Fool's problem of the day,
Four identical dice are tossed simultaneously. What is the probability that at least three of the four nos. shown are different?

- anonymous

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- anonymous

1-(1/6)(1/6)(5/6)(4/6)

- experimentX

4*(6*5*4)/6^4 + 6*5*4*3/6^4

- anonymous

@CoCoTsoi: Doesn't seem right.

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## More answers

- anonymous

oops I will think again :D

- Callisto

6x5x4x3/ 6^4 + 6x5x4x3/6^4 =5/9 @callisto doesn't seem right :S

- anonymous

\( \frac{35}{54}\) is not the right answer. @experimentX

- experimentX

is it (6*5*4)/6^3 + 6*5*4*3/6^4

- anonymous

Yes, Experimentx Congratz :D Now please post the explanation/solution :)

- experimentX

hahah .. sometimes you are lucky. :D

- Callisto

Nah , I was right if i had not changed my mind ?! (-sorry - neglect it-)

- anonymous

The harder I work, the luckier I get ~Samuel Goldwyn.

- experimentX

well, ... i guess i have to agree. thank you for the problem. i am looking forward for others.

- Callisto

The harder i work , the more mistakes i make ~ Me :P
Thanks too !!

- anonymous

I don't get the second part :-S

- experimentX

at least implies ... you have to consider the probability of 4 different also

- anonymous

the fourth dice becomes irrelevant, no?

- anonymous

The second part is self evident, in the first part he is considering 6^3 instead of 6^4. I think that's what he is referring as luck :P

- experimentX

yup ... you saw through it.

- Callisto

i don't understand the first and the second part
I thought
the first part
6x5x4x6 / 6^4
for the last one can be any numbers
the second part,
6x5x4x3 / 6^4
for the last number can only be the number that has been 'selected'
Is it like that?

- anonymous

6 / 6 cancel each other out, you could rewrite it 5*4/6^2. But I'm still confused about the second part

- experimentX

" .. probability that AT LEAST three of the four nos .." .. you also have to add the probability of getting 4 different nos

- Callisto

Ah, i got it reversed :S, thanks experimentX :)

- anonymous

ok, so, the important part is simply that 3 numbers are different, since the 4th can be whatever? so that would mean simply 6*5*4/6^3? (God I hate stats.)

- anonymous

No, that's not right, it's giving the right answer but it's the correct approach.

- anonymous

HINT:
For the part exactly 3 different:
\[ \frac 12 \binom{4}{3} \left( 6 \times 5 \times 4 \times 3 \right) \]

- anonymous

I would like to rephrase/edit "correct" to "clear"

- anonymous

ohhhh, ok, my piece of paper seems to be telling me something. I'm getting there lol

- Callisto

Sorry, I think I might be the one who still don't understand now :S

- anonymous

turns out I got the same result again... ok, if all 4 are different, the probability is 6*5*4*3/6^4. for exactly 3 different, it's 6*5*4*3/6^4. if you shave everything off of that, I still am getting 5*4/6^2

- anonymous

if you don't take into account the 4th dice, you get 6*5*4/6^3, but you can't then add 6*5*4*3/6^4, you're counting stuff twice

- experimentX

now i'm lost too

- anonymous

Ask Zarkon, he might help :)

- experimentX

I think @m_charron2 might be right ... the probability of getting exactly 3 different is 6*5*4*3/6^4 and of exactly 4 different is 6*5*4*3/6^4

- experimentX

adding up gives again 6*5*4/6^3

- anonymous

Meh, nah, I think I get it. I understand that, to get 3 different ones, you need 6*5*4/6^3, and that is because your fourth dice is hidden in there as 6/6 which you don't put. the fourth dice is already accounted for in the first part of the equation, because the 3/6 where it's different and the 3/6 where it's the same. Looks like a conceptual error. Either you don't check the 4th dice, which gives you 6*5*4/6^3 or you check it and you get 2(6*5*4*3/6^4)

- anonymous

Now, I'm getting myself a big coffee. That obscure part of my brain was all rusty.

- experimentX

for 4 rolls i guess

- Callisto

But then it's different from the answer given by FoolForMath :(

- anonymous

Okay, here are some more hints, if we consider two dices then (6,5) and (5,6) are considered two cases.
but (6,6) and (6,6 are considered to be the same.

- anonymous

oh... I,m completely boggled in trying to understand it, but yeah, you were right... is it something like :
(6*6*5*4+6*5*6*4+6*5*4*6+6*5*4*3)/6^4?

- anonymous

4*(6*5*4)/6^4 + 6*5*4*3/6^4

- anonymous

Case where 4 are different:
6 for first die,
*5 for second, only 5 different options left
*4 for third, same logic as above
*3 for fourth
Number of situations where 4 die rolled all have different readings. Dividing by the number of possible outcomes, 6^4 (6 for each die), it becomes 6*5*4*3/6^4 = 5*4*3/6^3 = 60/216 = 5/18
Case where 3 are different:
6 for first die,
*5 for second,
*4 for third,
*6 for fourth because it doesn't matter what it is.
Same as above, simplifies to 20/36 = 5/9
5/9 + 5/18 = 5/6

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