A probability problem by Tai Ching Kan

Let the number of coins removed from each of the stacks A, B and C after I stop, be $x$ , $y$ and $z$ respectively. Since we are interested in the case where stack A is depleted, we need only consider the different scenarios where $x=3$ .

For stack A to be the first stack to have all its coins removed, we also know that $y,z<x$ . This leads us to 9 possible cases: $(x,y,z)=(3,0,0), (3,1,0), (3,0,1), (3,2,0), (3,0,2), (3,1,1), (3,2,1), (3,1,2)$ and $(3,2,2)$ .

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We consider the probability of each case happening:

Case 1: $(3,0,0): 0.4^{3}=0.064$ [Each time we remove a coin from A: $0.4\times0.4\times0.4$ ]

Case 2: $(3,1,0): 3\times0.4^{3}\times0.3^{1+0}=0.0576$ [3 coins from stack A and 1 from B can be removed in 3 different orders: BAAA, ABAA and AABA, but not AAAB since after the third spin the last coin in stack A will be removed, therefore I stop and will not spin a fourth time. We can generalise the multiplier in front (the 3 in this case) to all of our cases: this situation is equivalent to arranging $(x+y+z)$ letters: $x$ A's, $y$ B's and $z$ C's, and that only A can take the last position for the reasons mentioned above. Fixing A in the last position, we can now apply the formula for Permutations with Repeated Items to obtain the multiplier as: $\frac{((x-1)+y+z)!}{(x-1)!y!z!}$

Case 3: $(3,0,1)$ is calculated in the same way as $(3,1,0)$ .

Case 4: $(3,2,0): \frac{((3-1)+2+0)!}{(3-1)!2!0!}\times0.4^{3}\times0.3^{2+0}=6\times0.4^{3}\times0.3^{2}=0.03456$

Case 5: $(3,0,2)$ is calculated in the same way as $(3,2,0)$ .

Case 6: $(3,1,1): \frac{((3-1)+1+1)!}{(3-1)!1!1!}\times0.4^{3}\times0.3^{1+1}=12\times0.4^{3}\times0.3^{2}=0.06912$

Case 7: $(3,2,1): \frac{((3-1)+2+1)!}{(3-1)!2!1!}\times0.4^{3}\times0.3^{2+1}=30\times0.4^{3}\times0.3^{3}=0.05184$

Case 8: $(3,1,2)$ is calculated in the same way as $(3,2,1)$ .

And finally, case 9: $(3,2,2): \frac{((3-1)+2+2)!}{(3-1)!2!2!}\times0.4^{3}\times0.3^{2+2}=90\times0.4^{3}\times0.3^{4}=0.046656$

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Adding the probabilities of each of the 9 cases,

$0.064+2\times0.0576+2\times0.03456+0.06912+2\times0.05184+0.046656=$ $\boxed{0.467776}$

Here is a crisp solution, based on Dynamic Programming (DP).

Let $P(m,n,l)$ denote the probability of emptying stack $A$ first before stacks $B$ and $C$ . We can easily write down the following recursion $P(m,n,l)=0.4P(m-1,n,l)+0.3P(m,n-1,l)+0.3P(m,n,l-1)$ With the boundary conditions $P(0,x,y)=1, P(x,0,y)=P(x,y,0)=0,\hspace{10pt}\forall x>0, y>0$ We need $P(3,3,3)=0.4678$ , which my computer calculates under a second.

Question: What if we start with $10$ coins in each stack (instead of $3$ )? The enumeration approach clearly becomes intractable. However the DP approach produces the result fairly quickly. How about 100 ?