Question

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?

Answer 1

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

Answer 2

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

Answer 3

@CoCoTsoi: Doesn't seem right.

Answer 4

oops I will think again :D

Answer 5

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

Answer 6

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

Answer 7

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

Answer 8

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

Answer 9

hahah .. sometimes you are lucky. :D

Answer 10

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

Answer 11

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

Answer 12

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

Answer 13

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

Answer 14

I don't get the second part :-S

Answer 15

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

Answer 16

the fourth dice becomes irrelevant, no?

Answer 17

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

Answer 18

yup ... you saw through it.

Answer 19

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?

Answer 20

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

Answer 21

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

Answer 22

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

Answer 23

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.)

Answer 24

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

Answer 25

HINT:

For the part exactly 3 different:

\[ \frac 12 \binom{4}{3} \left( 6 \times 5 \times 4 \times 3 \right) \]

Answer 26

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

Answer 27

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

Answer 28

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

Answer 29

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

Answer 30

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

Answer 31

now i'm lost too

Answer 32

Ask Zarkon, he might help :)

Answer 33

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

Answer 34

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

Answer 35

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)

Answer 36

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

Answer 37

for 4 rolls i guess

Answer 38

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

Answer 39

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.

Answer 40

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?

Answer 41

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

Answer 42

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