## 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

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{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}

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{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}

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{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}

The calculations are exacly the same for occurrences of the difits 6, 7 and 8.