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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.

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    When you say "the token is being removed", do you mean *one* token is removed, or do you mean *all* tokens on that number are removed? Thanks! – 2017-02-04
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    One token is being removed. This is like the game "bingo" that kids play – 2017-02-04
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    I´ll delete my answer till I get the right idea. So other users notice your question as unanswered. But I have to say that the exercise can be misunderstood very easily. – 2017-02-04
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    @user2899944 I rewrote your question a little bit to make it clearer. Please check that it describes the game correctly. – 2017-02-04
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    Perfect. Than you. – 2017-02-04
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    All on 15 takes 90 throws on average; with a simulation: 5 on each of 6, 7 and 8 takes $\sim 47.5$; 3 on each of 5,6,7,8,9 takes $\sim 40.9$; $2$ on each of 4, 5, 6, 8, 9, 10 and $3$ on $7$ takes $\sim 40.3$. – 2017-02-04

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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.