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.
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 $\mathbb{P}(\text{Flush})=\frac{6211433}{8388608}\approx0.74$.