Dice game

The players choose five numbers between 3-18. The dice is rolled three times and the numbers will be summed so the result will be between 3-18. However, in this game the dice will be rolled in total fifteen times but the same rule applies as told in the last statement, so we will have five summed up groups of numbers between 3 and 18. For example:

Dice is rolled fifteen times: 1 + 5 + 3 + 6 + 4 + 4 +2 + 1 + 6 + 6 + 5 + 1 + 4 + 2 + 5

These numbers will be now put in groups of three in the order as they were rolled and summed up:

(1 + 5 + 3); (6 + 4 + 4); (2 + 1 + 6); (6 + 5 + 1); (4 + 2 + 5) = 9; 14; 9; 12; 11.

The question is, how can I calculate the probability of the players that they have chosen five numbers between 3-18 so they are the same numbers as the summed up three number groups from dice?

pive said:

The players choose five numbers between 3-18. The dice is rolled three times and the numbers will be summed so the result will be between 3-18. However, in this game the dice will be rolled in total fifteen times but the same rule applies as told in the last statement, so we will have five summed up groups of numbers between 3 and 18. For example:

Dice is rolled fifteen times: 1 + 5 + 3 + 6 + 4 + 4 +2 + 1 + 6 + 6 + 5 + 1 + 4 + 2 + 5

These numbers will be now put in groups of three in the order as they were rolled and summed up:

(1 + 5 + 3); (6 + 4 + 4); (2 + 1 + 6); (6 + 5 + 1); (4 + 2 + 5) = 9; 14; 9; 12; 11.

The question is, how can I calculate the probability of the players that they have chosen five numbers between 3-18 so they are the same numbers as the summed up three number groups from dice?

Click to expand...

Think of each die roll as a random variable that can take on the values 1-6 each with probability 1/6.

Now form a new random variable that is the sum of 3 independent rolls of this die.

As you've noted this will take on values 3-18 each with a certain probability.

This probability is given by the number of ways you can form the sum with 3 dice rolls divided by (1/6)^3.

For example Pr[3] = 1/(6^3) = 1/216

I'll leave it to you to compute the rest. You might note that the distribution of a sum of independent random variables is a convolution of the two individual distributions.

So now you have the problem of having 5 people correctly choosing a number 3-18, that has this new probability distribution.

Look at the problem of one person correctly choosing. The probability of this is the sum of the probabilities that they roll the number they chose, i.e.

Pr[they get 3 | they chose 3]*Pr[they chose 3] +
Pr[they get 4 | they chose 4]*Pr[they chose 4]+
... +
Pr[they get 18 | they chose 18]Pr[they chose 18]

Think about this. Is the roll affected by their choice? Do the people choose completely at random?

Once you've figured this bit out all that's left to do is determine what the probability of a correct choice happening 5 independent times given that you know the probability of it happening once.

pive said:

The players choose five numbers between 3-18. The dice is rolled three times and the numbers will be summed so the result will be between 3-18. However, in this game the dice will be rolled in total fifteen times but the same rule applies as told in the last statement, so we will have five summed up groups of numbers between 3 and 18. For example:
Dice is rolled fifteen times: 1 + 5 + 3 + 6 + 4 + 4 +2 + 1 + 6 + 6 + 5 + 1 + 4 + 2 + 5
These numbers will be now put in groups of three in the order as they were rolled and summed up:
(1 + 5 + 3); (6 + 4 + 4); (2 + 1 + 6); (6 + 5 + 1); (4 + 2 + 5) = 9; 14; 9; 12; 11.
The question is, how can I calculate the probability of the players that they have chosen five numbers between 3-18 so they are the same numbers as the summed up three number groups from dice?

Click to expand...

It seems to me that whoever wrote this question was making busy work for some one.
Here is a useful webpage. On it we expanded \(\displaystyle {\left( {\sum\limits_{k = 1}^6 {{x^k}} } \right)^3}\).

Look at the expansion. It tells us all the possible sums from rolling three dice and the number of ways the sums can occur.
For example you see the term \(\displaystyle 10x^{15}\): tells us that there are ten ways to roll a sum of fifteen.
So \(\displaystyle P(\text{Sum}=15)=\dfrac{10}{6^3}\)

You can change the exponent 3 to 5 click the \(\displaystyle \boxed{=}\) see the distribution of rolling five dice.