

Die
Name: Charles
Status: N/A
Age: N/A
Location: N/A
Country: N/A
Date: N/A
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 nonLT3 dice. Total
number of possibilities, FOR ANY GIVEN PAIR OF LT31, LT32 DICE
W2a = 2 x 4^8 x 2 = 262,144
Additionally, 10 possible choices for LT31, and for each such
choice 9 possible choices for LT32, for a total of
W2b = 10 x 9 = 90
possible sets of LT31 and LT32. 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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Update: June 2012

