Poker Probabilities with a wild card Suppose you have a standard deck of 52 cards and add a 53rd card which is a “wild card” and can stand for any other card.


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*What’s the probability of getting a straight?

*What’s the probability of getting a flush?
I feel like this increases the complexity to a point at which I'm not sure how to do the problem. It's simple without the wild card.
 A: I’m assuming a five-card hand. There are $\binom{53}5$ possible hands. Flushes are a little easier to count than straights, so I’ll start with them.
To get a flush in a suit, you must get either a regular flush or a four-card flush plus the wild card, so there are $\binom{13}5+\binom{13}4=\binom{14}5$ ways to get a flush in any one suit and $4\binom{14}5$ ways to get a flush altogether. The probability of a flush is therefore $$\frac{4\binom{14}5}{\binom{53}5}=\frac{8008}{2869685}\approx 0.00279\;.$$ The figure without the wild card is about $0.00198$, so the wild card improves one’s chances by not quite $41$%.
Now I’ll count straights. A straight may begin with any value from ace (low) through ten. For any one of these starting points there are $4^5$ regular straights. The wild card may be substituted for any of the five values in the straight, and for each of those substitutions there are $4^4$ ways to fill out the ‘honest’ part of the straight, so there are $5\cdot 4^4$ four-card straights plus wild card. Thus, there are $4^5+5\cdot 4^4=9\cdot 4^4$ straights with each of the ten ranges of values and hence $10\cdot9\cdot4^4$ straights.
Edit: However, this overcounts when the wild card is at one end of the straight, since, for example, $2345W$ and $W2345$ are the same hand. (Thanks to the OP for pointing this out.) Specifically, we’ve counted an extra $4^4$ hands for each pair of adjacent ranges ($A2345$ and $23456$, $23456$ and $34567$, etc.). There are nine such pairs, so we’ve overcounted by $9\cdot4^4$, and the correct total is $10\cdot9\cdot4^4-9\cdot4^4=9^2\cdot4^4$. The probability of getting a straight is therefore $$\frac{9^2\cdot 4^4}{\binom{53}5}=\frac{20736}{2869685}\approx 0.00723\;.$$ The figure without the wild card is about $0.00493$, so in this case the improvement is a bit under $46.7$%.
