# After throwing $5$ dice, what is the probability that they don't have all the same number?

We have $$5$$ fair dice. What is the probability that they don't have all the same number?

Attempt

We must have at least one die that is different of the others. My argument is:

Let distinguish the dice: we give to each dice a letter $$a,b,c,d,e$$. So, we need to have either $$a$$ different of every other die, or $$b$$ different of every other die or $$c$$ different of every other die or $$d$$ different of other dice or $$e$$ different of other dice. So, for $$a$$ different of every other dice, we hace $$6$$ possibilities for $$b$$, $$6$$ for $$c$$, $$6$$ possibilities for $$d$$, $$6$$ possibilities for $$e$$ and $$5$$ possibilities for $$a$$. That make $$6^45$$ possibilities. For $$b$$ different of the other, we have the same: $$6^4\cdot5$$ possibilities as well, for $$c$$ the same... and so at the end, we get a probability of $$\frac{6^4\cdot5}{6^5}\cdot5>1$$, and thus it's wrong. The final result is $$\frac{6^4}{5}\cdot6^5$$ but I don't understand why my argument is not correct.

• hint. Think about the event they all have the same number. Then take the complement.
– EDZ
Commented Jan 10, 2019 at 21:56
• @EDZ: I want to understand why my argument is not correct. By the way, I would say the same : 6 possibilities for the first dice, the others are determinate, or 6 possibilities for the 2nd dice, the other are determinated,... at the end I find $\frac{6\cdot 5}{6^5}$ Commented Jan 10, 2019 at 22:01
• Well I would calculate probability all are the same, which is just 1/6^4, then subtract 1 from that. But I believe the more important lesson here is when you get a probability over 1, you look for where you counted 2 possibilities that are not mutually exclusive. In this case A being different from B automatically makes B being different from dice A, and you are counting that possibility twice. Not to mention fixing dice A gives A 6 possibilities but dice B 5 possibilities to be different. (and the rest of the die 5 possibilities each) Commented Jan 10, 2019 at 22:05
• @EDZ Rolling the same number with all die is not the complement event. You are missing the events where you roll $k$ die with the same number with $2\leq k\leq 5$ Commented Jan 10, 2019 at 22:21
• @calculus OP is asking for probability they don't ALL have the same number not ANY. So probability is 1 - probability all have the same number. Unless I'm misreading OP's question or OP worded it wrong.
– EDZ
Commented Jan 10, 2019 at 22:28

The answer is $$\boxed{7770/7776}$$
There are only six ways in which the five dice can each have the same number: they can all equal $$1$$, they can all equal $$2$$, they can all equal $$3$$, etc.
There are $$6^{5}$$ total ways to choose the outcome of the $$5$$ dice, since there are $$6$$ outcomes for the first dice, $$6$$ for the second dice, and so on.
Thus, the number of desirable outcomes equals $$6^{5} - 6 = 7770$$.
The probability of obtaining a desirable outcome is just the ratio of the number of desirable outcomes to total outcomes. This gives us $$7770/6^{5} = 7770/7776,$$ which is the answer.