MHB Probability: twice the same number

[mathmari]

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)

 

[I like Serena]

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?

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)

 

[mathmari]

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)

 

[I like Serena]

I understand that pattern! (Nerd)

Good! (Happy)

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)

 

[mathmari]

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)

 

[I like Serena]

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)

 

[mathmari]

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)

 

[I like Serena]

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)

 

[mathmari]

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)

 

[I like Serena]

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)

 

[mathmari]

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)