For three fair six-sided dice, the possible sum of the faces rolled can be any digit from 3 to 18.
For instance the minimum sum occurs when all three dices shows 1 (i.e. 1 + 1 + 1 = 3) and the maximum sum occurs when all three dces shows 6 (i.e. 6 + 6 + 6 = 18).
Thus, there are 16 possible sums when three six-sided dice are rolled.
Therefore, from the pigeonhole principle, <span>the minimum number of times you must throw three fair six-sided dice to ensure that the same sum is rolled twice is 16 + 1 = 17 times.
The pigeonhole principle states that </span><span>if n items are put into m containers, with n > m > 0, then at least one container must contain more than one item.
That is for our case, given that there are 16 possible sums when three six-sided dice is rolled, for there to be two same sums, the number of sums will be greater than 16 and the minimum number greater than 16 is 17.
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1.3 bc 8 miss 3 equals 24
Answer:
Step-by-step explanation:
These answers are in order btw!
1=2
2=4
4=6
8=10
Table 2:
1=3
2=6
3=9
4=12
5=15
I’ll finish table 3 and put it in the commenta since this one will take me longer
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