Calculate odds of getting all hight point tiles in single game of scrabble I am wondering how to best calculate the probability of getting the all of the following "high point" tiles in a single game of scrabble: Z, Q, J, X, and K
In a game of scrabble, there are 100 tiles for distribution. Of the Z, Q, J, X, and K tiles, only 1 each of these are in the bag of tiles. 
Let's assume that in a single game, each player gets 50 tiles each. I think of this as a bag of 100 marbles with 95 white marbles and 5 red ones. 
So, if you were to distribute them evenly to each of the two players, what are the odds that one person will receive all 5 red marbles/high point tiles.
Now, what are the odds of this happening twice in a row? 3 times in a row?
 A: The number of high-scoring tiles you get is governed by the Hypergeometric distribution. Under the assumptions you've laid out, the probability of a certain player getting all five tiles is about 2.8%. The probability of either player getting all five tiles is double that, about 5.6%.
The tiles drawn in each game are independent, so the probability of all five tiles going to one player or the other in two games in a row is about 0.3%, three games in a row is about 0.02%, and it shrinks geometrically from there.
A: Call the players A and B. There are $\dbinom{100}{50}$ ways to choose $50$ tiles from $100$, all equally likely. 
The number of ways for A to get the $5$ big point tiles is the number of ways to choose $45$ tiles from the remaining $95$, which is $\dbinom{95}{45}$.
So the probability A gets all the big point tiles is 
$$\frac{\binom{95}{45}}{\binom{100}{50}}.\tag{$1$}$$
For the probability that A or B gets all the big point tiles, multiply the above number by $2$.
Let $p=2\frac{\binom{95}{45}}{\binom{100}{50}}$.
The probability that there is a "bad split" twice in a row is $p^2$. The probability of a bad split $3$ times in a row is $p^3$. 
Remark: The numerical computation is not as bad as it looks. One can "cheat" and use say wolfram Alpha, or many other tools. It turns out that 
$p=\dfrac{1081}{19206}$, about $0.0563$.
It is not too bad with a simple calculator. Recall that $\dbinom{n}{m}=\dfrac{n!}{m!(n-m)!}$.  So for Expression $(1)$, we get
$$\frac{95!50!50!}{100!45!55!}.$$
Next we take advantage of the many cancellations. For example, $100!=(100)(99)(98)(97)(96)95!$. 
