# Probability: twice the same number

• MHB
• mathmari
In summary: Question)Do you mean at the initial example or the fractions of the table of...At the initial example.
mathmari
Gold Member
MHB
Hey!

We consider the following game:
We can roll a dice until a number appears twice. We can then write down as many points as the times that we rolled the dice.

Let $X$ be the random variable that describes the points that we get at each game. I want to calculate the probabilities $P(X=x_i)$, where $x_i\in \{2,3,4,5,6,7\}$.

We have the following:

1. $P(X=2)$ is the probability that we get $2$ points, i.e. that we roll twice a dice and get the same result.

When we roll a dice twice, we get the set $\Omega = \{ (a\mid b) \mid a,b \in \{1,2,3,4,5,6\} \}$ and so the number of possible results is equal to $6\cdot 6=36$.

$6$ of these $36$ possible results have twice the same number.

So we get the probability $P(X=2)=\frac{6}{36}$.
2. $P(X=3)$ is the probability that we get $3$ points, i.e., we roll three times a dice and a number appears twice and the second time is at the third place.

When we roll a dice three times, we get the set $\Omega = \{ (a\mid b\mid c) \mid a,b,c \in \{1,2,3,4,5,6\} \}$ and so the number of possible results is equal to $6\cdot 6 \cdot 6= 216$.

The favorable results are contained in the set $M=\{ (a\mid b\mid c) \mid a,b,c \in \{1,2,3,4,5,6\} , a\neq b, c=a \text{ oder } c=b\}$, right?

So $M$ contains $6\cdot 5= 30$ favorable results, or not?

So, we have $30$ favorable results.

Therefore, the probability is $P(X=3)=\frac{30}{216}$.
3. $P(X=4)$ is the probability that we get $4$ points, i.e., we roll four times a dice and a number appears twice and the second time is at the $4$th place.

When we roll a dice four times, we get the set $\Omega = \{ (a\mid b\mid c\mid d) \mid a,b,c,d \in \{1,2,3,4,5,6\} \}$ and so the number of possible results is equal to $6\cdot 6 \cdot 6\cdot 6= 1296$.

The favorable results are contained in the set $M=\{ (a\mid b\mid c\mid d) \mid a,b,c,d \in \{1,2,3,4,5,6\} , a\neq b\neq c, d=a \text{ or } d=b \text{ or } d=c\}$.

So $M$ contains $6\cdot 5 \cdot 4\cdot 3= 360$ favorable results.

Therefore, the probability is $P(X=4)=\frac{360}{1296}$.
Is everything correct so far? Do I continue in the same way for the remaining cases? (Wondering)

mathmari said:
The favorable results are contained in the set $M=\{ (a\mid b\mid c) \mid a,b,c \in \{1,2,3,4,5,6\} , a\neq b, c=a \text{ oder } c=b\}$, right?

So $M$ contains $6\cdot 5= 30$ favorable results, or not?

Hey mathmari!

Shouldn't that be $6\cdot 5 \cdot 2 = 60$ favorable results?
After all, dice c can be either equal to a or to c, can't it?

mathmari said:
Is everything correct so far? Do I continue in the same way for the remaining cases?

Yes, we can continue for the remaining cases. (Nod)

I do see a pattern. Note that the first dice can be anything, each following dice must be different from the previous ones, except the last, which can be any of the $(k-1)$ preceding dice.
So that:
$$P(X=k) = \frac{\overbrace{\Big(6\cdot ... \cdot (8-k)\Big)}^{k-1\text{ dice}} \cdot (k-1)}{6^k}$$
(Thinking)

I like Serena said:
Note that the first dice can be anything, each following dice must be different from the previous ones, except the last, which can be any of the $(k-1)$ preceding dice.
So that:
$$P(X=k) = \frac{\overbrace{\Big(6\cdot ... \cdot (8-k)\Big)}^{k-1\text{ dice}} \cdot (k-1)}{6^k}$$
(Thinking)

I understand that pattern! (Nerd) But I saw now the following table:

View attachment 7864

Can this table be correct? In this table we don't have that pattern. (Wondering)

#### Attachments

• table.JPG
61.3 KB · Views: 56
mathmari said:
I understand that pattern! (Nerd)

Good! (Happy)

mathmari said:
But I saw now the following table:

Can this table be correct? In this table we don't have that pattern.

Indeed. That table looks incorrect.
All probabilities must sum up to 1, which isn't the case in that table either. (Worried)

I like Serena said:
Indeed. That table looks incorrect.
All probabilities must sum up to 1, which isn't the case in that table either. (Worried)

So, that table is in general incorrect? Or just for this example? (Wondering)

mathmari said:
So, that table is in general incorrect? Or just for this example? (Wondering)

The table looks incorrect in general.
In theory it could still be correct, but only if the entries entries for 6 and 7 are unexpectedly high to compensate. (Nerd)

I like Serena said:
The table looks incorrect in general.
In theory it could still be correct, but only if the entries entries for 6 and 7 are unexpectedly high to compensate. (Nerd)

Ah ok! Do the numberators have a specific pattern? The denominators are of the form $6^k$. (Wondering)

mathmari said:
Ah ok! Do the numberators have a specific pattern? The denominators are of the form $6^k$.

Just the one that we've already found, which equals $\frac{6!\cdot (x_i-1)}{(7-x_i)!}$. (Thinking)

I like Serena said:
Just the one that we've already found, which equals $\frac{6!\cdot (x_i-1)}{(7-x_i)!}$. (Thinking)

Do you mean at the initial example or the fractions of the table of the picture? (Wondering)

mathmari said:
Do you mean at the initial example or the fractions of the table of the picture?

Ah, you meant the numerators in the table. (Tmi)

I don't see an obvious pattern.
Since $34=2\cdot 17$, and since $61$ is prime, it's clear that some kind of OR operation would be involved, but I don't see which one could apply.
We might see the binomial coefficients $6,15,20,15,6,1$ in there, but those don't really add up either. (Thinking)

I like Serena said:
Ah, you meant the numerators in the table. (Tmi)

I don't see an obvious pattern.
Since $34=2\cdot 17$, and since $61$ is prime, it's clear that some kind of OR operation would be involved, but I don't see which one could apply.
We might see the binomial coefficients $6,15,20,15,6,1$ in there, but those don't really add up either. (Thinking)

Ok, no problem.

Thank you very much for your help! (Smile)

## 1. What is the probability of rolling the same number twice in a row with a standard six-sided die?

The probability of rolling the same number twice in a row with a standard six-sided die is 1/6 or approximately 16.67%. This is because each roll is an independent event and the outcome of the first roll does not affect the outcome of the second roll.

## 2. How does the probability change if the number of sides on the die is increased?

If the number of sides on the die is increased, the probability of rolling the same number twice in a row decreases. For example, with a ten-sided die, the probability would be 1/10 or 10%.

## 3. What is the probability of flipping two coins and getting the same result?

The probability of flipping two coins and getting the same result is 1/2 or 50%. This is because there are two possible outcomes for each coin flip (heads or tails) and the outcome of one coin flip does not affect the outcome of the other.

## 4. Is the probability of getting the same number twice in a row the same for all games of chance?

No, the probability of getting the same number twice in a row can vary depending on the game of chance. For example, in a game of roulette, the probability of the ball landing on the same number twice in a row is 1/38 or approximately 2.63%.

## 5. How does the concept of probability apply to real-life situations?

The concept of probability is used in many real-life situations, such as in weather forecasting, risk assessment, and financial planning. It helps us to understand and predict the likelihood of certain events occurring and make informed decisions based on that information.

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