Geometric Distribution and the Minimum Number of Trials (Geometrische Verteilung)

15 min · 2 drills

Introduction

A trial is repeated under the same conditions each time: a die is rolled, a ball is drawn with replacement, a ticket is bought. Every trial succeeds with the same probability $p$, independently of the others. The geometric distribution (German: geometrische Verteilung) describes how long you wait for the first success.

The EPM tasks on this topic almost always ask the same question: how many times do you have to try at least, so that at least one success happens with a probability of more than $90\%$, $95\%$ or $99\%$? The answer comes from an inequality that you solve with a logarithm and then round up. Often you first have to find the $p$ of a single trial in a separate part of the task, for example for two dice or for a draw of two balls.

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The geometric distribution

Let $X$ be the number of the trial on which a success happens for the first time. $X$ can take the values $1,2,3,\dots$ The first success on the $k$-th trial means: the first $k-1$ trials fail and the $k$-th succeeds. Because the trials are independent, the probabilities are multiplied:

$$\boxed{P(X=k)=(1-p{)}^{k-1}\cdot p}$$

Example: You roll a die until the first 6 appears. The probability that it appears exactly on the third roll is

$$P(X==3)=={\left(\frac{5}{6}\right)}^2\cdot \cdot \frac{1}{6}==\frac{25}{216}\approx \approx 0.1157$$

For the minimum number you need two sums. "No success yet after $k$ trials" means: all $k$ trials fail. From this follows the complementary event (German: Gegenereignis) "the first success comes on the $k$-th trial at the latest":

$$\boxed{P(X>k)=(1-p{)}^k}\boxed{P(X\le k)=1-(1-p{)}^k}$$

$P(X\le k)$ is exactly the probability of having at least one success (German: mindestens einmal) in $k$ trials. It is the same number as $P(Y\ge 1)=1-P(Y=0)$ for the binomially distributed number $Y$ of successes in $k$ trials. Some solutions therefore set up the inequality with the number of hits; the calculation is the same in both cases.

The expected value (German: Erwartungswert) tells you how many trials you need on average until the first success:

$$\boxed{E(X)=\frac{1}{p}}$$

When rolling for a 6, that is $6$ rolls on average.

The tasks of this topic always count the trials including the successful one: they ask how many rolls, draws or tickets you need in total. This article therefore uses the first row of the table throughout.

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Minimum number of trials

We are looking for the smallest $n$ for which the probability of having at least one success in $n$ trials is greater than a threshold $s$:

$$1--(1--p{)}^n>>s$$

Bring the power to one side and take the logarithm of both sides:

$$(1--p{)}^n<<1--s$$ $$n\cdot \cdot \ln (1--p)<<\ln (1--s)$$

Both logarithms are negative, because $1-p$ and $1-s$ lie between $0$ and $1$. Dividing by the negative number $\ln (1-p)$ reverses the inequality sign:

$$\boxed{n>\frac{\ln (1-s)}{\ln (1-p)}}$$

The fraction is almost never a whole number. Because $n$ is a natural number, you take the next whole number above it, so you always round up, even at $18.01$. Instead of $\ln$ you may use any other logarithm, for example $\lg$: the quotient stays the same.

Procedure

1. Find $p$: the success probability of a single trial, if it is not given directly.
2. Take the complement $1-p$: the probability that a trial fails.
3. Write the inequality: $1-(1-p{)}^n>s$, rearranged to $(1-p{)}^n<1-s$.
4. Take logarithms and divide by $\ln (1-p)$, reversing the sign.
5. Round up and check with $n-1$ and $n$: at $n-1$ the probability is still below the threshold, at $n$ it is above.

A special case: if the quotient is exactly a whole number, the sign decides. With $p=\frac{1}{2}$ and "greater than $87.5\%$" you get ${\left(\frac{1}{2}\right)}^n<\frac{1}{8}$, so $n>3$ and $n=4$. With "at least $87.5\%$", $n=3$ is enough.

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The probability p of one trial

A "trial" can itself consist of several parts: two dice at once, two balls from an urn, a coin and a die. Then you first find the probability $p$ that one such round succeeds, and only then put it into the inequality. The rounds are independent of each other, because before each round everything is put back or thrown again.

For the counting rules see "Combinatorics", for the addition rule and total probability "Conditional Probability", for "exactly $k$ times" in several trials "Binomial Distribution".

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Examples

Example 1: The first success on the k-th trial

A spinner has five equal sectors, one of which wins. It is spun until you win. So $p=\frac{1}{5}$ and $1-p=\frac{4}{5}$.

The first win exactly on the fourth spin:

$$P(X==4)=={\left(\frac{4}{5}\right)}^3\cdot \cdot \frac{1}{5}==\frac{64}{625}==0.1024$$

No win yet after three spins, and the complementary event:

$$P(X>>3)=={\left(\frac{4}{5}\right)}^3==\frac{64}{125}==0.512,P(X\le \le 3)==1--\frac{64}{125}==\frac{61}{125}==0.488$$

As a check, add up the single values: $\frac{1}{5}+\frac{4}{25}+\frac{16}{125}=\frac{25+20+16}{125}=\frac{61}{125}$.

On average you need $E(X)=\frac{1}{p}=5$ spins.

Example 2: Just above a whole number

At a raffle stall each ticket wins with probability $12\%$. How many tickets do you have to buy at least to win at least once with a probability of more than $90\%$?

$$1--0.88^n>>0.9$$ $$0.88^n<<0.1$$ $$n>>\frac{\ln (0.1)}{\ln (0.88)}==\frac{-2.302585}{-0.127833}\approx \approx 18.01$$ $$\boxed{n=19}$$

The check shows why you have to round up here too: with $18$ tickets $1-0.88^{18}\approx 0.89984$, just below $90\%$. Only with $19$ tickets is $1-0.88^{19}\approx 0.91186$.

Example 3: p from two dice

Two dice are thrown at the same time. A round succeeds if the sum is 9. How many times do you have to throw at least, so that a sum of 9 appears at least once with a probability of more than $95\%$?

A sum of 9 comes from the four outcomes $(3,6)$, $(4,5)$, $(5,4)$ and $(6,3)$ out of 36:

$$p==\frac{4}{36}==\frac{1}{9},1--p==\frac{8}{9}$$ $${\left(\frac{8}{9}\right)}^n<<0.05$$ $$n>>\frac{\ln (0.05)}{\ln (8/9)}==\frac{-2.995732}{-0.117783}\approx \approx 25.43$$ $$\boxed{n=26}$$

Example 4: Two balls at once

An urn holds 5 white and 3 black balls. You draw two balls at the same time, note them and put both back. A round succeeds if both balls are black. We want the minimum number of rounds for a probability of more than $90\%$.

Within one round the draw is without replacement, so you count draws:

$$p==\frac{\left(\genfrac{}{}{0ex}{}{3}{2}\right)}{\left(\genfrac{}{}{0ex}{}{8}{2}\right)}==\frac{3}{28},1--p==\frac{25}{28}$$ $$n>>\frac{\ln (0.1)}{\ln (25/28)}==\frac{-2.302585}{-0.113329}\approx \approx 20.32$$ $$\boxed{n=21}$$

Between rounds the balls are put back. That is why every round has the same $p$ and the formula applies.

Example 5: "or" with an overlap

From a deck of 52 cards a card is drawn and put back. A round succeeds if the card is a heart or a face card (jack, queen, king). We want the minimum number for more than $99.9\%$.

There are 13 hearts and 12 face cards. The three heart face cards are in both numbers and are subtracted once:

$$p==\frac{13+12-3}{52}==\frac{22}{52}==\frac{11}{26},1--p==\frac{15}{26}$$ $$n>>\frac{\ln (0.001)}{\ln (15/26)}==\frac{-6.907755}{-0.550046}\approx \approx 12.56$$ $$\boxed{n=13}$$

Example 6: Total probability

A dealer gets $70\%$ of his goods from supplier A with $2\%$ rejects and $30\%$ from supplier B with $6\%$ rejects. He inspects randomly chosen pieces, each independently of the one before. How many pieces does he have to inspect at least to find at least one defective piece with a probability of more than $90\%$?

$$p==0.7\cdot \cdot 0.02++0.3\cdot \cdot 0.06==0.014++0.018==0.032$$ $$0.968^n<<0.1$$ $$n>>\frac{\ln (0.1)}{\ln (0.968)}==\frac{-2.302585}{-0.032523}\approx \approx 70.80$$ $$\boxed{n=71}$$

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EPM task: full solution

Task: Two fair dice are thrown at the same time. The throw counts as a success if at least one 6 is among them. The throw is repeated independently.

a) What is the probability that a throw is a success? b) How many times do you have to throw at least, so that at least one success happens with a probability of more than $99\%$? c) How many throws do you need on average until the first success? d) What is the probability that the first success happens exactly on the third throw?

Solution:

a) Via the complement: neither die shows a 6.

$$p==1--{\left(\frac{5}{6}\right)}^2==1--\frac{25}{36}==\frac{11}{36}\approx \approx 0.3056$$

As a check by counting: of the 36 outcomes, $6+6-1=11$ contain at least one 6, because $(6,6)$ is counted only once.

b) A throw fails with $\frac{25}{36}$:

$$1--{\left(\frac{25}{36}\right)}^n>>0.99$$ $${\left(\frac{25}{36}\right)}^n<<0.01$$ $$n>>\frac{\ln (0.01)}{\ln (25/36)}==\frac{-4.605170}{-0.364643}\approx \approx 12.63$$ $$\boxed{n=13}$$

Check: with $12$ throws the probability is $\approx 0.98742$, with $13$ throws $\approx 0.99127$.

c) $$E(X)==\frac{1}{p}==\frac{36}{11}\approx \approx 3.27\text{throws}$$

d) The first two throws fail, the third succeeds:

$$P(X==3)=={\left(\frac{25}{36}\right)}^2\cdot \cdot \frac{11}{36}==\frac{6875}{46656}\approx \approx 0.1474$$

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Typical EPM questions

The German task text almost always reads: "Berechnen Sie, wie oft man mindestens ... muss, damit die Wahrscheinlichkeit, mindestens einmal ... zu erhalten, größer als ... ist!"

• Minimum number without preliminary work: $p$ is given as a percentage or as a simple Laplace fraction (coin, die, spinner). The whole task is the inequality.
• Parts a) and b): a) asks for $p$ of one round, b) for the minimum number (German: Mindestanzahl). The work in a) is combinatorics, the addition rule or total probability, as in Examples 3 to 6.
• Two events in one task: for example "$A$ and $B$" in a) and "$A$ or $B$" in b), each with its own threshold. You run the same procedure twice with a different $p$.
• "with replacement", "independently", "at the same time ... then everything back": these phrases say that every round has the same $p$. Without them the formula $1-(1-p{)}^n$ would not apply.
• Stating the result: $n$ as a natural number, $p$ as a fraction or a decimal.

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Common mistakes

Mistake 1: The sign not reversed

From $n\cdot \ln (0.88)<\ln (0.1)$, dividing gives $n>18.01$, not $n<18.01$. If you leave the sign as it is, you get an upper bound and end up looking for a largest instead of a smallest $n$. The check with $n-1$ and $n$ reveals the mistake.

Mistake 2: Rounded down

The quotient $18.01$ in Example 2 tempts you to answer $18$. With $18$ tickets, however, the probability is $0.89984$, which is not greater than $90\%$. For minimum numbers you always round up.

Mistake 3: $p$ and $1-p$ or $s$ and $1-s$ swapped

The power takes the probability of failure, the right-hand side takes the rest $1-s$. If you set up $0.12^n<0.1$ or $0.88^n<0.9$, you are calculating a different event. For $0.88^n<0.9$, for example, you get $n=1$, which obviously cannot be right.

Mistake 4: Linear instead of with the power

With $p=0.15$ and a threshold of $90\%$, the approach $n\cdot 0.15>0.9$ gives the answer $n=7$. The correct answer is $n=15$, because the probabilities of the single trials overlap and must not be added. The expected value $\frac{1}{p}\approx 6.67$ is not a minimum number either: it says how long you wait on average, not how long you have to wait to be almost sure.

Mistake 5: Rounded too early

If in Example 6 you round the logarithm $\ln (0.968)\approx -0.032523$ to $-0.033$, you get $n>69.8$ and the answer $70$. With $70$ pieces, however, the probability is only $0.89737$. Calculate the logarithms with at least four significant digits, or form the quotient directly on the calculator.

Mistake 6: The overlap in "or" counted twice

In Example 5, $13+12=25$ cards instead of $22$ gives $p=\frac{25}{52}$, and from that $n=11$ instead of $13$. The addition rule subtracts the common part once.

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Check your knowledge

Your turn1 of 2

A basketball player scores with a probability of $20\%$ per shot. How many times does he have to shoot at least, so that he scores at least once with a probability of more than $95\%$?

A$n=5$B$n=13$C$n=14$D$n=1$

Check

Your turn2 of 2

You roll a die until the first 6 appears. What is the probability that it appears exactly on the third roll?

A$\frac{1}{216}$B$\frac{25}{216}$C$\frac{75}{216}$D$\frac{91}{216}$

Check

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Formulas to remember

First success on the $k$-th trial:

$$P(X==k)==(1--p{)}^{k-1}\cdot \cdot p$$

No success yet after $k$ trials, at least one success in $k$ trials:

$$P(X>>k)==(1--p{)}^k,P(X\le \le k)==1--(1--p{)}^k$$

Expected value:

$$E(X)==\frac{1}{p}$$

Minimum number for a threshold $s$:

$$1--(1--p{)}^n>>s⟺⟺n>>\frac{\ln (1-s)}{\ln (1-p)},\text{then round up}$$