(Will describe it better later) The cristhmas lottery has been celebrated today in Spain. So I decided to calculate the probability of having a ticket with a palindrome number. The tickets have 5 numbers each from 1 to 9. I got 1,11%. Is my result right?
2026-03-25 09:31:56.1774431116
probability of obtaining palindrome in lottery
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Since there are 9 digits and 5 numbers, the total number of possibilities is $9^5$
For a palindrome, there are 9 options for the first digit, 9 options for the second digit, 9 options for the third digit and 1 option for the last two. So there are $9^3$ palindromes.
So the probability is $\frac{9^3}{9^5}*100 = \frac{100}{9^2} = 1.23$% $(2dp)$
So no, 1.11% is not correct.