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Brokerage Fees and the Odds of Winning a Repeated Coin Toss

Article Quant Q&A · Author: Parikshit Bhinde
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Summary

The document examines a repeated fair coin toss game with a broker taking a fee on every toss. It compares an estimate attributed to Victor Niederhoffer with a binomial calculation: over 200 tosses, a player must get more than 120 heads to finish ahead after the fees. The stated calculation puts the chance of a positive result at about 0.003, rather than about one in 100,000.

A respondent reports a Monte Carlo simulation that gives a similar estimate, using net outcomes of +0.8 for a win and −1.2 for a loss. The exchange raises a key modeling caveat: these calculations assume enough capital to continue through all tosses, without stopping if funds run out. It does not resolve whether that assumption matches the original example, and the reported simulation uses 10,000 trials, so it is an approximate check rather than a proof.

Key ideas

  • Repeated transaction fees can make a fair underlying game unprofitable in expectation.
  • The stated break-even condition requires more wins than losses to cover the cumulative fees.
  • The binomial calculation and simulation both estimate a positive-result probability near 0.003.
  • Capital constraints and stop-out rules may change the probability compared with an unlimited-capital model.

Tags

StatisticsRisk management
Full text
# The Education of a Speculator - Gambling the Vig


# The Education of a Speculator - Gambling the Vig












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?

## Answer by nbbo2 (score 3)

https://quant.stackexchange.com/a/54988

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
```

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This summary was written by Stratmill's research agent from the original; it is not a copy of the source.