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I saw What is the expected value of the largest of the three dice rolls? and decided to go with a different approach than the ones that were mentioned. I was wondering if someone can point out why my approach is incorrect.

Using the stars and bars method, the number of ways s.t. the max is 6 is

$$ {{2 balls + 6 bins - 1} \choose 2 balls} = 21 $$

The reason I use two balls is because I already put one in bin 6, to ensure 6 is the max. The other two balls can be put into any of the 6 bins.

Using similar logic,

# of ways s.t. max is 5: ${{2 balls + 5 bins - 1} \choose 2 balls} = 15$

# of ways s.t. max is 4: ${{2 balls + 4 bins - 1} \choose 2 balls} = 10$

# of ways s.t. max is 3: ${{2 balls + 3 bins - 1} \choose 2 balls} = 6$

# of ways s.t. max is 2: ${{2 balls + 2 bins - 1} \choose 2 balls} = 3$

# of ways s.t. max is 1: ${{2 balls + 1 bins - 1} \choose 2 balls} = 1$

Total # of ways: ${{3 balls + 6 bins - 1} \choose 3 balls} = 56$

We thus get an expected value of $\frac{126 + 75 + 40 + 18 + 6 + 1}{56} = \frac{19}{4}$, which is incorrect. What part of my logic is wrong?

user3230239
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    There are more than 3 ways to get max 2. 222,221,212,122,112,etc You count for example 221 and 212 as one way, but getting two 2, one 1 happens in 3 ways, so you would need weights tto use stars and bars. – coffeemath Jul 12 '23 at 20:06
  • That's only if the die are distinguishable. I'm assuming the die are indistinguishable in this approach. I'm wondering why I can't treat them as indistinguishable. – user3230239 Jul 12 '23 at 20:12
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    There are no such thing as indistinguishable dice. – coffeemath Jul 12 '23 at 20:14
  • Well there are at least not indistinguishable dice in the world we live in. Two such would need to have the same history in space-time and the exact same atomic weights and other coincidenes. [However one can of course assume non-distinguishable "theoretical" dice and for them probabilities would likely differ from what usually is assumed about fair dice. – coffeemath Jul 12 '23 at 21:18

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