Getting $5$ in a row of same color card "mystery" I was wondering why this is the case... If you take a standard well shuffled deck of $52$ cards and draw cards randomly without replacement until you get $5$ of the same color (red or black) in a row (or run out of cards), my computer simulation is showing me that the most frequently occurring # of cards drawn when that happens is $5$ cards and it gets worse as more cards are drawn.  Can someone explain why this is?  For example, if I simulate $1$ million hands, it is telling me that $50{,}469$ hands (about $5$%) are "winners" with only $5$ cards drawn but then it drops off sharply at $6$ cards drawn down to only $27{,}826$ winners.  Why is that?  Also somewhat surprisingly (to me), it is telling me that about $2/9$ ($2$ out of every $9$ games) results in no "winner", even with all $52$ cards drawn.
I have the graph output from Wolfram Alpha and want to display it here so how do I embed an image here?
 A: On the 5/6 card question, note that winning in precisely 5 cards requires the 2nd, 3rd, 4th and 5th cards to all be the same colour as the first card, while winning in precisely 6 cards requires the 2nd, 3rd, 4th, 5th and 6th cards to all be the opposite colour to the first card. Being the opposite colour to the 1st card is slightly more likely than being the same colour, but not enough to cancel out the effect of needing one more card to be the right colour. We can work out these probabilities exactly quite easily:
$P(\text{win in exactly 5 cards}) = \frac{25}{51}\frac{24}{50}\frac{23}{49}\frac{22}{48} \approx 0.0506$ 
$P(\text{win in exactly 6 cards}) = \frac{26}{51}\frac{25}{50}\frac{24}{49}\frac{23}{48}\frac{22}{47} \approx 0.0280$ 
On the 52 card no-win question: you have 48 "chances" to win (i.e 48 blocks of 5 cards that could be all the same colour). The above suggests - though I don't have a full proof - that the probability of winning in any specific one is at most $1/20$, and often significantly smaller. If we assume that the probability of winning at any specific chance is $1/20$, then the probability of not winning would be $(19/20)^{48} \approx 0.085$. And this is almost certainly an underestimate - as we've seen above, the probability of winning at the second "chance" is significantly less than $1/20$. This suggests that the probability of not winning  being $2/9$ is not unreasonable.
