A gambler with 3 alike dice. Here's the problem:- We are given 3 alike dice and (this problem had multiple parts when I tried it, Here I am taking only one scenario) it is said that we will win the game if the sum of numbers appearing on the dice is less than 11(strictly).We are given only a single throw(in which you throw all 3 alike dice together). We are required to find the probability that we win.
Here is how I approached the problem.Lets 1st of all assume that the dice arn't alike but distinct so the 3 dice can show numbers like (113),(131),(311)
Then I thought that "simply because you cant distinguish between the dice doesn't mean that occuring of of the sum 5 should include only 113 and 122. ie the probability that the sum 5 occurs shouldn't be simply 2/216"
That made me realize that I should treat the dice as distinct So I went ahead with this idea of treating dice as distinct. This follows that I should calculate all possible combinations of nubers on 3 dices with sum less than 11. hence only possilibilties of sum is {3,4,5...10}
to calculate this i used beggars method:-
$$x_1+ x_2+x_3 =y$$ where $y\subset{3,4,5...10}$ and $x_1,x_2,x_3$ are all greater than 1
solving this for all y in the set yeilds:-
$${9\choose2}+{8\choose2}+{7\choose2}+{6\choose2}+{5\choose2}+{4\choose2}+{3\choose2}+{2\choose2}$$
Ok so this yeilds${10\choose3}$=120 ie the total no of ways =120 hence probability =120/216. But it comes out be 1/2. Kindly tell where am i wrong.
 A: I don’t understand where you’re getting the various $\binom k2$ components of your summand, but there are a couple of different ways to approach the problem.  The easiest is to use symmetry.  Note that for each die, $n_i$ and $7-n_i$ are equally likely.  Thus, $\sum n_i$ and $21-\sum n_i$ are equally likely.  That means the numbers $3$ through $10$ and the numbers $11$ through $18$ are equally likely, and those two cases exhaust the sample space, so your probability must be $\frac 12$.
Alternatively, the number of solutions in positive integers of $n_1+n_2+n_3 \leq 10$ with each $1 \leq n_i \leq 6$ is the same as the number of solutions in non-negative integers of $m_1+m_2+m_3 \leq 7$ with each $m_i \leq 5$, which is the same as the number of solutions with $m_i \leq 5$ of $m_1+m_2+m_3+z=7$.  Use stars-and-bars to see that there are $\binom{10}{3}=120$ unconstrained solutions to this equation.
From this, you have to subtract the number of solutions for which $m_i \geq 6$ for some $i$.  It’s easy to see that for each $i$ there are $4$ “forbidden” solutions, for a total of $4 \cdot 3=12$.  Thus, the total number of permissible solutions is $120-12=108$, which is exactly half of $216$.
It looks like you may have overlooked the constraint that no single die can give a result greater than $6$.
A: The roll of three dice $(a,b,c)$ where $1 \le a,b,c \le 6$ is equally likely as the roll $(7-a,7-b,7-c)$. So if $X=a+b+c$, then $P(X=x)=P(X=21-x)$; i.e., the distribution of the sum of the three dice is symmetrical about $21/2$.  Consequently, $P(X \le 10) = P(X \le 21/2) = 1/2$.
A: A very simple approach which struck me is literally to count desired cases since the cases are limited and easy to count, as below:
$1$. numbers on the dice are different:
$$(1,2,3),(1,2,4),(1,2,5),(1,2,6),(1,3,4),(1,3,5),(1,3,6),(1,4,5),(2,3,4),(2,3,5);$$
hence in this state we have $10\times6=60$ cases.
$2$. two of numbers on the dice are the same:
$$(1,1,2),(1,1,3),(1,1,4),(1,1,5),(1,1,6),(2,2,1),(2,2,3),(2,2,4),(2,2,5),(2,2,6),$$
$$(3,3,1),(3,3,2),(3,3,4),(4,4,1),(4,4,2);$$
hence in this state we have $15\times3=45$ cases.
$3$. all numbers on the dice are the same:
$$(1,1,1),(2,2,2),(3,3,3);$$
hence in this state we have $3\times1=3$ cases.
Therefore, the result is $\frac{60+45+3}{216}=0.5$.
