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Question:
If a die is rolled 10 times

1. how many ways can this be done such that the first and last roll results in a number less than three

2. in how many ways can this be done such that exactly two rolls result in a number less than three

3. how many ways can this be done such that at least one will result in a number less than three.



Replies:
What fun!

(1) 2 ways to roll the first die. 6 ways to roll the 2nd through 9th,
and 2 ways to roll the 10th. Total number of possibilities

W1 = 2 x 6^8 x 2 = 6,718,464

Total number of ways of rolling 10 dice

W = 6^10 = 60,466,176

Probability of this result on any one trial

P1 = W1/W = 1/9 = 0.11

(2) 2 ways to roll the first LT3 die, 2 ways to roll the second LT3 die, 4 ways to roll each of the remaining 8 non-LT3 dice. Total number of possibilities, FOR ANY GIVEN PAIR OF LT3-1, LT3-2 DICE

W2a = 2 x 4^8 x 2 = 262,144

Additionally, 10 possible choices for LT3-1, and for each such choice 9 possible choices for LT3-2, for a total of

W2b = 10 x 9 = 90

possible sets of LT3-1 and LT3-2. Total number of possibilities

W2 = W2a x W2b = 23,592,960

Probability of this result on any one trial

P2 = W2/W = 0.39

(3) Total number of ways of rolling 10 dice such that NONE ends up less than 3

W3a = 4^10 = 1,048,576

Total number of ways for rolling 10 dice such that at least one ends up less than 3:

W3 = W - W3a = 59,417,600

Probability of this result on any one trial

P3 = W3/W = 0.98

Dr. C. Grayce


All that matters here are the first and last roll. The eight intervening rolls have no effect. Now, the rest of your question is vague. Do you mean that the first and last roll are each less than three? In that case, there are two ways fir the first roll (1 and 2) and two for the last, for a total of four (2 x2). Do you mean that the SUM of the first and last rolls is less than three? In that case, there is exactly one way: (1,1).

2. in how many ways can this be done such that exactly two rolls result in a number less than three?

This means that two rolls are less than three and the remaining eight are three or higher. For any given roll, there are two ways to get a number less then three and four ways to get a number three or higher. If you pick any particular two rolls that must be less than three (as you did in the first question), there are four ways to do this. The number of ways available to the eight remaining rolls to all be three or higher is 4^8. So, the number of ways for the first and last rolls to be less than three and the remaining rolls to be three or higher is 4 x 4^8, or 4^9. Now, that's not quite your question. You want to know how many ways to get two rolls less than three and eight rolls three or higher, never mind which particular rolls get each score. One way to think about this is to say that we know the number of ways to have any particular two rolls < 3 and the rest „ 3. All that we then need to figure out is how many ways there are to arrange which particular rolls are the ones < 3. Once we get this number, we can multiply by 4^9 to find the answer to your question. Let's seehow this works. If the first roll is < 3, which other ones can also be < 3? Obviously, any of the remaining nine. If the first roll is „ 3 and the secons is <3, how many others can be <3? This is 8 (3,4,5,6,7,8,9, or 10). And so on up to the first roll that is < 3 being roll 9 (one way). So this gives us (9 + 8 + 7 +6 +5 +4 +3 +2 +1) x 4^9, or 45 x 4^9.



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