The IRS Audit Blog

Posts Tagged ‘mathematics’

Question: Has anyone ever calculated the odds of being audited by the IRS?

Answer: The IRS uses a secret formula called a DIF score to select returns for audit. Certain features of a return will increase you chances of audit. Having income over $100,000 especially if you also have a side business that lost money. Owning a Sub-S corp. Showing a low income but having lifestyle requiring a higher income. Publicly revealing your true income can get you audited. Like having your name in the paper bragging how you make $5,000 per week in the latest MLM direct sales program while showing only $20,000 on your tax return for the whole year.

Nurse Outduels IRS Over M.B.A. Tuition

A Maryland nurse defended herself against the IRS’s lawyers and won — getting a ruling to deduct the cost of an M.B.A. degree.

Taxes: Odds of an Audit

Question: Question about poker winnings?

Hi, I just recently turned 21 yrs old, and I have been playing Texas hold’em since i have been around 16 or so. I have quite a bit of experience, and I am looking to start playing more often. I went to a casino nearby the day after my 21st birthday and won minimal (like 30 dollars or so), and just out of curiosity, around 2 days ago I started playing online poker. I only bought in for around 25 dollars, but I got up to around 180 before falling back to around 115$. I was just wondering if this is major for not filing taxes on, and what I need to start doing for future play? Also, I was wondering what one must do before getting audited by the IRS? Any information is welcome because I’m new to the whole gambling/paying taxes on it thing. Thanks in advance for your input.

Answer: As a general rule, you don’t have to worry about gambling winnings until you hit $10k in the US. All gambling winnings over 10k are reported to the IRS, and if you deposit a check over 10k, the bank also reports that to the IRS.

Once you are over that threshold, and winnings are seriously large, you should start documenting your gambling more carefully. At this stage, you should consult an accountant.

Don’t be fooled by tax code myths

The U.S. tax code is massive and complicated — the perfect fodder for myth-making.Perhaps the most well-worn fallacy — shot down by many courts despite the best efforts of tightfisted taxpayers — is that federal taxes are actually illegal.It’s an argument often used by people who have turned to Tax Masters, which helps filers in trouble with the IRS, says company president Patrick Cox. Some …

Tips for Filing Taxes : Gambling Winnings Tax Tips

Question: Statistics: Suppose that the Internal Revenue Service (IRS) believes that the percentage of tax returns in …

Error is 10%. IF IRS is correct, what is the probability that in a random audit of 1000 tax returns, between 70 and 90 tax returns will be in error?

thanks

Answer: Let Xb be the number of tax returns in error. Xb has the binomial distribution with n = 1000 trials and success probability p = 0.1

In general, if X has the binomial distribution with n trials and a success probability of p then
P[Xb = x] = n!/(x!(n-x)!) * p^x * (1-p)^(n-x)
for values of x = 0, 1, 2, …, n
P[Xb = x] = 0 for any other value of x.

To use the normal approximation to the binomial you must first validate that you have more than 10 expected successes and 10 expected failures. In other words, you need to have n * p > 10 and n * (1-p) > 10.

Some authors will say you only need 5 expected successes and 5 expected failures to use this approximation. If you are working towards the center of the distribution then this condition should be sufficient. However, the approximations in the tails of the distribution will be weaker espeically if the success probability is low or high. Using 10 expected successes and 10 expected failures is a more conservative approach but will allow for better approximations especially when p is small or p is large.

In this case you have:
n * p = 1000 * 0.1 = 100 expected success
n * (1 – p) = 1000 * 0.9 = 900 expected failures

We have checked and confirmed that there are enough expected successes and expected failures. Now we can move on to the rest of the work.

If Xb ~ Binomial(n, p) then we can approximate probabilities using the normal distribution where Xn is normal with mean μ = n * p, variance σ² = n * p * (1-p), and standard deviation σ

Xb ~ Binomial(n = 1000 , p = 0.1 )
Xn ~ Normal( μ = 100 , σ² = 90 )
Xn ~ Normal( μ = 100 , σ = 9.486833 )

I have noted two different notations for the Normal distribution, one using the variance and one using the standard deviation. In most textbooks and in most of the literature, the parameters used to denote the Normal distribution are the mean and the variance. In most software programs, the standard notation is to use the mean and the standard deviation.

The probabilities are approximated using a continuity correction. We need to use a continuity correction because we are estimating discrete probabilities with a continuous distribution. The best way to make sure you use the correct continuity correction is to draw out a small histogram of the binomial distribution and shade in the values you need. The continuity correction accounts for the area of the boxes that would be missing or would be extra under the normal curve.

P( Xb < x) ≈ P( Xn < (x - 0.5) )
P( Xb > x) ≈ P( Xn > (x + 0.5) )
P( Xb ≤ x) ≈ P( Xn ≤ (x + 0.5) )
P( Xb ≥ x) ≈ P( Xn ≥ (x – 0.5) )
P( Xb = x) ≈ P( (x – 0.5) < Xn < (x + 0.5) )
P( a ≤ Xb ≤ b ) ≈ P( (a - 0.5) < Xn < (b + 0.5) )
P( a ≤ Xb < b ) ≈ P( (a - 0.5) < Xn < (b - 0.5) )
P( a < Xb ≤ b ) ≈ P( (a + 0.5) < Xn < (b + 0.5) )
P( a < Xb < b ) ≈ P( (a + 0.5) < Xn < (b - 0.5) )

In the work that follows X has the binomial distribution, Xn has the normal distribution and Z has the standard normal distribution.

Remember that for any normal random variable Xn, you can transform it into standard units via: Z = (Xn - μ ) / σ

P( 70 ≤ Xb ≤ 90 ) =

90
∑ P(Xb = x) = 0.1578600
x = 70

≈ P( 69.5 < Xn < 90.5 )
= P( ( 69.5 - 100 ) / 9.486833 < Z < ( 90.5 - 100 ) / 9.486833 )
= P( -3.214982 < Z < -1.001388 )
= P( Z < -1.001388 ) - P( Z < -3.214982 )
= 0.1583196 - 0.0006522629
= 0.1576674

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“Hon. Isaac Osei is humble, God-fearing, affable, charismatic, competent and above all without any controversies whatsoever”. This is according to Richard Ntim Frimpong of Subin.