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We’ve all been confounded by an all-vowel “hand” in Scrabble. What is the likelihood of it happening 3 turns in a row?

Some assumptions are necessary. Let’s use these:

  • Two-player game
  • English tile distribution
  • Vowels are AEIOU (ie, exclude Y)
  • The situation is drawing all vowels on the first pull, followed by being in that situation on the subsequent two turns.
  • The subject draws first.
  • The opponent plays first (this probably contradicts the game rules but is convenient for this experiment).
  • The opponent plays a four-letter word on every turn, and that word is two consonants and two vowels.
  • The subject is able to play two of their vowels on each turn, requiring a draw of two new tiles.

(If any of the above assumptions is logically impossible, please correct.)

An interesting bit is that blanks can be counted as either a consonant or vowel.

Distribution of English words and letter usage is disregarded for this calculation. Though if you want to throw it in for extra interest, great.

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In general I think all-vowel hands come up because we don't use our vowels as quickly as we use our consonants. I think it's very rare that anyone actually draws a all-vowel hand on the first turn. But if you tend to play 2 consonants to every vowel it won't take long before you reach an all-vowel hand (Typical since it's easy to play off your opponents vowels). Furthermore every time you draw a consonant thereafter, your temptation is to play it instantly so you get stuck in this position. Most of my friends rarely (if ever) use the exchange option. I use it more than anyone else I know. – dspyz Aug 4 '12 at 5:19
I'm not too happy about the second-to-last assumption. It changes the question from "What are the chances of three consecutive hands of all vowels?" to "What are the chances... given that the opponent receives at least two consonants and two vowels on their first turn, and at least four each over two turns?" Michael Lugo's answer is quite elegant, but it only works without this assumption. – Rahul Aug 4 '12 at 6:02
+1 for taking a real world problem and making it specific enough to have an answer. – Ross Millikan Aug 4 '12 at 17:39

The identity of the letters doesn't matter, so let's just note that there are 42 vowels, 56 consonants, and 2 blanks.

Imagine that the game starts by putting these tiles in random order, 1 through 100.

The subject draws tiles 1 through 7, and the opponent draws tiles 8 through 14.

The opponent then plays four tiles and draws tiles 15 through 18; the subject plays two tiles and draws tiles 19 and 20. This repeats, with the opponent drawing tiles 21 through 24, and the subject drawing tiles 25 and 26.

So under your assumptions, you're asking for the probability that positions 1 through 7, 19, 20, 25, and 26 are all among the vowels. (Blanks don't count as vowels here; if I had six vowels and a blank I don't think I'd say "I have all vowels", although I might be unhappy with my hand.) This is the probability that, if I draw eleven tiles at random, they will all be vowels.

There are ${100 \choose 11}$ ways to choose eleven tiles; of these ${42 \choose 11}$ consist entirely of vowels. So the answer is ${42 \choose 11}/{100 \choose 11}$ which is about $3 \times 10^{-5}$.

Note that your assumptions on how many tiles the opponent is going to play don't matter at all, but your assumptions on how many tiles the subject is able to play matter quite a bit.

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Actually, knowing that 8 of the opponents tiles are 4 vowels and 4 consonants reduce the set you're drawing from to 38 vowels and 52 consonants, so knowing what letters your opponent plays matters a little bit. – dspyz Aug 4 '12 at 5:15
Good point. So we should have ${38 \choose 11}/{92 \choose 11} \approx 2.2 \times 10^{-5}$. – Michael Lugo Aug 4 '12 at 14:28
My answer is slightly different, as I didn't consider that you know the opponent's first play when drawing the first rack or the second play when drawing the next two. It doesn't change much. – Ross Millikan Aug 4 '12 at 17:53

The tiles are 42 vowels and 58 consonants (counting blanks as consonants, which you would if vower-rich). The chance drawing seven vowels is $\frac {42!93!}{35!100!}\approx 0.001685$ The next turn you need to draw 2 vowels from a set of 33 vowels and 56 consonants, for a chance of $\frac {33\cdot 32}{89\cdot 88}\approx 0.1348$. The third turn you need to draw 2 vowels from a set of 29 vowels and 54 consonants, for a chance of $\frac {29\cdot 28}{83\cdot 82}\approx 0.1193$. The total is then $\frac {42!93!\cdot 33 \cdot 32 \cdot 29 \cdot 28}{35!100!\cdot 89 \cdot 88 \cdot 83 \cdot 82}\approx 0.00002711$ or about $1$ in $36886$

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