Double double word score: part I

This post was originally posted on my attempt at a science communication blog, Probably Interesting. I moved it over here in March 2026 so I could have everything I’ve blogged about in one place. It also explains why this wasn’t posted on a Sunday.

I was playing Scrabble GO yesterday, which, despite what the name might suggest, is just the new mobile app for Scrabble. (It isn’t like Pokémon Go, but where you’ve gotta catch all 26 letters instead, and then battle your eight-point J against someone’s three-point G.)

Anyway, the first tiles I drew were ADEINST, in some order. I was first to play, and noticed the seven-letter anagram STAINED in my rack, and played it across the centre double word score square, with the D on a double letter—that’s (1+1+1+1+1+1+(2×2))×2=20 points, plus 50 for using all my tiles. I pressed “Submit”, feeling rather pleased with myself.

While waiting for the other player, I looked to see what I might be able to do on my next turn. It didn’t take me long to notice that I had ADEINST again. “What are the chances?” I asked one of my Scrabble-GO-playing friends. “I feel like this is a question you are more qualified to answer,” she replied.

She’s probably right: in fact, this could probably be a probability question for the first year of undergrad. At least, one version of this question is. Because it depends what we’re asking: do we really want the probability of being able to spell the word STAINED twice? There’s nothing particularly special about that word: had I got SLENDER twice, I’d have been equally amazed. What we really want is the probability of drawing any valid seven-letter word, and then of drawing the same word again.

That question is much harder, but a classic way of attacking maths problems is to start with a simpler version of the problem, and work up. So if we are able to work out the probability of getting the word STAINED twice, we might be able to build on that to find the answer to the problem we’re really interested in. Indeed, STAINED is a good word to choose, because it has no repeating letters; this will make our calculations simpler.

Okay, so I originally gave the contents of my opening rack in alphabetical order: ADEINST. Let’s simplify the problem further: supposing I drew one tile at a time, what are the chances that I should get those letters, in that order? Well, for that, we need to know what letters are in the bag. Fortunately, Wikipedia has that info for us. (And not just for English: if you want to attack this problem in Klingon, Lojban or L33tspeak, Wikipedia is here to help.)

There are nine As in the bag out of 100 tiles in total, so our first probability, of drawing an A, is $\frac{9}{100}$. Supposing that works, we next want one of the four Ds, but now we only have 99 tiles to pick from, so that’s $\frac{4}{99}$. We then, in turn, want one of the twelve Es, one of the nine Is, one of the six Ns, one of the four Ses, and finally one of the six Ts; each time, the total number of tiles decreases by one, so that in the last trial we are trying to choose a T from the 94 remaining tiles.

Each of these probabilities was calculated conditionally on all the previous events happening. So, to find the probability that all of them happen, we can multiply them together; we get:

which works out at about 0.000 000 007. Or, to use the technical term, “very small”.

But we don’t actually care about seeing the tiles in that order. What if we picked them in the order STAINED? Well, then, our probability is

But all that that’s done is rearrange the tops of the fractions. And if you remember the rule from primary school about multiplying fractions (top times top, bottom times bottom), you’ll see that that doesn’t change the answer.

So all we have to do is multiply this probability (the very small one) by the number of ways to rearrange the letters in STAINED. So let’s count those: we have seven choices for the first letter (any of them can go first), then six for the next, then five, and so on. So there are 7×6×5×4×3×2×1=5040 ways to rearrange the letters. (Mathematicians call this 7!, pronounced “seven factorial”. The notation causes confusion when you want to talk about the number 7, but you’re just really excited about it.)

So our probability of drawing the word STAINED is

and we get about 0.000 0350. Still small, but a lot more likely.

To finish up, we now need the probability of drawing the word STAINED a second time, given that we drew it the first time. (One objection you might have: in a real game, wouldn’t the other player have drawn some tiles before we draw our second set? Well, yes, but until they play them, we don’t know what they are, so it makes no difference. Or, put another way: if I drew played and drew again before my opponent drew, it wouldn’t make any difference to the situation we’re interested in.)

In this second draw, we of course have seven fewer tiles in total, and the numbers available of each letter are reduced by one. So our new probability is

or 0.000 0167. Remember, this is a conditional probability that our second draw contains the right letters, given that the first one did. So we can multiply them together to get our overall answer: the probability of my getting STAINED twice was about 0.000 000 000 58, or about 1 in 1.7 billion. For comparison, your chance of winning the UK National Lottery is about 1 in 45 million, so about two hundred times more likely.

But of course, this is one specific word: when we don’t care about the word we get, our chance is much higher. As a rough idea, Google suggests there are a five-digit number of seven-letter words, so we might expect a probability more like 1 in 10,000: still unlikely, but still more likely than becoming a Lotto millionaire, and the sort of thing that will eventually happen, given enough Scrabble games (and I have enough Scrabble games: I have about twenty going at once).

But maybe STAINED was a particularly probable word? The “multiply by the number of words” method can give us a rough idea of the order of magnitude (i.e., how many 0s are there after the decimal point), but it can’t really go beyond that. So, in my next post, I’ll talk about how we might go about scaling up our problem, and find out how lucky I really was.