I’m sure a stats major will be able to do a better job at this, but I play D&D which uses 20 sided dice and have been through the odds of rolling 2 20s or 1s. I’m in a stats class but admittedly still working on it and have some spotty knowledge.

So odds of rolling a 6 is 1/6 (one in 6). I believe the odds of doing that twice in a row are multiplied fractions. So 1/6*1/6 = 1/36. So you have a one in 6 of rolling a 6, but only a 1/36 of doing it twice.

That much I know. This is where I start to be unsure a bit and am making some assumptions. The odds of rolling 2 - 4s would be the same. So presumably you could take the 1/36 of rolling the 2 4s and say that the other 35/36 is that the second number is anything but a 4. The 35 would then be split between the other numbers evenly since any of them are equally as likely in this scenario, just not a 4. The 6 would then receive 7 of those shares and so a 7/36 chance of rolling a 6 after rolling the 4 or any other number.

It can be a 4 or anything really, but presumably the odds of rolling any specific number right after rolling a different number would be 7/36 on a d6.