Probability, 3 dice
The solution's doing some multinomial thing I don't understand
14 Replies
@Apu
Note for OP
+solved @user1 @user2... to close the thread when your doubt is solved. Mention the users who helped you solve the doubt. This will be added to their stats.Uhhh yeah multinomial is how I learnt it.
Coefficient of x^k in the expansion of (x¹+x²+x³+x⁴+x⁵+x⁶)³
Basically, the number of ways you can get k by choosing from 1,2,3,4,5,6 a total of three times
well ok then
do you know a good source to learn multinomial
Wait for Lance
Ig
fair enough :kekw:
Or someone else
This i actually learnt from Cengage
oh lmao valid
Cengage Maths theory is unironically in depth
The problems, I have no clue, I never solved them
everyday i wish i didnt have a shit math teacher 👼
bet bet
ill look it up
This can be done without multinomial by beggar's method also.
Let the number on dice be a+1,b+1,c+1 where a,b,c>=0
Then a+b+c=k-3, so number of ways=(k-1)C2 which is (k-1)(k-2)/2, number of ways of 3 distinct numbers on dice are 216.
So probability is (k-1)(k-2)/432.
Oh right. Another way to think about it is: If there are some extraneous cases (for example a case where k=9), we would need to subtract those cases, which would occur as a separate term, which doesn't happen here. [Also notice that for numbers being non-zero that max. value of one of a,b and c can be 6]
(Not directly related to question, but just to think about if we can use beggar's method, prior to solving)
Yeah that's what I was thinking. Using just the simple beggar's method, you can't set upper bounds for a,b,c to 6, and it'll be up to a max of k.
ah yep
revised beggar's method
+solved SirLancelotDuLac Opt
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