The simplest proof is to use a combinatoric tree. Suppose you have a coin, what's the probability of tossing it twice and getting heads both times? The first toss has 2 possibilities (heads or tails), then each of these tosses has two branches for the second toss (again, heads or tails). The total sample space is therefore 4. Heads, heads, represents one event out of the 4. Hence probability =1/4 or 1/2 * 1/2

What is interesting is that although you may think having the lottery numbers 1,2,3,4,5,6 has just as much chance as any other combination, in fact because it is a run of consecutive numbers it is a far less likely outcome

>>7776271
Is this true? How so?

>Why the hell is this the case?
Because that's what it *means* for two variables to be independent.

>>7776307
>Is this true?
No. The payoff on that sequence may be lower than average because it's picked by a disproportionally large amount of people, but the odds of the combination happening is the same as for all other combinations.

>>7776202
>The simplest proof is to use a combinatoric tree

Retarded CS majors don't belong here

>>7776271
That is not true whatsoever.
>>7776195
>They tell me that the intersection of independent probabilities is equivalent to the multiplication of those probabilities.
This is like the very basis of probability. 12-year-olds understand this. If you don't understand this, then you know literally nothing about probability.

>>7776195
Event 1 happens A different ways out of B total different ways

Event 2 happens C different ways out of D total different ways

Events 1 and 2 are independent, so one has no effect on the other happening simultaneously. Therefore the probability of them happening simultaneously is the number of ways they can happen simultaneously divided by the total different ways:

(A*C) / (B*D)