The Education of a Speculator - Gambling the Vig

External question — Quantitative Finance Stack Exchange Author: Parikshit Bhinde Original post: https://quant.stackexchange.com/questions/54986 License: CC BY-SA 4.0 — https://creativecommons.org/licenses/by-sa/4.0/ Adaptation: HTML converted to plain text; contact email addresses removed. In his autobiography, The Education Of A Speculator, Victor Niederhoffer gives the following example: "For perspective on why frequent payment of rakes on speculative trades leads to ruin, consider playing the following game with a brokerage house. Each day, you flip a coin. If it comes up heads, you win 1 dollar. if it comes up tails, you lose 1 dollar. But on every toss the broker takes out 20 cents. What are the chances of ending a winner after 200 tosses? The answer: About 1 in 100,000." For 200 tosses the brokerage is 200 * 0.2 = 40. So, 120 heads should help to break-even (120x1 + 80x-1 - 40=0). Hence anything above 120 heads leads to being a 'winner'. Using the binomial formula to calculate the probabilities of getting 121 heads + 122 heads + ... + 200 heads turns out to be around 0.003 which is more like 1 in 333 instead of 1 in 100,000. Am i making a serious mistake here? Or has the author seriously erred in his calculation?

External answer — Quantitative Finance Stack Exchange Author: nbbo2 Original post: https://quant.stackexchange.com/a/54988 License: CC BY-SA 4.0 — https://creativecommons.org/licenses/by-sa/4.0/ Adaptation: HTML converted to plain text; contact email addresses removed. I tried a MonteCarlo simulation with 10000 iterations and it seems to confirm a probability of about 0.0033 However I wonder if we are interpreting the problem correctly, we are assuming infinite capital, perhaps we should take into account the gambling has to stop when you hit zero capital? Any other issues I may have missed? With the vig taken into account the wins are worth +0.8 and the losses -1.2, so I wrote the following: import numpy as np np.random.seed(1) l= [] count = 10000 win = .8 lose = -1.2 for _ in range(count): z = np.random.choice([win, lose], size=200, replace=True, p=None) a = np.sum(z) l.append(a) wins = sum(i > 0 for i in l) print (wins/count) 0.0033