The number 1 – 49 supposedly occur with equal probability when the six numbers are picked in the British National Lottery. It does not follow that the digits 0 – 9 occur with equal probability.
The numbers 1 – 49 are listed below.
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1 |
2 |
3 |
4 |
5 |
|
6 |
7 |
8 |
9 |
10 |
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11 |
12 |
13 |
14 |
15 |
|
16 |
17 |
18 |
19 |
20 |
|
21 |
22 |
23 |
24 |
25 |
|
26 |
27 |
28 |
29 |
30 |
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31 |
32 |
33 |
34 |
35 |
|
36 |
37 |
38 |
29 |
40 |
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41 |
42 |
43 |
44 |
45 |
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46 |
47 |
48 |
49 |
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The digits 1 to 4 each appear 15 times, but 1 appears twice in 11, 2 appears twice in 22, 3 appears twice in 33 and 4 appears twice in 44. Counting each of these as one occurrence, the digits 1 to 4 each appear 14 times.
The numbers 5 to 9 each appear 5 times.
The digit 0 appears 4 times.
There are 4*14+5*5+4=85 occurrences altogether.
The probability that 0 will appear just once in six balls is\[6 \times (\frac{4}{85})(\frac{81}{84})(\frac{80}{83}) (\frac{79}{82})(\frac{78}{81}) (\frac{177}{80})\]
.The probability that 0 will appear twice in six balls is
\[\frac{6 \times 5}{2} (\frac{4}{85})(\frac{3}{84})(\frac{81}{83}) (\frac{80}{82})(\frac{79}{81}) (\frac{178}{80})\]
.The probability that 0 will appear three times in six balls is
\[\frac{6 \times 5 \times 4}{2 \times 3} (\frac{4}{85})(\frac{3}{84})(\frac{2}{83}) (\frac{81}{82})(\frac{80}{81}) (\frac{179}{80})\]
.The probability that 0 will appear four times in six balls is
\[\frac{6 \times 5 \times 4 \times 3}{2 \times 3 \times 4} (\frac{5}{85})(\frac{4}{84})(\frac{3}{83}) (\frac{2}{82})(\frac{80}{81}) (\frac{179}{80})\]
.The probability that 0 will appear five or six times in six balls is 0 since there are only four balls with 0 on them
The expected number of 0's is then
\[\begin{equation} \begin{aligned} \sum_1^6 x \times P(x 1's) &= 1
\times 6 \times (\frac{4}{85})(\frac{81}{84})(\frac{80}{83}) (\frac{79}{82})(\frac{78}{81}) (\frac{177}{80}) \\ &+ 2 \frac{6 \times 5}{2} (\frac{4}{85})(\frac{3}{84})(\frac{81}{83}) (\frac{80}{82})(\frac{79}{81}) (\frac{178}{80}) \\ &+ 3 \times \frac{6 \times 5 \times 4}{2 \times 3} (\frac{4}{85})(\frac{3}{84})(\frac{2}{83}) (\frac{81}{82})(\frac{80}{81}) (\frac{179}{80}) \\ &+ 4 \times \frac{6 \times 5 \times 4 \times 3}{2 \times 3 \times 4} (\frac{5}{85})(\frac{4}{84})(\frac{3}{83}) (\frac{2}{82})(\frac{80}{81}) (\frac{179}{80}) \end{aligned} \end{equation}\]
The probability that 1 will appear just once in six balls is
\[6 \times (\frac{14}{85})(\frac{71}{84})(\frac{70}{83}) (\frac{69}{82})(\frac{68}{81}) (\frac{167}{80})\]
.The probability that 1 will appear twice in six balls is
\[\frac{6 \times 5}{2} (\frac{14}{85})(\frac{13}{84})(\frac{71}{83}) (\frac{70}{82})(\frac{69}{81}) (\frac{168}{80})\]
.The probability that 1 will appear three times in six balls is
\[\frac{6 \times 5 \times 4}{2 \times 3} (\frac{14}{85})(\frac{13}{84})(\frac{12}{83}) (\frac{71}{82})(\frac{70}{81}) (\frac{169}{80})\]
.The probability that 1 will appear four times in six balls is
\[\frac{6 \times 5 \times 4 \times 3}{2 \times 3 \times 4} (\frac{14}{85})(\frac{13}{84})(\frac{12}{83}) (\frac{11}{82})(\frac{71}{81}) (\frac{170}{80})\]
.The probability that 1 will appear five times in six balls is
\[\frac{6 \times 5 \times 4 \times 3 \times 2}{2 \times 3 \times 4 \times 5} (\frac{14}{85})(\frac{13}{84})(\frac{12}{83}) (\frac{11}{82})(\frac{10}{81}) (\frac{171}{80})\]
.The probability that 1 will appear six times in six balls is
\[\frac{6 \times 5 \times 4 \times 3 \times 2 \times 1}{2 \times 3 \times 4 \times 5 \times 6} (\frac{14}{85})(\frac{13}{84})(\frac{12}{83}) (\frac{11}{82})(\frac{10}{81}) (\frac{19}{80})\]
.The expected number of 1's is then
\[\begin{equation} \begin{aligned} \sum_1^6 x \times P(x 1's) &= 1 \times 6 \times (\frac{14}{85})(\frac{71}{84})(\frac{70}{83}) (\frac{69}{82})(\frac{68}{81}) (\frac{167}{80}) \\ &+ 2 \times \frac{6 \times 5}{2} (\frac{14}{85})(\frac{13}{84})(\frac{71}{83}) (\frac{70}{82})(\frac{69}{81}) (\frac{168}{80} \\ &+ 3 \times \frac{6 \times 5 \times 4}{2 \times 3} (\frac{14}{85})(\frac{13}{84})(\frac{12}{83}) (\frac{71}{82})(\frac{70}{81}) (\frac{169}{80}) \\ &+ 4 \times \frac{6 \times 5 \times 4 \times 3}{2 \times 3 \times 4} (\frac{14}{85})(\frac{13}{84})(\frac{12}{83}) (\frac{11}{82})(\frac{71}{81}) (\frac{170}{80}) \\ &+ 5 \times \frac{6 \times 5 \times 4 \times 3 \times 2}{2 \times 3 \times 4 \times 5} (\frac{14}{85})(\frac{13}{84})(\frac{12}{83}) (\frac{11}{82})(\frac{10}{81}) (\frac{171}{80}) \\ &+ 6 \times \frac{6 \times 5 \times 4 \times 3 \times 2 \times 1}{2 \times 3 \times 4 \times 5 \times 6} (\frac{14}{85})(\frac{13}{84})(\frac{12}{83}) (\frac{11}{82})(\frac{10}{81}))(\frac{9}{80}) \end{aligned} \end{equation}\]
The calculations are exacly the same for occurrences of the difits 2, 3 and 4. The probability that 5 will appear just once in six balls is
\[6 \times (\frac{5}{85})(\frac{80}{84})(\frac{79}{83}) (\frac{78}{82})(\frac{77}{81}) (\frac{176}{80})\]
.The probability that 1 will appear twice in six balls is
\[\frac{6 \times 5}{2} (\frac{5}{85})(\frac{4}{84})(\frac{80}{83}) (\frac{79}{82})(\frac{78}{81}) (\frac{177}{80})\]
.The probability that 1 will appear three times in six balls is
\[\frac{6 \times 5 \times 4}{2 \times 3} (\frac{5}{85})(\frac{4}{84})(\frac{3}{83}) (\frac{80}{82})(\frac{79}{81}) (\frac{178}{80})\]
.The probability that 1 will appear four times in six balls is
\[\frac{6 \times 5 \times 4 \times 3}{2 \times 3 \times 4} (\frac{5}{85})(\frac{4}{84})(\frac{3}{83}) (\frac{2}{82})(\frac{80}{81}) (\frac{179}{80})\]
.The probability that 1 will appear five times in six balls is
\[\frac{6 \times 5 \times 4 \times 3 \times 2}{2 \times 3 \times 4 \times 5} (\frac{5}{85})(\frac{4}{84})(\frac{3}{83}) (\frac{2}{82})(\frac{1}{81}) (\frac{180}{80})\]
.The probability that 1 will appear six times in six balls is 0 since there are only 5 balls with a five on them.
The expected number of 1's is then
\[\begin{equation} \begin{aligned} \sum_1^6 x \times P(x 1's) &= 1 6 \times (\frac{5}{85})(\frac{80}{84})(\frac{79}{83}) (\frac{78}{82})(\frac{77}{81}) (\frac{176}{80}) \\ &+ 2 \times \frac{6 \times 5}{2} (\frac{5}{85})(\frac{4}{84})(\frac{80}{83}) (\frac{79}{82})(\frac{78}{81}) (\frac{177}{80}) \\ &+ 3 \times \frac{6 \times 5 \times 4}{2 \times 3} (\frac{5}{85})(\frac{4}{84})(\frac{3}{83}) (\frac{80}{82})(\frac{79}{81}) (\frac{178}{80}) \\ &+ 4 \times \frac{6 \times 5 \times 4 \times 3}{2 \times 3 \times 4} (\frac{5}{85})(\frac{4}{84})(\frac{3}{83}) (\frac{2}{82})(\frac{80}{81}) (\frac{179}{80}) \\ &+ 5 \times \frac{6 \times 5 \times 4 \times 3 \times 2}{2 \times 3 \times 4 \times 5} (\frac{5}{85})(\frac{4}{84})(\frac{3}{83}) (\frac{2}{82})(\frac{1}{81}) (\frac{180}{80}) \end{aligned} \end{equation}\]
The calculations are exacly the same for occurrences of the difits 6, 7 and 8.