# Probability of drawing three letters out of four

We would like to organize the following prize game (simplified):

We print five letters, say, $10 \times A$, $10 \times B$, $5\times C$, $1 \times D$, $50\times E$. A customer draws a letter at each purchase.

1) What is the probability that after three visits, he has $A, B$ and $C$?

2) How does this probability change if he makes not $3$, but $4, 5$ or $6$ purchases?

3) What is the probability that after $4$ ($5,6,7,\ldots$) visits he has $A,B,C,D$?

4) How many purchases he needs to make in order to achieve given probability, say for $ABC$?

Of course in reality the numbers will be higher.

Thank you

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That's a lot of questions. We need to use multivariate hypergeometric calculations for this. I'm depending on someone checking my work.

1. The probability of choosing 3 items out of the urn of 76 total items and choosing A, B, and, C is:

$\frac{{{10}\choose{1}}{{10}\choose{1}}{{5}\choose{1}}}{{{76}\choose{3}}} = \frac{10*10*5}{70,300} = \frac{500}{70,300} = 0.01$

This is the product of the number of ways to get each item divided by the total number of combinations you could have chosen.

2. To add more visits, the numerator should be multiplied by the number of ways you could pick any remaining letter:

$\frac{{{10}\choose{1}}{{10}\choose{1}}{{5}\choose{1}}{{76}\choose{1}}}{{{76}\choose{4}}} = \frac{10*10*5*76}{1,282,975} = \frac{38,000}{1,282,975} = 0.03$

So your chances are much better with a fourth visit in terms of percentage, but still not very good. Just out of curiosity I'll do another one for 5 visits:

$\frac{{{10}\choose{1}}{{10}\choose{1}}{{5}\choose{1}}{{76}\choose{1}}{{76}\choose{1}}}{{{76}\choose{5}}} = \frac{10*10*5*76*76}{1,282,975} = \frac{2,888,000}{18,474,840} = 0.16$

That is quite a bit better. I'm sure it increasingly increases with each visit.

3. I won't type this up right now, but you can see that how you can add combinations in the numerator and increase the $k$ value of the combination in the denominator to figure it out.

4. In this case, you can probably guess and check. You using only a few discrete values, so it would not be difficult.

By the way, I am not sure what you mean by, "Of course in reality the numbers will be higher." Numbers never lie.

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