What are the odds a random 6-digit string would have at least one digit repeat? That is, not just the same number twice (eg 123453) but paired (123345)?
What are the odds a random 6-digit string would have at least one digit repeat? That is, not just the same number twice (eg 123453) but paired (123345)?
For adjacent digits, 123345 not 123453, I think it's 1−((10×9^5)÷10^6) = 41% chance of at least one adjacent pair. That is, there's 10 choices for the first digit, but then only 9 choices for each succeeding digit.
There are 1000000 6-digit strings, 409510 of those contain at least one pair of adjacent identical digits, so the probability is exactlt 40.951%.
(With numbers this small, you can literally just check every case to be sure. Anything less than about 4 billion is "small".)
a little lower than 50-50.
Because I see that all the time in confirmation codes.
so the odds that a number is the same as the one before it is 1/10. and you've got to dodge that 5 times, so that's 1 minus 9/10 to the 5th power
0.9**5 is the chance of no doubled characters = 59%, so 1-0.9**5 = 41% should be the chance of doubled characters. I *think*. Even if I'm rongg, which is quite likely, it's very roughly 50-50 chance of doubled characters.
aahhhh you're right, i had it right the first time
need to write this stuff down and not try to do it in my head lmao
so like 41%
sorry, 41% that there's *no* repeats, so 59% that there is at least one repeat
Yup. People have terrible intuitions about probability, so things like this are surprising, but true
I'm getting about 85%. There are 10⁶ 6-digit strings and only 10×9×8×7×6×5 = 151200 strings with no duplicates, so a 15.12% chance that you'll get a no-duplicate string and an 84.88 chance of seeing at least one duplicate.
That's the answer to "not just the same number twice (eg 123453)". Adjacent same digit is surprisingly not that much harder.