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Two Dice With A Twist AKA Throwing Two Dice Three Times |
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Studying the probability distribution of scores that
are obtained by throwing two dice is a very common topic. Here we add a small twist
by asking about the highest score from several throws of two dice. The first
part of this unit can be taught in any grade, because all of the necessary
computations can be done even on a four-operation calculator. But Part 2 uses a
graphing calculator, the TI-84.
Part 1
1.1. You are going to throw two dice three times and to record
the highest of the three results, each one of which will vary between 2 and 12.
Before each trial, write down your guess. So the written record may look like
this:
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Trial number : |
1 |
2 |
... |
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My guess : |
11 |
8 |
... |
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Highest of 3 throws: |
8 |
9 |
... |
(1) What is
the best guess? (Which value occurs most often as the highest of 3 throws?)
(2) What is
the distribution of the scores?
(3) What is
the average of this distribution?
Students should work individually or
in small groups. Later their results and conclusions should be discussed. All
of the distributions that students create can be combined into a single
distribution.
1.2.
Now we are going
to compute the theoretical distribution of the best score from three throws of
two dice.
Distribution of the probability p(s) of a score s of one
throw of two dice:
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s |
2 |
3 |
4 |
5 |
6 |
7 |
8 |
9 |
10 |
11 |
12 |
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p(s) |
1/36 |
2/36 |
3/36 |
4/36 |
5/36 |
6/36 |
5/36 |
4/36 |
3/36 |
2/36 |
1/36 |
The cumulative distribution
C(s), which is the probability that the score is smaller than or equal to s. (It is computed by adding the values of p(s).)
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S |
2 |
3 |
4 |
5 |
6 |
7 |
8 |
9 |
10 |
11 |
12 |
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C(s) |
1/36 |
3/36 |
6/36 |
10/36 |
15/36 |
21/36 |
26/36 |
30/36 |
33/36 |
35/36 |
36/36 |
The probability H(s) that s is the highest score out
of three is the probability that all three tosses give scores smaller than or
equal to s, which is C(s)3, minus the probability that all tosses
give scores less than or equal to s -1, which is C(s-1)3. Thus H(s)
= C(s)3 - C(s-1)3.
1.3 Finally we are going to compare the theoretical
distribution with the results of our experiments.
Students should compute H(s) using calculators; they
should round the values to .01, namely, 1%, and create a table for H(s), and
compare it with the results of the experiments that they performed earlier.
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s |
2 |
3 |
4 |
5 |
6 |
7 |
8 |
9 |
10 |
11 |
12 |
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H(s) |
0 |
0 |
0 |
.02 |
.05 |
.13 |
.18 |
.20 |
.19 |
.15 |
.08 |
So, theoretically, the best guess is 9, with 8 and 10 being
almost as good. The mean value of H(s) is 9.058, a little bit more than 9.
Part 2
In this part we are going to investigate how the
distribution H(s) of the highest score depends on the number of times, n, that
we throw the two dice.
We would especially like to know how big n is for
which the best guess is 12, the highest possible score. And we might know how
big n is for which the theoretical probability is 1 that 12 is the highest
score
From 1.2 above we
can see that the general formula for H(s), when two dice are tossed n times, is
H(s) = C(s)n – C(s-1)n.
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Store p(s) in list L1 {0,1,2,3,4,5,6,5 ,4,3,2,1}/36→ L1 ENTER |
Notice the leading 0, now p(s) is stored in L1 (s). |
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Store C(s) in L1 cumSum(L1)→L1
ENTER |
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For any given n, store H(2), H(3), ...,
H(12) in L2, and display its values rounded to .01. |
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1→N: ∆List(L1^N)→ L2:round(L2,2)
ENTER |
Now the leading 0 is important. |
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Edit this line to store other
values in N, and create a table of values of H for different values of n. Below are a few examples |
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A
table for H, for different n from 1 to 189, and for s from 2 to 12
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S |
2 |
3 |
4 |
5 |
6 |
7 |
8 |
9 |
10 |
11 |
12 |
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n=1 |
.03 |
.06 |
.08 |
.11 |
.14 |
.17 |
.14 |
.11 |
.08 |
.06 |
.03 |
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n=2 |
0 |
.01 |
.02 |
.05 |
.10 |
.17 |
.18 |
.17 |
.15 |
.10 |
.05 |
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n=3 |
0 |
0 |
0 |
.02 |
.05 |
.13 |
.18 |
.20 |
.19 |
.15 |
.08 |
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... |
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n=10 |
0 |
0 |
0 |
0 |
0 |
0 |
.03 |
.12 |
.26 |
.34 |
.25 |
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n=11 |
0 |
0 |
0 |
0 |
0 |
0 |
.03 |
.11 |
.25 |
.35 |
.27 |
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… |
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n=17 |
0 |
0 |
0 |
0 |
0 |
0 |
0 |
.04 |
.18 |
.39 |
.38 |
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n=18 |
0 |
0 |
0 |
0 |
0 |
0 |
0 |
.03 |
.17 |
.39 |
.40 |
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n=19 |
0 |
0 |
0 |
0 |
0 |
0 |
0 |
.03 |
.16 |
.39 |
.40 |
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… |
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n=40 |
0 |
0 |
0 |
0 |
0 |
0 |
0 |
0 |
.03 |
.29 |
.68 |
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… |
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n=60 |
0 |
0 |
0 |
0 |
0 |
0 |
0 |
0 |
.01 |
.18 |
.82 |
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… |
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n=188 |
0 |
0 |
0 |
0 |
0 |
0 |
0 |
0 |
0 |
.01 |
.99 |
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n-189 |
0 |
0 |
0 |
0 |
0 |
0 |
0 |
0 |
0 |
.005 |
.995 |
So the
lowest n for which the best guess is 12 is n=18 (see the red .40 above). And
with 189 rolls , the theoretical probability is
approximately .995 that 12 will be rolled.
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