Frequency With Which Digits Appear in National Lottery Balls
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.
1 | 2 | 3 | 4 | 5 |
6 | 7 | 8 | 9 | 10 |
11 | 12 | 13 | 14 | 15 |
16 | 17 | 18 | 19 | 20 |
21 | 22 | 23 | 24 | 25 |
26 | 27 | 28 | 29 | 30 |
31 | 32 | 33 | 34 | 35 |
36 | 37 | 38 | 29 | 40 |
41 | 42 | 43 | 44 | 45 |
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.
