are the past outcomes relevant in calculating future odds.

Question

If I rolled a dice and got on a $6$ the chance of that was obviously $1/6$. So if I now roll again, will my chance of rolling another 6 be again 1/6 or $1/36$?

Another example, lets say I rolled yesterday $100$ times a $6$ in a row the odds of that would have been $1,5306467074865063414445284410446\times 10^{-78}$.

So if i then go to bed wake the next day up and wanna role a 6 are the odds gonna be for that throw $1/6$, or $1,5306467074865063414445284410446\times 10^{-78}$ times $1/6$ ?

Thanks for anyone who responds :)

Answer 1

The events are independent - meaning they do not affect one another.

The word that's key here is "both", just as @user1001001 pointed out. Let's say I flip a coin $3$ times consecutively. What are the possible outcomes? ${HHH}$, ${HHT}$, ${THT}$... etc etc. How many possible outcomes are there? Well, it's

$${2\times 2\times 2=8}$$

There's $8$ possible outcomes. But only $1$ out of these $8$ outcomes is a sequence where I simply get three heads (${HHH}$). So the probability of getting all three as heads consecutively is "$1$ out of $8$" = ${\frac{1}{8}}$.

We can actually show that we have no predictive power. We can use the formula

$${P(\text{A given B})=\frac{P(\text{A and B})}{P(\text{A})}}$$

If you replace ${A}$ with "the third toss being heads" and ${B}$ with "the first two tosses were heads" you end up with

$${P(\text{the last toss is heads GIVEN the first two were heads})}$$

$${=\frac{P(\text{the last toss is heads and the first two tosses were heads})}{P(\text{the first two tosses were heads})}=\frac{\frac{1}{8}}{\frac{1}{4}}=\frac{1}{2}}$$

In other words - just because you got two heads to begin with, the chance of you getting another heads is still ${\frac{1}{2}}$.

The point to take away is context. If we are looking at the overall situation (the three tosses as a sequence ${THT}$ etc) the probability of getting the sequence ${HHH}$ is low. But even though this is the case - it doesn't allow you to predict at runtime what will happen, no matter how counter-intuitive it may seem, since if you've already gotten two heads for example - the event has already happened. Since the flips are independent, it still is not going to affect your chances on the last coin flip being heads of ${\frac{1}{2}}$. The "unlikely" part of the sequence has already occurred. Hope this helps.

Answer 2

As long as you claim it is a fair die the chance of a $6$ is $\frac 16$ regardless of the history. Period. End. If you roll $100$ sixes in a row maybe you doubt the assertion that it is a fair die. That would make your first statement false-it is not true that the chance of a $6$ is $\frac 16$. There are a number of unfair dice in the world. Maybe you have found one. In that case the odds are quite high that the next roll will be a $6$ as well. The exact number is difficult to calculate as it depends on your assessment of the population of dice you may be using, but that is not important.