Asked and Answered 4.0
I am one of those annoying persons who is kind and cooperative in everyday interactions but whose competitive bloodlust rises to the fore when Scrabble is involved. Ask my wife. She stopped playing Scrabble with me years ago, even after I began to spot her 100 points. That was little help, as far as she was concerned, because it only made me dig my heels in that much harder to overtake her.
I play online now, as I assume most players do, because who stops by your place to play Scrabble these days? I am a good casual player but not tournament-level by any means — not that I care to be. I have only recently begun to learn the two-letter-words and I am constantly surprised by the odd three-letter-words my opponents play.
Things were different in the analog days. Back then, diligent players had to memorize the accepted words, because consulting a dictionary was verboten except to answer challenges. (VERBOTEN is an official Scrabble word, by the way.) But there are no challenges in the online version — instead, players are allowed (and expected) to check whether words are valid before playing them. And the two-letter-word list is now just a feature of the game. So no one really needs to memorize much.
That said, there is still a place for vocabulary skills in online Scrabble, tempered as always by the luck of the draw of one’s letters. And that, at long last, brings me to the point of this article. I just completed a game with an anonymous online opponent named Micki. Here is the board we played:
You can see that Micki defeated me, 339 to 308. What you do not see is that Micki drew and played all five of the highest-scoring letters (Q, Z, J, X, K), along with seven of the ten next-highest-scoring letters (two each of F, H, V, W, Y) as well as the two blanks, each of which confer a 15 to 30 point advantage [Thomas, 2011]. Now perhaps I am biased, but this seemed to me to represent an especially good string of luck — for my opponent.
The competitive person that I am asked, what are the chances of such a lopsided draw? The nerd that I am set about to answer it.
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At first I thought I would need to program a computer simulation of thousands of games with random word lengths, to see how likely it would be that a real game would have such a one-sided letter draw. But then I made a few simplifying assumptions. (That is what old engineers eat for breakfast, simplifying assumptions. For us, they go down as easily as, say, eggs Benedict did for the last Pope.)
My first simplifying assumption was that each player will likely draw half of the stockpile of letters over the course of the game. This assumption allows us to neglect the number of tiles drawn on a play-by-play basis. Once we stipulate that each player will draw 50 of the 100 tiles over the course of the game, we can focus on the distribution of tiles between the two players, as if all the tiles were shuffled and dealt out to the players like a deck of cards.
To further simplify the problem, we can imagine that the special, high-scoring letters are dealt out first, followed by the low-scoring letters, using the following process. Start with the Q. To see which player gets the Q, the dealer flips a coin. If the result is heads, the Q is dealt to my pile, otherwise it is dealt to my opponent.
The dealer repeats this process for the remaining high-scoring letters and the two blanks, seven tiles in all. Since the chance of my being dealt a tile after a coin flip is 1/2, then the chance that I would be dealt all seven of the special tiles is 1/2 to the 7th power, or 1/128. Needless to say, this is also the chance that my opponent would be dealt all seven tiles.
Though not common, landing all the best-scoring tiles is not as rare as I had first assumed. But wait — my opponent not only also played the top seven tiles but also seven of the ten four-point tiles. So we need to do a bit more math.
In the second phase of the process, the dealer takes the ten four-point tiles and again flips a coin to determine which player is dealt each tile. What is the chance that I will be dealt exactly three of the ten tiles? Interestingly, I found that this specific problem was asked and answered on Mathematics Stack Exchange. The general formula is
prob(k heads in n tosses) = C(n,k) * p k * (1-p) n-k
where C(n,k) is the number of combinations of n things taken k at a time (explained here) and p is the probability of a single event, which in our case is 1/2. For n = 10 and k = 3, the answer is 15/128, or a little better than one chance in nine that I would be dealt three of the ten four-point tiles.*
So the overall likelihood of Micki’s lopsided draw is the product of the two probabilities we calculated, that is, 1/128 * 15/128 = 15/16384. Expressed as odds, this would be 1091 to 1, or about the same likelihood as getting bumped from an airline flight due to overbooking. If one looks at it this way, yes, I was a bit unlucky (or Micki lucked out, take your pick).
The big difference between one-sided Scrabble games and overbooked airline flights is that my bad luck did not result in my getting beaten up and hauled out of my recliner, kicking and screaming — instead, I wrote this post. So there. I’m vindicated. But I still lost.