[pop000]

there a keno game that if you buy card that there are number with 6 digit between 1-6 (include 1 and 6).
every of the digit can get any value between 1-6 (include 1 and 6).
what is the probability to get Palindrome number ? (for example 111111)


thanks.

[ebaines]

For a palindromic number that is 6 digits long you don't care what the first three digits are, only digit 4 = digit 3, digit 5 = digit 3 (**EDIT: this should be "digit 5 = digit 2" as noted in subsequent posts**), and digit 6 = digit 1. The chance that any one of the last 3 digits equals its counterpart is 1 out of 6, right? So now - what is the probabiliy that all 3 of the last 3 digits equals its counterpart? Post back with your answer and we'll check it for you.

[pop000]

Thanks for answer but if I understand you right , you told "6 digits long you don't care what the first three digits are, only digit 4 = digit 3, digit 5 = digit 3, and digit 6 = digit 1" so for example I did this number:
124441, this is not an palindromic number.
What do you say?


Thanks.

[pop000]

well tell me if I get closer:
for the 3 first digit I can get any of the 6 digit so is will be 6*6*6.
for the 3 last digit if I want it to be an palindromic number I have only 1 correct digit in each digit.
so together I get 216 combination and I know that the sum of the combination for a number with 6 digit is 6^6
so the probability to get Palindrome number will be 216/6^6 ?

thank you.

[ebaines]

thanks for answer but if i understand you right , you told "6 digits long you don't care what the first three digits are, only digit 4 = digit 3, digit 5 = digit 3, and digit 6 = digit 1" so for example i did this number:
124441, this is not an palindromic number.
what do you say?


thanks.

Sorry - that was a typo on my part. A palindromic number is a "symmetric" number, where:
Digit 4 = digit 3
Digit 5 = digit 2
Digit 6 = digit 1

Thus 124441 is NOT palindromic, but 124421 is.

[ebaines]

so the probability to get Palindrome number will be 216/6^6 ?

That's right - your answer of 216/(6^6) is equal to (6^3)/(6^6) = 1/(6^3). But I think there's a simpler thought process to go through:

The probability that digit 4 = digit 3 is 1/6.
The probability that digit 5 = digit 2 is 1/6.
The probability that digit 6 = digit 1 is 1/6.
Hence the probability that digit 4 = digit 3 AND digit 5 = digit 2 AND digit 6 = digit 1 is (1/6)*(1/6)*(1/6) = (1/6)^3

[pop000]

OK thank you for helping me :)