# Is Scrabble's method of determining turn order fair?

At least the way my family plays, turn order is determined by drawing tiles and seeing who has the letter closest to A (blanks taking precedence). If I recall correctly, there are 100 unevenly distributed letters.

Intuitively, it seems like it would be unfair, though I can't come up with a way to prove it. Obviously the distribution matters: my thoughts are if there are more tiles in the first half of the alphabet (including blanks), then the starting player holds an advantage since they're more likely to get a tile earlier in the alphabet than the next (and vice versa if there are fewer tiles in the first half). That doesn't seem too right, though. I'm probably just forgetting some basic prob stats.

I imagine this is likely a duplicate, but I couldn't find anything relating to it (maybe since I've been searching for Scrabble). My apologies if it is.

Please feel free to edit in appropriate tags.

• Suppose you father picks out, from a pile of face down tiles, enough tiles for the whole family. Then he passes them out, one to each family member. Before he passes them out, he tries to remember the first tile he happened to touch and keeps it for himself. Does your father get any advantage this way? Maybe this changes your intuition?
– Mark
Aug 29, 2017 at 4:12
• "Obviously the distribution matters." Why would you think that? The only way the distribution matters is by making ties more likely or less likely.
– bof
Aug 29, 2017 at 4:33
• I appreciate your comments -- indeed, it seems my intuition was off the mark. I got a bit caught up in the finer details without looking at the big picture. I especially appreciate that I didn't receive any sort of hostility for my rather simple question.
– cole
Aug 29, 2017 at 5:00
• Why would the first player be more likely than the second to draw a lower tile? Aug 29, 2017 at 6:02
• You didn't specify how ties are broken, e.g. if player 1 and 2 both draw an A. If the rule is that the first player to draw wins, then drawing first will have an advantage that will depend on the distribution. If the resolution is to redraw, then it's fair. Aug 29, 2017 at 16:00

Because of symmetry, no player has an advantage. That is, if $\mathscr A$ is the set of outcomes where player $A$ gets the letter closest to A and $\mathscr B$ is the set of outcomes where player $B$ gets the letter closest to A, there is a one-to-one correspondence between $\mathscr A$ and $\mathscr B$ obtained by interchanging the letters players A and B get. Since all outcomes are equally likely, they have the same probability of getting to go first.

• That makes sense, so I'm marking this as the accepted answer. I can see how it generalizes to more than two people. Thanks!
– cole
Aug 29, 2017 at 4:52

Your intuition would make sense if you could choose individual tiles with unequal probabilities. For example, imagine that there are just two tiles, 'A' and 'B', and the selection process is such that tile 'A' is chosen with probability $0.75$ and 'B' with $0.25$. Then obviously the player who draws first has a higher probability of being the winner.

Now imagine that instead of two we have four tiles, and all are equiprobable to pick (much like your scrabble scenario). Three 'A's and one 'B'. Now you still have a $0.75$ probability of drawing 'A', but unlike the first case, you do not pick all the 'A's from the bag if you pick an 'A'.

What's the probability of winning then? Let's name the outcome X and give it a value $+1$ for winning and $-1$ for losing. What is the expected value $E(X)$ if you go first? If you are unlucky to draw the 'B' you certainly lost (there are only 'A's for the second person to pick). If you pick 'A', you might win (with probability 1/3 the second person will pick 'B'), or you might replay the whole game (because the second person will pick another 'A' and it's a draw). Let's write these down:

$$E(X) = 0.25\cdot (-1) + 0.75\cdot(\frac{1}{3}\cdot 1 + \frac{2}{3}\cdot E(X)) \iff \\ E(X) = -\frac{1}{4} +\frac{1}{4} + \frac{1}{2}E(X) \iff \\ E(x) = \frac{1}{2}E(X) \iff\\ E(X) = 0$$

You see how the good cases are balanced out by the bad ones and the net result is zero gain. Granted, this is a specific example to illustrate a point, but you can easily see how this generalises to any probabilities.

There is of course the symmetry argument outlined in the answer by Robert Israel, that generalises this for any number of players. This works because tiles are equiprobable, and I think this approach here helps us a little to get a better intuition.

How do the players draw their tiles? If the tiles are in a bag, and the players reach into the bag and pull out a tile, they may cheat by feeling for a blank tile. Or, if the tiles are spread out face down on the table, players may be able to recognize them by marks on the back. If this kind of cheating is allowed, the player who draws first has an advantage. However, IF WE ASSUME RANDOM DRAWING, no player has an advantage.

Suppose that the first player has a better chance of drawing a blank than the second player. Then, by the same token, the first player has a better chance of drawing a Z than the second player, because (under our assumption of random drawing) there is no difference between a blank and a Z (except that there are two blanks and only one Z in the standard English set). And the same goes for every other letter of the alphabet: the first player has a better chance of drawing it than the second player. But this is plainly absurd. It would mean that, over a long series of games, the first player would draw every tile in the set more often than the second player.

• Upvoted for considering the actual practical aspect of drawing from a bag :) However the OP's intuitive argument is that if the distribution of tiles is uneven, the first player has a higher chance of drawing a more popular tile than the second player. In this case, it's plausible that first player could have an advantage.
– Mark
Aug 29, 2017 at 4:27
• Unlike @Mark, I don't think the 'practical aspect of drawing from a bag' is helpful here. This is mathematics, and therefore we expect simplifying assumptions, not a painstaking examination of how one might cheat.
– jwg
Aug 29, 2017 at 16:19
• @jwg I agree completely. That's why, in answering, I explicitly ruled out the possibility of any kind of cheating, and made the simplifying assumption that tiles were drawn at random.
– bof
Aug 30, 2017 at 3:29

Look at a toy problem. Suppose there are two players and three tiles: $A_{1}$, $A_{2}$, and $B$. The subscripts on the $A$,s indicate there are two $A$ tiles. Look at all the possible outcomes: the first letter is what the first player draws, the second letter is what the second player draws and the third letter is the remaining letter.

1. $A_{1}A_{2}B$
2. $A_{1}BA_{2}$
3. $A_{2}A_{1}B$
4. $A_{2}BA_{1}$
5. $BA_{1}A_{2}$
6. $BA_{2}A_{1}$

The first and third outcomes are draws. The first player wins the second and fourth outcomes. The second player wins the fifth and sixth outcomes. In this example; if the first player draws a good tile, an $A$, the first player can do no worse than draw; if the first player draws a bad tile, a $B$, the second player has to win.

• I like this concrete example (+1), but am left a bit unsatisfied by the possible "draw" conclusion (in the given game, someone must go first). Perhaps if you expanded to another tile "C", and initial draws then use the 3rd and 4th tiles as a tiebreaker, so that every determination is conclusive? Aug 29, 2017 at 18:49
• If, in the case of a draw, the players select another tile from the remaining tiles the players are repeating the drawing game, only with a smaller, somewhat different collection of tiles. Since the number of letters being used for repeated draws is being reduced every time there is a draw eventually someone will have to a letter for which there is only one tile. At that point one player or the other will win.
– Jay
Aug 29, 2017 at 19:35

Whether the method is fair or not depends on what you do in case of a tie. (That is, several players draw the "best" letter.)

If you let each player draw another letter until there's a winner, everything is entirely symmetric and there is no advantage to anyone. You could as well forget in which order the letters got drawn once you compare them to each other.

If you just let the first person who drew the best letter start, there goes their advantage.

If after someone draws a tile then replaces it, the events are independent, therefore the probability is the same.

However, if someone draws a tile and holds it, the events are dependent. Example:

Player 'A' draws R. They do not replace it.

Player 'B' now has a slightly larger chance of drawing an A or a B or another "high" letter.

Now, look at this scenario:

Player 'A' draws A. They do not replace it.

Player 'B' now has a much smaller chance of drawing an A. They have a slightly higher chance of drawing a B or another "high" letter, but it doesn't matter since they'll lose anyway.

TL;DR: Make sure all players are clear on whether to hold or replace tiles, as it has an effect on probabilities of tying (but not overall fairness).

• While drawing without replacement does make the draws dependent events, drawing with and without replacement are both fair. You cansee this by noticing that any pair of tiles is just as likely to come in either order. Aug 30, 2017 at 0:34
• In the second scenario, I talk about a tie. If the tiles are not replaced, the chances of a tie are significantly smaller. If they are replaced, the chances of a tie are larger. I guess that as long as the method is determined beforehand, then it would be fair either way. Editing. Aug 30, 2017 at 1:43
• Who cares if there are ties? Someone will win eventually. Whether I or anyone else at the table keeps their tile or replaces it is irrelevant to the final outcome. Your tl;dr focuses on a part of the answer that doesn't matter. Aug 30, 2017 at 13:04