What are the odds of rolling 4, 5, 6, and 7 of a kind with 7 dice?

[Peterconfused]

Could somebody please tell me, in layman's terms, the answer to the following:

When 7 standard six sided dice are thrown what are the odds of 4 showing the same number, 5 showing the same number, 6 and 7. Also the odds of 7 sixes being thrown. This is all to do with a game used to raise money for charity - people sometimes ask what the odds are and it would be good to know.

Hope someone can help.

Many thanks

Peterconfused

 

[MarkFL]

Let's look first at 4 of the 7 dice showing the same number. We have 6 choices for the value shown on the quadruple. There are \(\displaystyle {7 \choose 4}=35\) ways to choose 4 dice from 7, and there are 5 values the remaining 3 dice may have. The total number of outcomes is $6^7$, and so putting all this together, we find:

\(\displaystyle P(\text{quadruple})=\dfrac{6\cdot{7 \choose 4}\cdot5^3}{6^7}=\frac{4375}{46656}\)

Can you work out the others in a similar fashion?

 

[Peterconfused]

Let's look first at 4 of the 7 dice showing the same number. We have 6 choices for the value shown on the quadruple. There are \(\displaystyle {7 \choose 4}=35\) ways to choose 4 dice from 7, and there are 5 values the remaining 3 dice may have. The total number of outcomes is $6^7$, and so putting all this together, we find:

\(\displaystyle P(\text{quadruple})=\dfrac{6\cdot{7 \choose 4}\cdot5^3}{6^7}=\frac{4375}{46656}\)

Can you work out the others in a similar fashion?

Many thanks for the quick response. I am not a mathematician and unfortunately don't understand the equation. I simply cannot work out how you reach the final result. On the top line, how do you arrive at 4375? On the bottom line 6 to the power of 7 I thought should be 279,936.

Yours gratefully but in puzzlement

Peterconfused

 

[MarkFL]

Many thanks for the quick response. I am not a mathematician and unfortunately don't understand the equation. I simply cannot work out how you reach the final result. On the top line, how do you arrive at 4375? On the bottom line 6 to the power of 7 I thought should be 279,936.

Yours gratefully but in puzzlement

Peterconfused

There is a 6 in the numerator, and so they can be divided out...I will add an intermediary step:

\(\displaystyle P(\text{quadruple})=\dfrac{6\cdot{7 \choose 4}\cdot5^3}{6^7}=\dfrac{\cancel{6}\cdot35\cdot125}{\cancel{6}\cdot6^6}=\frac{4375}{46656}\)

 

[Peterconfused]

There is a 6 in the numerator, and so they can be divided out...I will add an intermediary step:

\(\displaystyle P(\text{quadruple})=\dfrac{6\cdot{7 \choose 4}\cdot5^3}{6^7}=\dfrac{\cancel{6}\cdot35\cdot125}{\cancel{6}\cdot6^6}=\frac{4375}{46656}\)

Thank you again. I am starting to get there (slowly!). The only thing I still don't understand is how 7 over 4 equals 35.

Your patience is really appreciated

Peterconfused

 

[MarkFL]

Thank you again. I am starting to get there (slowly!). The only thing I still don't understand is how 7 over 4 equals 35.

Your patience is really appreciated

Peterconfused

That's not 7 divided by 4, that is a binomial coefficient or combination...it represents the number of ways to choose 4 things from a group of 7 things...it is defined as follows:

\(\displaystyle {n \choose r}\equiv\frac{n!}{r!(n-r)!}\)

where:

\(\displaystyle n!\equiv\prod_{k=1}^{n}(k)=n(n-1)(n-2)\cdots3\cdot2\cdot1\)

Thus, we find:

\(\displaystyle {7 \choose 4}\equiv\frac{7!}{4!(7-4)!}=\frac{7\cdot6\cdot5\cdot4\cdot3\cdot2\cdot1}{(4\cdot3\cdot2\cdot1)(3\cdot2\cdot1)}=\frac{7\cdot\cancel{6}\cdot5\cdot\cancel{4\cdot3\cdot2\cdot1}}{(\cancel{4\cdot3\cdot2\cdot1})(\cancel{3\cdot2\cdot1})}=7\cdot5=35\)

 

[Peterconfused]

That's not 7 divided by 4, that is a binomial coefficient or combination...it represents the number of ways to choose 4 things from a group of 7 things...it is defined as follows:

\(\displaystyle {n \choose r}\equiv\frac{n!}{r!(n-r)!}\)

where:

\(\displaystyle n!\equiv\prod_{k=1}^{n}(k)=n(n-1)(n-2)\cdots3\cdot2\cdot1\)

Thus, we find:

\(\displaystyle {7 \choose 4}\equiv\frac{7!}{4!(7-4)!}=\frac{7\cdot6\cdot5\cdot4\cdot3\cdot2\cdot1}{(4\cdot3\cdot2\cdot1)(3\cdot2\cdot1)}=\frac{7\cdot\cancel{6}\cdot5\cdot\cancel{4\cdot3\cdot2\cdot1}}{(\cancel{4\cdot3\cdot2\cdot1})(\cancel{3\cdot2\cdot1})}=7\cdot5=35\)

Thank you very much for your time and patience. We are into areas that are totally outside my knowledge or understanding. We never did anything like that when I was at school (very many years ago!). If you could be very kind and tell me the probabilities for 5, 6 and 7 numbers out of 7 being the same, and also of 7 sixes being thrown I would be very, very grateful. We use this game in my local Rotary Club to raise money, and it works well. We keep being asked the odds and we simply don't know.

I am so grateful for your help.

With best wishes

Peterconfused

 

[MarkFL]

Okay, this is what I get:

4 of a kind:

\(\displaystyle \frac{4375}{46656}\)

5 of a kind:

\(\displaystyle \frac{175}{15552}\)

6 of a kind:

\(\displaystyle \frac{35}{46656}\)

7 of a kind:

\(\displaystyle \frac{1}{46656}\)

7 sixes:

\(\displaystyle \frac{1}{279936}\)

 

[Peterconfused]

Okay, this is what I get:

4 of a kind:

\(\displaystyle \frac{4375}{46656}\)

5 of a kind:

\(\displaystyle \frac{175}{15552}\)

6 of a kind:

\(\displaystyle \frac{35}{46656}\)

7 of a kind:

\(\displaystyle \frac{1}{46656}\)

7 sixes:

\(\displaystyle \frac{1}{279936}\)

Hi

Thank you so much for taking the time to help me with this - I really appreciate it. It has answered the questions very well.

With best regards and good wishes (writing from Bonny Scotland by the way)

Peterconfused