Answer: Approaches to 1.
Step-by-step explanation:
If there are only 3 dice used, then the only chance that the player has to win is when the 3 dice have the same outcome, 6.
The probability will be:
p = (1/6)^3 = 1/216.
Now, if we add one more dice, we still need at least 3 sixes to win, but the other dice can have any other value. so now the probabilities are:
dice 1---- outcome = 6, prob = 1/6.
dice 2---- outcome = 6, prob = 1/6.
dice 3---- outcome = 6, prob = 1/6.
dice 4---- outcome = any number, prob = 1.
The probability for this arrangement is still:
p = 1/216.
But now we have permutations!.
The dice that can be any number has 4 possible positions, so the actual probability will be:
P = 4*p = 4/216.
Now remember that if we have N elements, the total number of combinations of K elements ( N ≥ K) is:
[tex]C(N, K) = \frac{N!}{(N - K)!K!}[/tex]
if we add other dice, then we will have 5 dices, and 2 of them that can not be 6 that can take any position, then the number of combinations will be:
[tex]C(5, 2) = \frac{5!}{(5 - 2)!2!} = \frac{5*4}{2} = 10[/tex]
Then the probability will be:
P = 10*p = 10/216.
So we can start to see a pattern here, if we have N dices, we still only need 3 of them to be strictly 6, then we have (N - 3) dices that can be any number.
Then the probabilty of winning if you have N dices is:
P = C(N, N - 3)*p = C(N, N - 3)*(1/216)
Then as N increases, we will see that the probability tends to 1, (it actually grows larger than that, but remember that the probability is a number between 0 and 1, so the maximum is 1)
Why? well... if you roll a lot of dice, suppose 1000 of them, is really likely to have at least 3 sixes in there, so as the number of dice increases, also does the probability.
