Bunuel wrote:
Aki wrote:
BDSunDevil wrote:
Say: only 1st and 2nd show the same face
possible outcome 6^4
1st dice = 6c1=6
2nd dice= 1c1 [same as 1st]=1
this can happen in 6 different ways
3rd dice= 5c1=5
4th dice= 4c1=4
(6*1*6*5*4)/6^4=5/9
I also did it this way. Now, the only difference btw Bunuel's method and this one is that we use 6C1 instead of 4C2. Can anybody help me understand if this approach is wrong in any way?
It's not clear what does the red part in the above solution mean.
If it means that there are total of 4C2=6 ways to chose which two show the same face then the logic is correct.
Well I'm not sure about BDSunDevil, but for me the red part means : there are 6 different ways the 1st and 2nd dice can correlate.
i.e
if both D1 and D2 :1
OR if both D1 and D2 :2
and so on till 6.
Let me further elaborate:
1. We know that the both die can have one of the following numbers i.e 1-6.
So possible combinations for D1 and D2: 6C1*1C1 (since D1 and D2 will have to be the same number)
2. Now, the 3rd dice can be any of the remaining 5 numbers. So, possible combinations: 5C1
3. Finally, the 4th dice can be any of the remaining 4 numbers. So, possible combinations: 4C1
From 1,2 and 3: Total Favorable Combinations: 6C1*1C1*5C1*4C1
Total Outcomes(as per your logic): 6^4
So, P(F)= (6C1*1C1*5C1*4C1)/ (6^4)
I have no problems with your logic either, just want to know if this one is possible also.