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By "unique results" I mean that the order of the two resulting numbers does not matter - a 1 and a 2 is the same as a 2 and a 1. I have used a spreadsheet to brute force the answer of 21, and inspection of the spreadsheet makes it clear why this is the correct answer. However, 21 = 7C2; given the nature of the problem, I cannot believe this is a coincidence, but for the life of me I cannot figure out where the 7 comes from when you are rolling two 6-sided dice. Thanks in advance for shining light on my darkness...

SB Silverhammer
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    Look up "combinations with repetitions". If you are counting the number of ways to choose $r$ times from $d$ options, repetitions allowed and order does not matter, the formula is $(r+d-1)Cr$. Here, $d=6$, $r=2$. A proof is in Wikipedia, "Theorem 2". – Arturo Magidin Sep 23 '23 at 02:00
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    Please clarify your specific problem or provide additional details to highlight exactly what you need. As it's currently written, it's hard to tell exactly what you're asking. – Community Sep 23 '23 at 02:04
  • A proof also here. – Arturo Magidin Sep 23 '23 at 02:06
  • The unique results from rolling two dice are ${2,3\dots,12}$ which is a set of size $11$. Where are you getting $21$? – John Douma Sep 23 '23 at 02:55
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    @JohnDouma In effect you do not treat rolls $(x,y)$ and $(y,x)$ as different. So one count method is $6+\frac{6^2-6}2 =21$. Since this is ${n \choose 1}+{n\choose 2}={n+1 \choose 2}$ when $n=6$, there is no surprise here – Henry Sep 23 '23 at 03:03
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    @JohnDouma Most people consider rolling a double 5 a different outcome than rolling a 4 and a 6. Just ask the Craps table attendant... Or someone playing backgammon. – Arturo Magidin Sep 23 '23 at 03:09
  • I see. Since we only count combinations we get $(1,1)\dots (1,6)$ which equal six rolls where a one appears and $(2,2)\dots (2,6)$ which is five all the way to $(6,6)$ which is one to get $6+5+\dots +1=\frac{6\cdot 7}{2}$ is the result you got. – John Douma Sep 23 '23 at 04:33
  • And yet another proof – Arturo Magidin Sep 23 '23 at 11:28
  • I see. This is a multi-subset problem. I did not make the connection. – SB Silverhammer Sep 23 '23 at 13:12

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