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I've seen a few of these questions of the form $max$(X,constant) but they get more math heavy than I know mine should get. Could someone help me understand how to solve this?

What we are given is that the Distribution of our random variable X is {$1,2,3,4,5$} with probabilites {$1,\frac{39}{51},\frac{26}{50},\frac{13}{49},1$} respectively. Given this, I am told to find $E(X,3)$. How would I do this? I know there shouldn't be any integrals involved.

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Let us define $Z:=\max\left\{X,3\right\}.$ Now, first let us consider the values of $Z$ conditioned on the values of $X$. That is, it should be clear that $$[Z|X=x] = \left\{\begin{matrix} 3, & x = 1,\\ 3, & x =2,\\ 3, & x = 3,\\ 4, & x = 4,\\ 5, & x = 5. \end{matrix}\right.$$

Now, because $[Z|X=x]$ is always constant, it's expectation is simply its constant value, i.e., $$[Z|X=2] = 3 \Rightarrow \mathbb{E}(Z|X=2) = 3.$$ Therefore, by conditioning we can simply compute the expectation in the normal way, i.e., $$\mathbb{E}(Z) = \sum_{x=1}^{5}\mathbb{E}(Z|X = x)\mathbb{P}(X=x) \\ = 3\left(\mathbb{P}(X=1) + \mathbb{P}(X=2) + \mathbb{P}(X=3)\right) + 4\mathbb{P}(X=4) + 5\mathbb{P}(X=5).$$

Where you can insert the PMF appropriately (it seems like in the question there is a problem with the PMF you gave as it doesn't sum to 1).

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A simpler method would be to use the following fact: for a discrete random variable $X$ and function $f$

$$\mathbb{E}[f(X)] = \sum_x f(x)\mathbb{P}(X=x).$$

In our case, $f(X)=\max \{X,3\}$, thus the solution is $$\mathbb{E}(\max \{X,3\}) =\sum_{x=2}^5\max\{x,3\}\mathbb{P}(X=x).$$ The technique in the original answer is more general though and will aid in getting the expectation of more complicated random variables, e.g., $\max\{X,Y\}$ where $X$ and $Y$ are both random variables.

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    the random variable and corresponding probabilities are meant to be the number of draws from a deck before you see a suit you've already drawn. So I figured the first draw can be $\frac{52}{52}$ for any suit, then the next draw must be of the 3 remaining suits and so forth until the 5th card forces a match (since there are only 4 suits). Do the probabilities seem correct then? – 2017-02-12
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    I'm not sure that I understand. What I believe you are saying is this, you plan on drawing cards from a deck until you get two cards in your hand of the same suit. Once, you stop, call the number of cards in your hand $X$. Does that sound right? If so, we certainly have $\mathbb{P}(X = 1) = 0$. I will never stop drawing at one card. Now, the $\mathbb{P}(X = 2)$ is the probability that I've drawn the first two cards from the same suit. This should be $\frac{12}{51}$. The reason for this is that, in order to stop, I must redraw from the one suit I have already drawn from. There are $12$ ... – 2017-02-12
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    ... cards left in this suit, and $51$ cards left in total. You proceed in this way for $X=3,4$ and then take $1-\text{sum of those other probabilities}$ to find the probability $X=5$. – 2017-02-12
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    You are correct, I interpreted the question incorrectly but having read your comment and the question again I understand what is going on. So, for X = 3, that means of the first 3 cards I have drawn, the 3rd card is of the same suit as one of the previous 2 cards. Does that mean $ℙ(X = 3) = \frac{52}{52} \cdot \frac{39}{51} \cdot \frac{24}{50}$? – 2017-02-12
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    Indeed, I believe that the whole distribution looks something like $$\mathbb{P}(X = x) = \left\{\begin{matrix} 0 & x = 1,\\ 0.235 & x =2,\\ 0.367 & x = 3,\\ 0.292 & x = 4,\\ 0.105 & x = 5. \end{matrix}\right.$$ Additionally, if this answer has sufficiently helped you, feel free to accept it as correct (little check mark). Otherwise, feel free to wait for additional answer. – 2017-02-12
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    How would you solve the max function for this distribution? I am getting 3.26 but am told it's incorrect – 2017-02-15
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    Using the above formula it looks like you'll get 3.5. Is that the correct answer? Show me your work and I'll help you sort it out. – 2017-02-15
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    Yes it is. My work is $2*0.235+3*0.367+4*(0.292)+5*0.105$ – 2017-02-15
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    @jayant.M look more carefully at what I've written in my answer, the first term should be $(3\cdot 0.235)$ because $\max \{2,3\}= 3$. – 2017-02-15
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    @jayant.M I added an additional comment in my answer that might help you better understand the problem. – 2017-02-15