# How do chances change each time?

This is a beginner question but bear with me, let's say i have thrown a dice 5 times, and every time the result was 1. The next time i throw it what are the chances/probabilities it will be 1 again? Is it still 1/6? If not how is this calculated?

What i am trying to understand is whether the past is relevant for the next throws or not. If someone could point out a place where i can further read about it, it would very much appreciated.

Thanks.

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no, the past is not relevant; each roll is independent. the probability of rolling a 1 is 1/6. – yoyo Apr 5 '11 at 1:51
If you're totally confident that the die is fair, then each roll is independent. In fact, as long as you believe that the die lands on 1 with some probability $p$, the past is irrelevant, and each roll result in a 1 with the same probability $p$. But I'm thinking: what if the probability itself is not certan, i.e., $p$ is random?? – GWu Apr 5 '11 at 1:55
@GWu I don't think you can have a random probability, because of the Law of Averages. – Leonardo Fontoura Apr 5 '11 at 2:00
As Wikipedia says: Law of Averages is a lay term''. It's not a mathematical principle and does not make sense to me. – GWu Apr 5 '11 at 2:09
It's a lay term, but there is a theorem backing it up. If you can, refer to the book in my answer, chapter 2.2, there is a proof to it there. – Leonardo Fontoura Apr 5 '11 at 2:23

If you threw a fair dice $5$ times and got a $1$ every single time, then the event comprised of the $5$ throws together has a probability of $\frac{1}{6^5}$. The odds that the next throw results in a $1$ is $\frac{1}{6}$, because the events are independent.

In the case of independent events, it doesn't matter what has happened in the past. Throwing the dice now doesn't affect how it will behave in the future, unlike dealing cards from a deck without putting cards back in it and then shuffling.

A good book on the subject is Probability and Random Processes, by Geoffrey R. Grimmett and David R. Stirzaker.

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Let me emphasize that your calculation is based on the assumption that the die is a fair one. If I'm gambling in real life and I don't have a chance to examine the die, I'd guess that it's unfair and bet on 1 :-) – GWu Apr 5 '11 at 2:11
Well, I'd wait out and watch more than 5 throws. – Leonardo Fontoura Apr 5 '11 at 2:26
Geoffrey with a why. Grimmett with two teas. But no R. in either name. – Did Apr 5 '11 at 10:05
@Didier I used my new powers as editor to correct the spelling. In fact, both authors have an "R" as middle initial:) – Byron Schmuland Apr 5 '11 at 14:53
@Byron Powabunga, then. :-) – Did Apr 6 '11 at 5:30

This is not a mathematical question. Mathematically speaking, independence is an assumption. That is, we can model the experiment you describe mathematically by several independent events, but nothing mathematically forces us to model the events as independent; the motivation for that comes from extra-mathematical sources, namely a) the fact that this model turns out to be good at predicting instances of this experiment and b) most of us have an overall world view which suggests that events in the physical world occur due to physical causes, and since, as Qiaochu pointed out, no physical mechanism is known or easily imaginable that would make these events depend on each other, the assumption of independence makes physical sense.

None of this, however, proves that the events are actually independent. Whether they are or not is a question for physics, philosophy and religion, not for mathematics. If you believe that there's a karma account and no-one is entitled to more than a fair share of luck, or if you believe that there exists a god who determines whether you win or lose in a dice game, or if you believe that there are morphogenetic fields that cause a tendency for things to happen as they've happened in the past, then all those beliefs imply that the events are in fact not independent, and nothing in mathematics can disprove any of those beliefs.

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Very interesting way of thinking, I like it. – Leonardo Fontoura Apr 6 '11 at 0:13

If you are not certain that the dice is fair you might use Bayes' theorem. If you think there is some chance that the dice is loaded and you can quantify that chance as a probability, Bayes' theorem tells you how to update your estimate of the chance that the dice is loaded after seeing the result of some number of rolls.

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