Mr. Sammi

Actually TRON, what sounds good to me at any time, is usually what is best. I am mildly curious as to why you keep repeating certain structures from within your mind. Is there some tiny element of probabilistic tension?

I am not a schoolboy, who has to go around proving I am right. What I say, is said. If you cannot understand, then you should occupy a junior position in philosophical matters. I consider this a senior debating board AND I fully expect an elevated level of comprehension.

* * *

The math for the answer given is only half the answer. That given answer must be multiplied by the probability of the event itself. All your answers assume a probability of 1, for the event to occur. Perhaps it was a little premature to assign such a definitive initial probability to so unknown a cause.

Sammi Na Boodie (my brain does not hurt)

tronvillain

Sammi, I think that if we polled people about the likelyhood of what sounds good to you actually being the best, we would find that it is quite low. Anyone want to side with Sammi one this one? Anyone?

So, you do not have to go around proving that you are right, because you are right, and other people should be smart enough to be able to tell? What if your fevered little brain only thinks you are right and other people are smart enough to tell that you're just spouting gibberish?

Sammi:
Even if that were the case, the answer would never be "infinity" since probability is measured from zero to one. Obviously you are so enamoured with the word that you cannot think clearly. Anyway, the question specifies (implicility) that the events have a probability of one of happening within the week:

Any other reading would make the answer "unknown", or perhaps "one" but such other readings have little justification.

Mr. Sammi

Sorry again Tron,

you are wrong again, gone again,

infinity in probability theory means infinitesimal a probability so small that you could have never guessed it...

I don't know why you bother debating me, you will lose every single time, unless you cheat OR lie.

Sammi Na boodie ()

Technos

I guess it depends on what I do all day. If I drive in traffic every day the chances of me getting a red light every day, even more than once a day, increase a great deal do they not?

tronvillain

So, it's "Mr. Sammi" now is it? Unless you can back up your assertion about the use of "inifinity" in probability theory with something, I am going to have to say "I don't know why you bother debating me, you will lose every single time, unless you cheat or lie."

Mr. Sammi

Technos, the dependence is on "mutually exclusive events", when certain experiences in life are assigned probabilities. Events which are in reality unrelated, but one tries to relate them using a probabalistic function, usually ends up with "an odd chance that the predicted result carries into reality", in other words, a double dose of probability.

I do not think today's drive to work depends on yesterday's road rage, but, I could not be so sure seeing I am mostly walking in the city.

Sammi Na Boodie ()

sandlewood

Ok, I’ll give it a shot. Thanks to Michael, we already know that the total number of ways that the events could be assigned to days is 7^7 = 823543.

Next count the number of ways in which it’s possible to assign the events as you describe. Start by picking an event for Sunday. There are 7 possible ways to pick an event for Sunday. Next pick 3 events for Monday. Because there are 6 events left and we want to pick 3 of them, there are C(6,3) = 20 events. Likewise, pick 2 events for Friday and there are C(3,2) = 3. Then there is only 1 possibility left for Saturday.

Therefore the probability is:

(7 x 20 x 3 x 1) / 823543 = 420/823543 ~= .00051

Abacus

I believe that is correct Sandlewood, though I used a slightly different, but equivalent, method.

If you write out the formula for the combinations explicity, you get:

C(7,1)*C(6,3)*C(3,2)*C(1,1)=7!/(1!(7-1)!)*6!/(3!(6-3)!)*3!/(2!(3-2)!)*1!/(1!(1-1)!)

After some cancellations, this reduces to:
7!/(1!3!2!1!)

This is the formula I used. You compute that and then divide the result by 7^7.

petrachor

This is a distribution problem. A fairly common way of dealing with such problems is to consider boxes and balls. in this case we are distributing 7 distinct balls into 7 distinct boxes. We want to know what is the probability that each box will get exactly one ball.

There are 7^7 total distributions.

We can find the number of distributiosn such that each box gets one ball using a multinomial coefficient. In this case teh coefficient is 7!/(1!)^7=7!

So the probability is 7!/(7^7)= 0.611989 . . . %

If you'd like a deeper explanation, I can give one, as all I'm doign here is applying models.

petrachor

I'll apply the same model I used before.

There are 7^7=823543 total distributions.

This time the multinomial coefficeint is 7!/(3!*2!*1!*1!)=420

So the probability is 420/823543=0.05099 . . .%