Sorry if this topic has been discussed before. I hate looking through forums randomly for an answer to a question. So, maybe one of you could help me out.

Here is the situation:
Myself and five other friends order some take out and need to decide on who will go pick it up (no one wants to leave and miss part of the game on the tube). So, we agree that we will have a single roll using a six sided die to decide who will go. Now it comes down to each of us picking a number. If one's number comes up, your are the one who will go for the pick up.
But, before any of us pick a number, I casually through the die on the coffee table and it lands on a four. Thereby intiating the arguement.
The arguement is that the roll-off has been jeopardized because the die was already rolled once without us picking our numbers first.
One side of the arguement is that the individual who chooses four as their number will have an advantage since the odds of the same number coming up on the next roll (the real roll that will decide who leaves) will be in their favor. Remember that whoever's number comes up must leave to get the take out and miss the game.
The other side of the arguement (my side, btw) is that the next roll will be a separate event and the die "doesn't care" what happened 5 minutes ago.
I tried to explain that this is no different than picking 1, 2, 3, 4, 5, and Mega Ball 6 in the Mega Millions lotto. The numbers on the lotto balls are nothing more than shapes, only understood by us humans.

So, my questions are: Mathematically speaking, who wins the arguement?

Please, if you can, support your answer with an expression. And by the way, this is irl, not a homework problem. Although i can imagine it is a homework problem somewhere. I will tell you how this all panned out if you reply.

Thanks!

Edit: I will be happy to clarify if anything doesn't make sense

You win. The outcome of a single roll has absolutely no impact on outcomes of future rolls.

The other side of the arguement (my side, btw) is that the next roll will be a separate event and the die "doesn't care" what happened 5 minutes ago.

So, my questions are: Mathematically speaking, who wins the arguement?





Each roll, of the fair die, is independent of previous rolls. That is, the probability of rolling a 4 is 1/6 = 0.16666...for each trial.

That said, if the case were; "What is the probability of rolling two consecutive 4s?", then we would have

Pr{4,4} = (1/6) * (1/6) = 1/36 = 0.0277777...

by the multiplication rule for independent events.

Each roll, of the fair die, is independent of previous rolls. That is, the probability of rolling a 4 is 1/6 = 0.16666...for each trial.

That said, if the case were; "What is the probability of rolling two consecutive 4s?", then we would have

Pr{4,4} = (1/6) * (1/6) = 1/36 = 0.0277777...

by the multiplication rule for independent events.

So, by the multiplication rule for independent events, I am wrong? Sorry if I'm missing the picture, but how can the multiplaction rule for independent events be just that? What I mean is, if it is truly independent, than why would one include the previous event in the calculation?

Also, thanks for your responses (all of you) thus far, it is greatly appreciated!

Well,
I'm eargerly looking forward to further discussions and answers. But in the meantime, I owe those of you that replied w/ how we resolved our arguement. But, I can't give it all away yet. I decided to go into my draw and grab a completely different die than the original, and to make sure we all picked our numbers before the die was ever thrown, casually or not (this was all just to prove a point). Also, please excuse my spelling. I'm not the best with the American-English language!

So, by the multiplication rule for independent events, I am wrong? Sorry if I'm missing the picture, but how can the multiplaction rule for independent events be just that?

Okay, the key word you need to understand is AND.

What I mean by that can be explained by the following example. Suppose I want to consider 3 rolls of a fair die. I then ask what is the probability of rolling: 4 on the first trial and a 4 on the second trial and 4 on the third trial?

This can be expressed as:

Pr{4 and 4 and 4}= Pr{4}*Pr{4}*Pr{4} = 1/6 * 1/6 * 1/6 = 1/210.

where the probability of rolling a four on the last (third) trial is still 1/6 - irrespective of the previous trials.

Thus, if you want to ignore any previous trials (rolls) i.e. eliminate the "and"s and consider just a single subsequent roll the probability is simply Pr{4} = 1/6.

Does this help?

Note: I would point out is that the issue we are discussing is related to the so-called "Gambler's Fallacy."