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?

  • $\begingroup$ One note: your assumption that each player gets 50 tiles/game is a pretty big one, and deserves some scrutiny. This is, in fact, one of the reasons why 'fishing' (playing off one or two tiles per turn looking for the ones to turn your rack into a bingo) can be a bad idea: even if the plays themselves are valuable, by giving your opponent more tile turnover you're giving them more opportunities to draw the key tiles. $\endgroup$ Commented Dec 6, 2012 at 21:49
  • $\begingroup$ For instance, a first approximation for the odds in the scenario where players split the tiles evenly is to compute the odds of winning five 'coin flips' (each representing a different high-value tile); this leads to a probability of $\left(\frac{1}{2}\right)^5$ or $1$ in $32$; roughly $3\%$. (Compare to the exact values in the answers). By contrast, if one person gets $\frac{2}{3}$rds of the tiles, then the probability that they get all five goes up to approximately $\left(\frac{2}{3}\right)^5 = \frac{32}{243}$ - roughly $1$ in $8$, or about $12\%$ - four times as high. $\endgroup$ Commented Dec 6, 2012 at 21:54
  • $\begingroup$ If you're me, the probability is $1-2^{-N}$ for a very large value of $N$. $\endgroup$
    – MJD
    Commented Dec 6, 2012 at 22:17

2 Answers 2


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.


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, $100!=(100)(99)(98)(97)(96)95!$.


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