game are just fitting to pseudo random number generator patterns in hard coded data.

First you should use other random number generation stream like http://www.random.org/ (true random numbers), and using that data you should converge in first algorithm to 50% (50/50, equal gain/loss), after that you can start testing hypothesis.

## Answer by user name (score 0)

https://quant.stackexchange.com/a/80531

Note that the 'medium' of payouts after 10 nodes is $(1+a)^{10} * (1-a)^{10} = (1-a^2)^{10} < 1$, and the probability of achieving higher than medium equals to that of achieving lower than medium (by standard binomial distribution properties), and hence the probability of losing will be larger than that of winning by at least a margin of $P($medium$)$.