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2m
comment probability of not getting same number twice in a row after n die rolls
They lead to the same answer. Since the probability $p_n$ is $a_n/6^n$, you can let $a_k=6^kp_k$ and then my recurrence becomes $6^np_n=(5)(6^{n-1})p^{n-1}+(5)(6^{n-2})p_{n-2}$. Divide both sides by $6^n$ and we get your recurrence. For no good reason, I thought a counting recurrence might be intuitively clearer than a probability recurrence.
15m
answered Linear Recurrence - Form not familiar
56m
revised probability of not getting same number twice in a row after n die rolls
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1h
comment probability of not getting same number twice in a row after n die rolls
I will add to the remark in $10$ minutes or so.
1h
comment probability of not getting same number twice in a row after n die rolls
The proper analogy is then I think the problem I mentioned in the remark. The no two in a row (which two unspecified) is as you saw much simpler. If you experience trouble, I can add to the remark.
1h
revised probability of not getting same number twice in a row after n die rolls
added 199 characters in body
1h
answered probability of not getting same number twice in a row after n die rolls
1h
comment If given the limit that is a derivative, how do I find it's function and the point?
Hint: The derivative of $f(x)$ at $a$ is $\lim_{x\to a}\frac{f(x)-f(a)}{x-a}$. Compare.
1h
comment Linear Recurrence - Form not familiar
@NobleMushtak: The problem indeed can be solved with no machinery. However, from the way the question is worded, in particular the first part, I think they expect the use of characteristic polynomials.
2h
comment Linear Recurrence - Form not familiar
Have you used the method of characteristic polynomials on other linear recurrences? It may be tough to give hints that are not close to a solution!
2h
comment Suppose you draw a five-card hand randomly from the deck and get four cards that that would make a straight if you could replace the fifth card…
If you have for example a $5,7,8,9$ there are $4$ "good" cards. If you have a $6,7,8,9$ there are $8$ good cards.
2h
comment Suppose you draw a five-card hand randomly from the deck and get four cards that that would make a straight if you could replace the fifth card…
Depends on whether you are trying to fill an inside straight or an outside straight. Depends also in some cases on whether Ace can count as low.
3h
comment Number of integral solutions to $y_1 + y_2 + y_3 + y_4 =30$ with $y_i \ge 2$
Same as the number of solutions of $x_1+x_2+x_3+x_4=22$ in non-negative integers. Is that a familiar problem?
3h
comment Is there any value for $x$ that would make the statement $(x+3)^3 = x^3+3^3$ true?
By inspection, $x=0$ and $x=-3$ work. There can be no more solutions. For imagine expanding (it is safer than expanding). We will get a quadratic equation, which can have no more than $2$ roots.
6h
awarded  Enlightened
7h
comment Maximizing the probability of a poll prediction
You are welcome. You were on the right path, but there were issues of detail.
8h
comment What number comes next in the sequence $7, 16, 8, 27, 9,…$?
$40$. I am sort of joking. There is a semi-plausible pattern that does yield $40$, but the sample size is too small.
9h
answered Maximizing the probability of a poll prediction
9h
comment What is the probability that none of the cans of soup are next to each other?
If you prefer, we can let the denominator be $\binom{12}{3}3!9$, that is, $12!$, and the numerator to be $\binom{10}{3}3!9!$. The ratio is unchanged.
9h
comment What is the probability that none of the cans of soup are next to each other?
I am not assuming that the cans of soup are indistinguishable. In fact I am assuming that all permutations of the cans are equally likely.