Card Game Probability 13 Card Hand Me and my friends play a four person poker style card game. Each person is dealt 13 cards, and it is a standard trump card game. Now, as the standard, a five card flush beats a five card straight, but it seems, and we all seem to agree, that we get far more flushes than straights. 
The rest of the hands seem to occur an appropriate number of times... So, I am wondering if someone knows, or could help me figure out if the probability of being dealt a straight in a 13 card hand is higher than a flush.
 A: I'll post the solution for a flush now. @Awkward came up with a solution for the straight.
(The probability for a straight is about 0.38)
Probability for Flush (including royal flush and straight flush):
Let $p_i$ be the probability that you get a flush in $i$ given suits. Note that $p_3=p_4=0$.
By the inclusion-exclusion principle:
$$\mathbb{P}(\text{Flush})=4p_1-6p_2+4p_3-p_4=4p_1-6p_2$$
Now let's calculate $p_1$:
$$p_1=1-\mathbb{P}(\text{4 or less hearts})$$$$=1-\frac{\binom{13}{0}\cdot 3^{13}+\binom{13}{1}\cdot 3^{12}+\binom{13}{2}\cdot 3^{11}+\binom{13}{3}\cdot 3^{10}+\binom{13}{4}\cdot 3^{9}}{4^{13}}$$$$=1-\frac{3^9}{4^{13}}\cdot(81+351+702+858+715)=1-\frac{2707\cdot 3^9}{4^{13}}=\frac{13826983}{4^{13}}$$
And $p_2$:
$$p_2=\frac{1}{4^{13}}\sum_{i=5}^{8}\sum_{j=5}^{8}\left [1[i+j\le13]\cdot\binom{13}{i}\binom{13-i}{j}\cdot 2^{13-i-j}\right ]$$
$$=\frac{13!}{4^{13}}\cdot(\frac{1\cdot 2^{3}}{5!5!3!}+\frac{2\cdot 2^{2}}{5!6!2!}+\frac{2\cdot 2^{1}}{5!7!1!}+\frac{2\cdot 2^{0}}{5!8!0!}+\frac{1\cdot 2^{1}}{6!6!1!}+\frac{2\cdot 2^{0}}{6!7!0!})=\frac{936078}{4^{13}}$$
So $\mathbb{P}(\text{Flush})=\frac{6211433}{8388608}\approx0.74$.
A: Ok after trying to tackle this problem for a while, let's just go with the easy 'intuitive' way since we're just comparing probabilities, keep in mind this is probably wrong, and I gave up about halfway through
So both involve 5 cards. 
Let's assume they both start with an ace of spades 
For a flush, there are 12 options for card 2, 11 options for card 3, 10 options for card 4 and 9 options for card 5 = 11880
For a straight, there are 32 options for card 2, 12+16+20+24/4 options for card 3 (depending on how close card 2 is to card 1), at least 8 options for card 4 and 4 options for card 5. = at least 18432 
So according to this shifty research a straight is at least 55% more likely than a flush. But feel free to rip this up and suggest your own way. 
tbh i see many errors with this method.
A: There are some tricky aspects to this problem.  For example, it's possible for a single thirteen-card hand to contain both a straight and a full house.  In that case, according to the rules of poker, it should be counted as a full house, not a straight.  It's also possible for a single thirteen-card hand to contain two flushes, which makes it easy to accidentally overcount. 
So rather than attempting an analytic solution, here are the results of a Monte Carlo simulation.  I simulated 10^6 hands of thirteen cards each and classified the hands according to the highest-ranking five card subset.  Here are the results:
$$\begin{align}
0 & \qquad\text{high card (i.e., nothing)}\\
0 & \qquad\text{1 pair}\\
56,031 & \qquad\text{2 pairs}\\
31 & \qquad\text{3 of a kind}\\
110,243 & \qquad\text{straight}\\
338,621 & \qquad\text{flush}\\
444,528 & \qquad\text{full house}\\
33,794 & \qquad\text{4 of a kind}\\
16,752 & \qquad\text{straight flush}
\end{align}$$
Divide each of these numbers by 10^6 for an approximate probability.
Evidently, flushes are more common than straights.
A: A flush is dealt in a 13-card hand iff some suit has length 5 or more.  I.e. a flush is not dealt iff the longest suit has length 4.  This is something that Bridge players care about, and the probabilities are widely available.  E.g. according to:
http://web2.acbl.org/documentLibrary/teachers/celebritylessons/handpatternsrevised.pdf
https://www.bridgehands.com/P/Probability_Hand_Distribution.htm
http://www.bridgeguys.com/MGlossary/mathematical_tables.html
The probability that the longest suit is length 4 (patterns 4-4-3-2, 4-3-3-3, 4-4-4-1 - arbitrary ordering onto the 4 suits are already included) is $\approx 0.35$, and so $P(flush) \approx 1 - 0.35 = 0.65$.  
Not sure how they got those numbers but since multiple websites agree, and Bridge is a pretty popular game, and acbl = American Contract Bridge League = the governing body, I would tentatively trust the numbers.
Obviously this doesn't answer the question of how often a flush is dealt and it is actually the best poker hand available among the 13 cards (i.e. there are no full house, etc).
