Shuffles a deck of r ranks and k suits; looks for m cards of the same rank dealt in a row This is a question that has arisen in my work.
Game the First: the dice game. Suppose I have a fair die with r faces. Given some k, I roll the die r*k times; call that sequence of rolls a game. 
Call a game 'm-good' if it is not the case that m rolls of the same face are thrown in a row at any point in the game.
Q1: What is the probability that a game is m-good?
Game the Second: the card game. Suppose I have a deck of cards with r ranks and k suits. So a standard bridge deck has (r,k) = (13,4); and would contain r*k = 52 cards.
Take such a deck and randomly shuffle it. Then flip out the cards, one after another. 
Call a shuffle 'm-good' if it is not the case that m cards of the same rank are turned over in a row.
Q2: What is the probability that a random shuffle is m-good?
While it probably isn't the most elegant approach, Q1 is certainly solvable by treating the problem like a state machine, with states for none in a row, 1 in a row, 2 in a row, ..., m in a row; and a Markov transition matrix. Then exponentiating the transition matrix r*k times yields the desired probability.
But I can't see a 'nice' way to evaluate the card version. 
And while my gut feeling is that for any given r,k the probability of m-good is always smaller in the dice case than the card case (which would be an equally helpful conclusion), I struggle to prove this.
Your thoughts?
 A: Think of the dice rolls as coming from a (typically unbalanced) card deck. To show that the probability of $m$-good dice rolls is lower than the probability of $m$-good card draws, it suffices to show that the probability of $m$-good card draws is highest when the deck is maximally balanced. This you can show by induction. Assume that a deck is unbalanced, i.e. there is a pair of ranks such that there are at least two more cards of one rank than of the other. (If the maximal difference between ranks is one card, the deck is as balanced as it can be given the total number of cards and the number of ranks.) Now convert one card of the more frequent rank into a card of the less frequent rank, and in all possible sequences in which the deck can be drawn, consider only the positions of these two ranks. It shouldn't be too difficult to show that the conversion increases the number of $m$-good sequences. Since each such conversion reduces the imbalance of the deck (as measured e.g. by the variance of the rank frequencies), the process ends in a maximally balanced deck.
(I wrote "lower" and "increases", which is correct for sufficiently small $m$; for larger $m$ it should be "not higher" and "doesn't decrease".)
