Two Dice Game - Decision Making

Question

Imagine the following game: Each player gets a board with the numbers 2-12 on it, and 15 tokens. The player need to spread the tokens on the board. It is legal to put all on the same number, and it is possible to spread in any way. In each round, two dice are thrown and the sum is calculated. If a player has any tokens on the number of the observed sum, one token is removed from that number. The first player to remove all tokens wins.

My question, what is better: Putting all tokens on 7, or spreading them between 6,7,8. I do know how the probability function of the sum looks like (i.e., the probability of the sum being 7 is 6/36, etc...). Thank you.

Answer 1

If we put all $15$ tokens on $7$, then the expected number of throws needed until all tokens are removed is $$ \frac{15}{1/6} = 90. $$ Suppose we start with $5$ tokens on each of $6,7$ and $8$. As we throw the dice, we remove at most one token. We let the triple $\langle a,b,c \rangle$ represent the current state, where $a$ is the number of tokens left on $7$, $b$ is the number of tokens left on $6$ and $c$ is the number of tokens left on $8$. We then have a Markov chain: with each throw, we either stay in the same state (when we throw something other than a $6$,$7$ or $8$, or throw a number with no tokens left on it), or we move to another state based on the throwing a $6$, $7$, or $8$. By creating the transition matrix for this chain, we can calculate the expected number of throws until we are in the $\langle 0,0,0 \rangle$ state.

Here is some GP/PARI code:

{
startval=5;
state(i,j,k) = 1+i*(startval+1)^2+j*(startval+1)+k;
}
{
sizeo=(startval+1)^3;
A=matrix(sizeo,sizeo);
for(a=0,startval,for(b=0,startval,for(c=0,startval,\
r=state(a,b,c);\
print(a," ",b," ",c," ",r);\
notstay=0;
if(a>=1,A[r,state(a-1,b,c)]=1/6;notstay=notstay+1/6);\
if(b>=1,A[r,state(a,b-1,c)]=5/36;notstay=notstay+5/36);\
if(c>=1,A[r,state(a,b,c-1)]=5/36;notstay=notstay+5/36);\
A[r,r]=1-notstay;
)));
A[1,1]=1;
}

This creates the $216 \times 216$ transition matrix $A$, from which we can extract $B$, the matrix we get by truncating the first row and column of $A$.

Calculating $N=(I-B)^{-1}$ yields a matrix with lots of useful info. In particular, the sum of the $215$th row of $N$ yields the expected number of rolls until absorbtion, which is exactly $$ \frac{3944780087497523445776943}{82962908926146261286912} \approx 47.5487195248803794\dots $$ Since this is much less than $90$, we conclude that this is a much better stategy than placing all tokens on $7$.

A similar approach can be used for other starting token placements. My earlier comments suggests that we can do even better by placing the tokens uniformly and symmetrically around $7$, at least to a point.