Scrabble probabilities
Mark Spahn
you start with a bag of 100 square tiles, on each of which a letter
of the alphabet is printed. Among the 100 tiles, E appears on 12
tiles; A, I on 9 tiles each; O on 8 tiles; N, R, T on 6 tiles each;
D, L, S, U on 4 tiles each; G on 3 tiles; B, C, F, H, M, P, V, W, Y, ?
on 2 tiles each; and J, K, Q, X, Z on 1 tile each. (The letter "?"
represents a blank tile, which a player can use for any desired letter.)
Let's calculate the probability of drawing a given set of tiles
when randomly drawing 7 tiles from the full bag at the beginning
of the game.
As an example, let's calculate P{MINIMAL}, which is defined as
the probability, upon randomly drawing 7 tiles from the 100-tile
bag, of drawing, in any order, tiles that can be rearranged to
spell out the word MINIMAL.
There is C(2,2)=1 subset of 2 M's from the 2 M's in the bag.
There are C(9,2)=36 subsets of 2 I's from the 9 I's in the bag.
There are C(6,1)=6 subsets of 1 N from the 6 N's in the bag.
There are C(9,1)=9 subsets of 1 A from the 9 A's in the bag.
There are C(4,1)=4 subsets of 1 L from the 4 L's in the bag.
Thus there are 1*36*6*9*4=7776 distinct subsets of
2 M's, 2 I's, and 1 each N, A, L. The 7 distinct tiles can
be arranged in 7! ways. Thus there are 7776*7! distinct
permutations of tiles that can be rearranged to spell MINIMAL.
The probability of drawing such a set of tiles from the bag is
this number of permutations, divided by the number of all
permutations of 7 tiles drawn (without repetition) from the
bagful of 100 tiles, which is P(100,7) = 100*99*98*97*96*95*94.
So P{MINIMAL} = 7776*7!/P(100,7) = 7776/(P(100,7)/7!)
= 7776/C(100,7).
C(100,7) = 2^5*3*5^2*7*11*19*47*97 = 16,007,560,800.
P{MINIMAL} works out to 81/166,745,425.
Here is an annotated excerpt from the book "Word Freak:
Heartbreak, Triumph, Genius, and Obsession in the World
of Competitive Scrabble Players", by Stefan Fatsis (p. 44-45):
Serious players learned that the word game was really
a math game, as the pages of the renamed _Scrabble Players
Newspaper_ reinforced. In the February 1980 issue, an expert
named Albert Weissman, a Connecticut psychologist, conducted
the game's first computerized mathematical experiments.
[What follows is computation, not computer-simulation experiments.]
Weissman calculated the probability of drawing certain racks
to start play. The least likely combination of letters was
BBJKQXZ, where the B could be replaced by any tile of which
there are two in the bag (B, C, F, H, M, P, P, W, Y, and the blanks);
the probability of drawing such a rack was about 1 in 16 billion.
[Yup. It works out to 1/C(100,7) = 16,007,560,800.]
The single most probable rack, he found, was AEINORT, with
an expected frequency of 1 in about 9,530 draws from a fresh bag.
[I get 9*12*9*6*8*6*6/C(100,7) = 17,496/166,745,425 = 1/9530.488.
I thought it might be that P{EEAIONR} > P{AEINORT}, but it is not.]
But there was no acceptable seven-letter word in AEINORT
(there still isn't), so Weissman figured out the most and least
probable bingo-producing draws. [A "bingo" means forming a
word that uses up all 7 of a player's tiles.]
The former was (and still is) AEEINRT, which made RETINAE and
TRAINEE, with a probability of 1 in about 13,870 opening draws.
[I get P{AEEINRT} = P{EEAINRT} = C(12,2)*9*9*6*6*6/C(100,7)
= 2,187/30,317,350 = 1/13,862.53. Where does the "13,870"
come from? Or at least, why is it not rounded off to 13,860?]
A word like ERRATIC would come up 1 in 91,743 opening pulls.
[I get P{ERRATIC} = 12*C(6,2)*9*6*9*2/C(100,7)= 729/66,698,170
= 1/91,492.69. Did I miscalculate, or did Weissman?]
POACHES, 1 in 588,235.
[I get P{POACHES} = 2*8*9*2*2*12*4/C(100,7)
= 288/166,745,425 = 1/578,977.2. Again, why the discrepancy?]
MUZJIKS, 1 in 55,555,555.
[No. I get P{MUZJIKS} = 2*4*1*1*9*1*4/C(100,7)
= 3/166,745,425 = 1/55,581,808.33. Why all the 5's in
Weissman's result?]
Finally, MUUMUU?, where the blank is an S, could be expected
to appear once every 8 _billion_ opening turns, making it the
least probable opening bingo.
[Yep. I too get P{MUUMUU?} = C(2,2)*C(4,4)*C(2,1)/C(100,7)
= 1/8,003,780,400.]
-- Mark Spahn (West Seneca, NY)
Mark Spahn
is a passage I do not understand. What does an oddsmaker
do, and what does it mean to say that someone is
"with odds of 5 to 2" or "is 3 to 1"? What do these
numbers represent? Is this gambling jargon?
Excerpt (p. 78):
The oddsmakers have cofavorites: Logan and another
former whiz kid, Brian Cappelletto, an options trader in
Chicago, with odds of 5 to 2. G. I. Joel is 3 to 1 and
Edley is 7 to 2. Matt Graham is 10 to 1, Marlon is 25 to 1.
Frederick Williams
>
> ...
> former whiz kid, Brian Cappelletto, an options trader in
Doesn't one stop reading at this point?
--
Remove "antispam" and ".invalid" for e-mail address.
"He that giveth to the poor lendeth to the Lord, and shall be repaid,"
said Mrs Fairchild, hastily slipping a shilling into the poor woman's
hand.
Brian M. Scott
<msp...@localnet.com> wrote in
<news:13empk5...@corp.supernews.com> in
alt.math.recreational:
> In the same book about Scrabble I referred to last time
> is a passage I do not understand. What does an oddsmaker
> do, and what does it mean to say that someone is
> "with odds of 5 to 2" or "is 3 to 1"? What do these
> numbers represent? Is this gambling jargon?
Yes, though certainly not exclusively. It's a way of
expressing probabilities. To say that the odds of event E
occurring are m to n is to say that the probability of E is
m/(m + n). More often odds are given *against* the
occurrence of E, in which case m/(m + n) is the probability
that E does *not* occur. For instance, if you toss two fair
coins, the odds against getting two heads are 3 to 1.
> Excerpt (p. 78):
> The oddsmakers have cofavorites: Logan and another
> former whiz kid, Brian Cappelletto, an options trader in
> Chicago, with odds of 5 to 2. G. I. Joel is 3 to 1 and
> Edley is 7 to 2. Matt Graham is 10 to 1, Marlon is 25 to 1.
Here it's understood that the odds given for a contestant
are the odds against his winning. The oddsmakers reckon
that Marlon is a long shot, with a probability of winning of
only 1/26; Logan, on the other hand, is reckoned to have a
2/7 probability of winning.
In this betting context, however, what I said obviously
can't be quite correct, because
2/7 + 2/7 + 1/4 + 2/9 + 1/11 + 1/25 > 1.
In that setting the odds refer to the payoffs: a bet of $2
on Logan will get you a $5 payoff if Logan wins, a bet of $1
on G.I. Joel will get you $3 if G.I. Joel wins, and so on.
In other words, your payoff if your pick wins is what would
be a fair payoff if the probability of his winning were in
fact given by the odds. Suppose that there is a constant k
such that the amounts bet on Logan, Cappelletto, Joel,
Edley, Graham, and Marlon are respectively (2/7)k, (2/7)k,
(1/4)k, (2/9)k, (1/11)k, and (1/25)k dollars. Then no
matter who wins, the oddsmakers will have to pay out k
dollars, but they will have taken in more than k dollars (a
bit over 1.17k, in fact). Of course the amounts bet may not
be distributed in this way; if the deviation is too great in
the early betting, the odds offered to later bettors are
likely to be adjusted to compensate.
Brian
Mark Spahn
"Brian M. Scott" <b.s...@csuohio.edu> wrote in message
news:xq8gv0h9r1jv$.1j3fhtp33opgn.dlg@40tude.net...
Brian,
Thank you for this clear explanation. So one meaning [1] of
"odds" is in expressing the ratio of the probability that a
certain event will happen, to the probability that it will not happen.
So Logan having odds of 5 to 2 mean that (in an oddsmaker's
estimation), the probability p of Logan winning is 2/(5+2) = 2/7.
But, you report, there is another meaning [2] of "odds" too.
In this case, Logan having odds of 5 to 2 means that
getting a payoff of $5 if you bet $2 that Logan wins, will
be, on average, a no-win, no-loss proposition. Letting
p be the probability that Logan wins, your expected payoff
on a $2 bet will be $(-2)*(1-p) + $5p. For this to be an even
bet, we need this expected payoff to be $0. That is, p must
be such that $(-2)*(1-p) + $5p = $0, from which
-2 + 2p + 5p = 0
7p = 2
p = 2/7.
Yup, we get the same value for p.
So these two meaning are consistent.
(By the way, I think you meant 1/26 where you wrote 1/25.)
Thanks again,
Mark Spahn
Brian M. Scott
<msp...@localnet.com> wrote in
<news:13ep0u3...@corp.supernews.com> in
alt.math.recreational:
[...]
> (By the way, I think you meant 1/26 where you wrote 1/25.)
Yep; either a typo or a slip of the mind, and at this point
I've no idea which it was.
Brian
Mark Spahn
To review,
in the game Scrabble (see http://en.wikipedia.org/wiki/Scrabble),
you start with a bag of 100 square tiles, on each of which a letter
of the alphabet is printed. Among the 100 tiles, E appears on 12
tiles; A, I on 9 tiles each; O on 8 tiles; N, R, T on 6 tiles each;
D, L, S, U on 4 tiles each; G on 3 tiles; B, C, F, H, M, P, V, W, Y, ?
on 2 tiles each; and J, K, Q, X, Z on 1 tile each. (The letter "?"
represents a blank tile, which a player can use for any desired letter.)
On page 127 of the book "Word Freak: Heartbreak, Triumph,
Genius, and Obsession in the Worldof Competitive Scrabble
Players", by Stefan Fatsis, is the statement:
"There are 3,199,724 unique combinations of seven tiles that
can be plucked from a virgin Scrabble bag of ninety-eight
letters and two blanks."
(A few sentences later, the author writes, "Mathematicians
have determined tht the possibility of choosing an acceptable
seven-letter word from a fresh bag is 12.63 percent, or just
over one in eight.")
I have been unable to verify this 3,199,724 statement;
I get a different answer.
I am not sure whether this author is using the word
"combinations" in the sense it is used in combinatorics.
What I think he means is that the seven tiles
GRAVART and TRAGARV are considered the same
(i.e., the order of the tiles does not matter), and
that these are also considered the same even if
different tiles of the same letter are used (there are
three tiles labeled G, so the G in GRAVART might
be a different tile from the G in TRAGARV).
Thus these two "racks" are considered identical, because
they yield the same set of words for a player.
Let's try to count how many different patterns
(what the author calls "combinations") can be
drawn from a full bag of the 100 tiles.
How many patterns are there of the form aaaaaaa,
that is, in which all seven tiles are of the same type
(have the same letter printed on them)?
There are only 4 letters that can yield 7 copies:
E, A, I, O. Thus there are four distinct patterns of
the form aaaaaaa, namely, EEEEEEE, AAAAAAA,
IIIIIII, OOOOOOO.
How many patterns are there of the form aaaabbb ?
The role of a can be played only by a letter that
has at least four copies, namely by the 11 letters
EAIONRTDLSU. The role of b can be played by any
of the 12 letters EAIONRTDLSUG that can appear
3 times, except the letter that plays the role of a.
Thus there are 11 choices for a, and 11=12-1 choices
for b, so there are 11*11 = 121 distinct patterns of
the form aaaabbb.
One more example: Let's count the number of patterns
of the form aabbcde. In choosing letters to play the
roles of a and b, we must choose 2 letter from among
the 22 letters that can appear twice, namely, from
all 27 letters (we count the blank ? as a letter) except
the letters (JKQXZ) that appear on only one tile each.
There are C(22,2) ways to choose the pair of letters
to play the roles of a and b, and for the remaining 3 letters
cde, we can choose any of the 27 letters except the two
letters that were chosen for a and b. So we have C(25,3)
choices for the set of three letters {c,d,e}.
Thus the number of patterns of form aabbcde is
C(22,2)*C(25,3).
In going through all the possible patterns of seven tiles
(did I forget any?), I get the following counts.
aaaaaaa: C(4,1) 4
aaaaaab: C(7,1)*C(26,1) 182
aaaaabb: C(7,1)*C(21,1) 147
aaaabbb: C(11,1)*(C11,1) 121
aaaaabc: C(7,1)*C(26,2) 2,275
aaaabbc: C(11,1)*C(21,1)*C(25,1) 5,775
aaabbbc: C(12,2)*C(25,1) 1,650
aaabbcc: C(12,1)*C(21,2) 2,520
aaaabcd: C(11,4)*C(26,3) 858,000
aaabbcd: C(12,1)*C(21,1)*(25,2) 75,600
aabbccd: C(22,3)*C(24,1) 36,960
aaabcde: C(12,1)*C(26,4) 179,400
aabbcde: C(22,2)*C(25,3) 531,300
aabcdef: C(22,1)*C(26,5) 1,447,160
abcdefg: C(27,5) 80,730
Total: 3,221,824
This is not the 3,199,724 that the author reports.
Where did I (or he?) go wrong? And is there an easier
way count these "combinations" than considering the
various patterns?
Mark Spahn
"Mark Spahn" <msp...@localnet.com> wrote in message
news:13eubfi...@corp.supernews.com...
>
> To review,
> in the game Scrabble (see http://en.wikipedia.org/wiki/Scrabble),
> you start with a bag of 100 square tiles, on each of which a letter
> of the alphabet is printed. Among the 100 tiles, E appears on 12
> tiles; A, I on 9 tiles each; O on 8 tiles; N, R, T on 6 tiles each; D, L,
> S, U on 4 tiles each; G on 3 tiles; B, C, F, H, M, P, V, W, Y, ? on 2
> tiles each; and J, K, Q, X, Z on 1 tile each. (The letter "?" represents
> a blank tile, which a player can use for any desired letter.)
>
> On page 127 of the book "Word Freak: Heartbreak, Triumph, Genius, and
> Obsession in the Worldof Competitive Scrabble Players", by Stefan Fatsis,
> is the statement:
>
> "There are 3,199,724 unique combinations of seven tiles that
> can be plucked from a virgin Scrabble bag of ninety-eight
> letters and two blanks."
>
> (A few sentences later, the author writes, "Mathematicians
Here's a correction of my own errors. I miscomputed the
number of ways to get the patterns aaaabcd and abcdefg.
The correct counts are
aaaabcd: C(11,1)*C(26,3) 28,600
abcdefg: C(27,7) 888,030
When these corrections are made, the total does indeed
come to 3,199,724. All is well, all is well.
The author reports, "Mathematicians have determined that
the possibility of choosing an acceptable seven-letter word
from a fresh bag is 12.63 percent, or just over one in eight."
It is not clear exactly what this means. It could mean that
404,125 the 3,199,724 possible patterns (12.63% of them) can be
rearranged to form a 7-letter word, like EEEGRUX can be
rearranged to form the legitimate word EXERGUE (= the
space on a coin around the picture or design, where the
year and mint mark are placed). But not every pattern
is equally likely. So maybe the quoted statement means that
over the long run, 12.63% of the 7-letter racks that are drawn
from a full bag will yield a 7-letter word.
Dan in NY
>
> "Mark Spahn" <msp...@localnet.com> wrote in message
> news:13eubfi...@corp.supernews.com...
>>
>> To review,
>> in the game Scrabble (see http://en.wikipedia.org/wiki/Scrabble),
>> you start with a bag of 100 square tiles, on each of which a letter
>> of the alphabet is printed. Among the 100 tiles, E appears on 12
>> tiles; A, I on 9 tiles each; O on 8 tiles; N, R, T on 6 tiles each; D, L,
>> S, U on 4 tiles each; G on 3 tiles; B, C, F, H, M, P, V, W, Y, ? on 2
>> tiles each; and J, K, Q, X, Z on 1 tile each. (The letter "?" represents
>> a blank tile, which a player can use for any desired letter.)
>>
>> On page 127 of the book "Word Freak: Heartbreak, Triumph, Genius, and
>> is the statement:
>>
>> "There are 3,199,724 unique combinations of seven tiles that
>> can be plucked from a virgin Scrabble bag of ninety-eight
>> letters and two blanks."
>>
>> (A few sentences later, the author writes, "Mathematicians
>> have determined that the possibility of choosing an acceptable
>> seven-letter word from a fresh bag is 12.63 percent, or just
>> over one in eight.")
>>
>> I have been unable to verify this 3,199,724 statement; I get a different
>> answer.
>>
>> I am not sure whether this author is using the word "combinations" in the
>> sense it is used in combinatorics.
>> What I think he means is that the seven tiles GRAVART and TRAGARV are
>> considered the same
>> (i.e., the order of the tiles does not matter), and
>> that these are also considered the same even if
>> different tiles of the same letter are used (there are
>> three tiles labeled G, so the G in GRAVART might
>> be a different tile from the G in TRAGARV).
>> Thus these two "racks" are considered identical, because they yield the
>> same set of words for a player.
>>
> rearranged to form a 7-letter word, like EEEGRUX can be
> rearranged to form the legitimate word EXERGUE (= the
> space on a coin around the picture or design, where the
> year and mint mark are placed). But not every pattern
> is equally likely. So maybe the quoted statement means that
> over the long run, 12.63% of the 7-letter racks that are drawn
> from a full bag will yield a 7-letter word.
> --
> Mark Spahn (West Seneca, NY)
Greetings Mark,
I almost expected that someone would find some corrections before I posted
my results and here. It isn't a surprise but it is pleasant that I see my
correction is confirmed. I found two typographical errors plus one of the
corrections you did. Typos- missing and misplaced C in these two formulas:
aaabbcd: C(12,1)*C(21,1)*(25,2)
aaaabbb: C(11,1)*(C11,1)
I found the error abcdefg: C(27,7), but the pleasant part is that you found
another error that I missed, completing the verification. I also verified
that you have all the patterns. First I tried to find another pattern, then
I found that Sloane A130561 shows that there are 15 partitions of 7 (ways
to add to 7), which verified that you found them all. I have no explanation
for the percentages mentioned, or a reason to doubt yours. I presume it
uses a list of 7-letter words. It would be interesting to have percentages
for smaller words, also.
You mention that the patterns are not equally likely and I agree. Do you
(or other readers) have numbers and/or formulas for the likelihood of any or
all of these patterns? I have not (yet) tried to figure it out, but I might
do that.
I wonder a bit about the value of finding words on the rack. It has been
years since I played Scrabble. From memory, on the first play, a word off
the rack is placed on the board. However, after that, it is most usual to
place letters onto the board that also use a letter already there.
Sometimes two or more letters on the board are used. A combination of
letters on the rack and one (or more) letters on the board are used to form
the word.
Occasionally, a word from the rack can be placed on the board in a way that
one letter is added to the beginning or end of a word (or more) already on
the board. Does this happen often enough to look for a seven-letter word on
the rack? Instead, I start by looking for a letter on the rack that forms a
word when added before or after words on the board, then look for words on
the rack which use each such letter. (I can remember that squares that a
triple-word-score square often figured in how often I considered looking for
such letters.)
What other interesting math did the book contain?
--
Dan in NY
(for email, exchange y with g in
dKlinkenbery at hvc dot rr dot com)
Mark Spahn
"Dan in NY" <D...@NY.net> wrote in message
news:46f006fd$0$15359$4c36...@roadrunner.com...
Hi Dan,
I have read the book halfway, and I have listed here practically
all of its math content. You ask about formulas.
Let's compute the probability P{RECEDED} of drawing from
a full bag (all 100 tiles) exactly the letters needed to spell RECEDED.
Since the order in which the letters are drawn makes no difference,
P{RECEDED} = P{EEEDDRC}. Now let's count how many
distinct ways there are to have EEEDDRC; that is, we count how many
distinct sets of 7 tiles there are that contain 3 E's, 2 D's, 1 R, and 1 C.
Let's distinguish every tile from all the other tiles.
There are 12 E's, so we can distinguish them (mentally, at least)
by numbering them E1, E2, ..., E12, or by giving each a different color.
There are C(12,3) ways to choose which 3 of the 12 E's shall be
in our set {E,E,E,D,D,R,C}. Similarly, since there are a total of 4 D's,
there are C(4,2) ways to pick which 2 D's shall be in our set.
There are C(6,1) choices to pick which of the 6 R's shall be in our set,
and there are C(2,1) choices for which of the 2 C tiles to include.
Thus there are C(12,3)*C(4,2)*C(6,1)*C(2,1) = 220*6*6*2 = 15,840
distinct sets {E,E,E,D,D,R,C} of 7 tiles that can be rearranged to
spell out RECEDED. These can be arranged in any of 7! orders,
so there are C(12,3)*C(4,2)*C(6,1)*C(2,1) * 7! permutations
of tiles that spell out RECEDED. One such permutation is
D2 E5 E10 R3 C2 D4 E8, and another such permutation is
C1 E7 E6 D4 E4 R2 D1. How many permutations are there
of any 7 tiles chosen from the bag of 100 distinguishable tiles?
Answer: 100*99*98*97*96*95*94 = P(100,7).
Thus the probability of drawing a permutation of 7 tiles that
can be rearranged to spell out RECEDED is
(number of such permutations)/(number of all 7-tile permutations)
= C(12,3)*C(4,2)*C(6,1)*C(2,1) * 7! / P(100,7)
= C(12,3)*C(4,2)*C(6,1)*C(2,1) / [P(100,7)/7!]
= C(12,3)*C(4,2)*C(6,1)*C(2,1) / C(100,7)
= 220*6*6*2 / 2^4*3*5^2*7^2*11*19*97
= 12/451,535
= 1/37,627.9 = 26.576 micro (check my arithmetic!).
In general, this probability works out to be
(number of distinct sets of 7 tiles consisting of the desired
letters)/C(100,7).
I don't immediately see how to calculate the probability of
getting the pattern aabcdef (i.e., all letters different, except for one
pair)
when drawing 7 tiles from a full bag, except for considering
many cases (e.g., a=E, a=R, a=Z, ...).
On the other hand, this datum does not seem of any use in Scrabble.
The "Word Freak" book about Scrabble is interesting,
but it does not make the top levels of the Scrabble world
look very attractive. To play at the top level, you practically
have to devote your life to memorizing which strings of
letters are legitimate words, as listed in a dictionary used for
this purpose. Scrabble is like poker in that bluffing is allowed.
You can lay out a fake word on the board. If it is unchallenged,
it stands (and you gain the points from it), but if it is successfully
challenged, you have to take back the tiles and lose your turn.
In my opinion, Scrabble might be improved by a rule that
allows everyone to use a dictionary to check which strings
of letters are real words.
mensa...@aol.com
> "Dan in NY" <D...@NY.net> wrote in messagenews:46f006fd$0$15359$4c36...@roadrunner.com...
>
> > "Mark Spahn" <msp...@localnet.com> wrote in message
> >news:13eupos...@corp.supernews.com...
>
> >> "Mark Spahn" <msp...@localnet.com> wrote in message
> >>news:13eubfi...@corp.supernews.com...
>
> >>> To review,
Although I don't believe there is any such rule, we always forced
the challenger to lose a turn if the word DID exist in the dictionary.
We didn't want people challenging EVERY word, otherwise it would be
pointless to try to bluff.
Of course, then you don't just lay down the tiles and cross your
fingers.
You pick up the Scrabble dictionary, wave it under your opponent's
nose
shouting "Go ahead! Challenge my word! I DARE you!" which you could
use
also as an anti-bluff, hoping he'll bite and give you a double-turn
shot
at reaching that triple word score.
Mark Spahn
>> The "Word Freak" book about Scrabble is interesting,
>> but it does not make the top levels of the Scrabble world
>> look very attractive. To play at the top level, you practically
>> have to devote your life to memorizing which strings of
>> letters are legitimate words, as listed in a dictionary used for
>> this purpose. Scrabble is like poker in that bluffing is allowed.
>> You can lay out a fake word on the board. If it is unchallenged,
>> it stands (and you gain the points from it), but if it is successfully
>> challenged, you have to take back the tiles and lose your turn.
>
> Although I don't believe there is any such rule, we always forced
> the challenger to lose a turn if the word DID exist in the dictionary.
> We didn't want people challenging EVERY word, otherwise it would be
> pointless to try to bluff.
>
> Of course, then you don't just lay down the tiles and cross your fingers.
> You pick up the Scrabble dictionary, wave it under your opponent's nose
> shouting "Go ahead! Challenge my word! I DARE you!" which you could use
> also as an anti-bluff, hoping he'll bite and give you a double-turn shot
> at reaching that triple word score.
>
>>
>> In my opinion, Scrabble might be improved by a rule that
>> allows everyone to use a dictionary to check which strings
>> of letters are real words.
>> -- Mark Spahn (West Seneca, NY)
Yes, my description of bluffing in Scrabble was incomplete.
If you challenge a word that is found to be legitimate, you lose your turn.
-- Mark Spahn
Mark Spahn
from a Scrabble bag containing 100 tiles (exactly 12 of
which are labeled E), the 7 tiles you draw will contain at
least one E?
I have worked out the answer in two different ways,
and I get two different answers. Where am I going wrong?
In computing this probability P{Exxxxxx}, we distinguish
every tile from all the other tiles. How many distinct sets
of 7 tiles are there that include at least one E?
There are C(12,1) choices for picking the one E that
is guaranteed to be in this set. The other 6 tiles can
be selected from the other 99 tiles. This provides a
selection of C(99,6) distinguishable 6-tile sets.
Thus the number of distinguishable 7-tile sets
in which at least one E appears is C(12,1)*C(99,6).
Each of them can be arranged in 7! sequences,
so we have a total of C(12,1)*C(99,6)*7! sequences
of 1st-drawn, 2nd-drawn, ..., 7th-drawn tiles in which
at least one E appears. But there are P(100,7) different
sequences of 7 tiles selected from the 100.
Thus the desired probability is
P{Exxxxxx} = C(12,1)*C(99,6)*7!/P(100,7)
= C(12,1)*C(99,6)/C(100,7) = 21/25 = .84.
Another way to compute the probability P{Exxxxxx}
that at least one E will be drawn in 7 draws from a
full Scrabble bag, is to compute the probability P{no E}
of not drawing a single E in 7 draws from the bag.
P{no E} = P{1st-drawn tile is non-E} * P{2nd-drawn tile is non-E}
* ... * P{7th-drawn tile is non-E}
For the first draw, 100-12 = 88 of the 100 tiles in the bag are non-E,
so P{1st-drawn tile is non-E} = 88/100.
For the second draw, 87 of the remaining 99 tiles are non-E,
so P{2nd-drawn tile is non-E} = 87/99.
Similarly, P{3rd-drawn tile is non-E} = 86/98, and so on, so we have
P{no E} = (88/100)*(87/99)*(86/98)*(85/97)*(84/96)*(83/95)*(82/94)
= P(88,7)/P(100,7) = C(88,7)/C(100,7)
= 72,140,197/181,904,100 = .39654836779.
Therefore P{Exxxxxx} = 1 - P{no E} = .6034163221.
Why are these answers different?
Where am I going wrong?
In further thinking about htis, I think now that the first answer is wrong.
Why? Because we are counting the same sets of 7 tiles twice.
Let's say the guaranteed-to-be-there E tile is tile E5,
and that the other six tiles are D3 W1 A6 E10 T4 G2.
This yields the (alphabetized) 7-tile set A6 D3 E5 E10 G2 T4.
Suppose the guaranteed-to-be-there E tile is E10.
Then when the other six tiles are W1 E5 T4 D3 G2 A6,
we get the same 7-tile set A6 D3 E5 E10 G2 T4.
We are counting this 7-tile set twice.
A Monte Carlo simulation should provide further evidence
that the asked-for probability is .6034163221, not .84.
Another question then: What is the probability of drawing
PAUL among the 7 tiles drawn from a full Scrabble bag?
I computed that it is
P{PAULxxx} = C(2,1)*C(9,1)*C(4,1)*C(4,1)*C(96,3)/C(100,7)
= 2570.6 micro, but I now think this answer is wrong,
for the reason given above. But how to compute the right answer?
Mark Spahn
> Why? Because we are counting the same sets of 7 tiles twice.
> Let's say the guaranteed-to-be-there E tile is tile E5,
> and that the other six tiles are D3 W1 A6 E10 T4 G2.
> This yields the (alphabetized) 7-tile set A6 D3 E5 E10 G2 T4.
> Suppose the guaranteed-to-be-there E tile is E10.
> Then when the other six tiles are W1 E5 T4 D3 G2 A6,
> we get the same 7-tile set A6 D3 E5 E10 G2 T4.
> We are counting this 7-tile set twice.
> A Monte Carlo simulation should provide further evidence
> that the asked-for probability is .6034163221, not .84.
>
> Another question then: What is the probability of drawing
> PAUL among the 7 tiles drawn from a full Scrabble bag?
> I computed that it is P{PAULxxx} =
> C(2,1)*C(9,1)*C(4,1)*C(4,1)*C(96,3)/C(100,7)
> = 2570.6 micro, but I now think this answer is wrong,
> for the reason given above. But how to compute the right answer?
>
> -- Mark Spahn (West Seneca, NY)
In thinking further on how to compute the probability
that when 7 tiles are randomly chosen from a full 100-tile
Scrabble bag, at least one E tile will be drawn,
it is easy enough to calculate the probability that
exactly k E tiles will be drawn. This probability is
P{k} := P{exactly k E's are drawn}
= C(12 E's in the bag,k) * C(88 non-E's in bag,7-k)/C(100,7).
P{1} = 6,503,174,832/C(100,7)
P{2} = 2,585,599,632/C(100,7)
P{3} = 513,015,800/C(100,7)
P{4} = 54,319,320/C(100,7)
P{5} = 3,031,776/C(100,7)
P{6} = 81,312/C(100,7)
P{7} = 792/C(100,7)
--------------------------
P{1-7 E's} = the sum 9,659,223,464/C(100,7)
= 9,659,223,464/[2^5*3*5^2*7*11*19*47*97]
= 109,763,903/181,904,100 (in lowest terms)
= .6034163221
This is exactly the same as 1 - P{no E's in 7 draws}.
But I still do not see an easy way to calculate the
probability of drawing 7 tiles that allow you to spell out PAUL.
Calculating P{PAULxxx} where none of the x are P, A, U, or L
is easy enough:
C(2,1)*C(9,1)*C(4,1)*C(4,1)*C(81 non-PAUL tiles,3)/C(100,7)
= 24,572,160/C(100,7),
but it looks like we also have to consider about 20 possibilities
in which the three x's variously take on the values of
P, A, U, L, and non-PAUL (but no nore than 2 P's in all,
because there are only 2 P tiles).
For similar reasons, it looks difficult to calculate the
probability of drawing 7 tiles that will allow you to spell
out any given word of 6 or fewer letters.
Mark Spahn
a book about Scrabble by Stefan Fatsis, page 265:
A few years ago, holding a rack of BEEIORW, [Jim Geary]
determiined that playing off the B and an E would yield a
1 in 68 chance of drawing an A and a T that would give him
a rack of AEIORTW and allow him to play, through the
disconnected letters Z and O already on the board, the
word WATERZOOI. It happened. He will happily detail
his great plays and label them as such but then always
leave room to berate himself. (His Web site, jimgeary.com,
bears the slogan "Something to bore everyone.")
Question: At this point in the game, how many tiles remain
in the bag from which tiles are randomly drawn, and how
many A's and how many T's does the bag contain?
Originally the bag contains 100 tiles,
9 of which are A's and 6 of which are T's.
mensa...@aol.com
> Here's a nice little math problem from "Word Freak",
> a book about Scrabble by Stefan Fatsis, page 265:
>
> A few years ago, holding a rack of BEEIORW, [Jim Geary]
> determiined that playing off the B and an E would yield a
> 1 in 68 chance of drawing an A and a T that would give him
> a rack of AEIORTW and allow him to play, through the
> disconnected letters Z and O already on the board, the
> word WATERZOOI. It happened. He will happily detail
> his great plays and label them as such but then always
> leave room to berate himself. (His Web site, jimgeary.com,
> bears the slogan "Something to bore everyone.")
>
> Question: At this point in the game, how many tiles remain
> in the bag from which tiles are randomly drawn, and how
> many A's and how many T's does the bag contain?
total remaining: 17 tiles
either 1 A and 4 T
or 2 A and 2 T
or 4 A and 1 T
Mark Spahn
<mensa...@aol.com> wrote in message
news:1190422014.1...@o80g2000hse.googlegroups.com...
No, I get a different answer. And I should have said "Show your work".
The interest here is not in what the answer is, but how it is found.
Below, after some spoiler space, is how I got my answer.
Spoiler
Spoiler
The probability P{AT} of drawing an A and a T in two draws from a bag
contaning n tiles, a of them being A's and t of them being T's, is
P{1st-drawn tile is A, 2nd-drawn tile is T} + P{1st is T, 2nd is A}
= (a/n)*(t/(n-1)) + (t/n)*(a/(n-1)) = 2at/(n(n-1)). It is given that
this probability is 1/68, so we need to find the values of n, a, t for which
2at/(n(n-1)) = 1/68
n(n-1) = 136at
n^2 - n - 136at = 0
Solving this quadratic equation in n, we get
n = (1/2)[1 +- sqrt(1 + 4*136at)] = (1/2)[1 + sqrt(1+544at)]
In order for n to be an integer, we need sqrt(1+544at) to
be a square and [1 + sqrt(1+544at)] to be an even number.
Knowing that a is an integer between 1 and 9
and t is an integer between 1 and 6, we try all 9*6 = 54 values
(I wrote a small program) to see which values of a and t yield
an integer value for n. The only solutions are a=1, t=2, or
a=2, t=1, and in both these cases n = (1/2)[1 + sqrt(1+544*2)]
= (1/2)[1 + sqrt(1+1088)] = (1/2)[1 + 33] = 17.
If someone has a better solution (e.g., one where you don't
have to consider 54 cases), let's see it.
mensa...@aol.com
> <mensana...@aol.com> wrote in message
Hmm...I may have calculated it based on A being
the first drawn.
Mark Spahn
I drew a total of 3 blanks in 9 games --
about a 100 to 1 shot -- and lost every game."
For each 2-person Scrabble game, there are 2 blank tiles
in the bag, so in 9 games there are 18 opportunities, upon
drawing tiles from the bag, for a blank tile to be drawn by
a given player. Let the probability that each blank tile will
be drawn by this player be p, and define q = 1-p.
Then this player's probability of drawing exactly k of the
18 blanks that are drawn in the course of 9 games is
C(18,k)*p^k*q^(18-k).
Typically, p=q=1/2, so the probability that a given player
will draw exactly k blanks in 18 opportunities will be
C(18,k)*(1/2)^18, and the probability that he will draw
3 or fewer blanks will be
[C(18,0)+C(18,1)+C(18,2)+C(18,3)]*(1/2)^18
= [1 + 18 + 153 + 816]/262,144 = 988/262,144 = 247/65,536
= 1/265.33 = .00376892.
This is far from the stated probability of 1/101.
But it is not necessarily true that the probability that a
blank will be drawn by a given player is 1/2, because
over the course of the games, the player who places more
tiles on the board during his moves will draw more tiles
from the bag than the other player will.
So we need to find p such that
C(18,0)*1*q^18 + C(18,1)*p*q^17 + C(18,2)*p^2*q^16
+ C(18,3)*p^3*q^15 = 1/101, that is,
(1-p)^18 + 18*p*(1-p)^17 + 153*p^2*(1-p)^16 + 816*p^3*(1-p)^15 = 1/101.
I don't see a way to solve this short of siccing a search program on
this for finding the general root of f(x)=0.
-- Mark Spahn (West Seneca, NY)
for determining
Proginoskes
> Another little math problem from "Word Freak" (page 308):
>
> I drew a total of 3 blanks in 9 games --
> about a 100 to 1 shot -- and lost every game."
>
> For each 2-person Scrabble game, there are 2 blank tiles
> in the bag, so in 9 games there are 18 opportunities, upon
> drawing tiles from the bag, for a blank tile to be drawn by
> a given player. Let the probability that each blank tile will
> be drawn by this player be p, and define q = 1-p.
> Then this player's probability of drawing exactly k of the
> 18 blanks that are drawn in the course of 9 games is
> C(18,k)*p^k*q^(18-k).
> Typically, p=q=1/2, so the probability that a given player
> will draw exactly k blanks in 18 opportunities will be
> C(18,k)*(1/2)^18, and the probability that he will draw
> 3 or fewer blanks will be
> [C(18,0)+C(18,1)+C(18,2)+C(18,3)]*(1/2)^18
> = [1 + 18 + 153 + 816]/262,144 = 988/262,144 = 247/65,536
> = 1/265.33 = .00376892.
> This is far from the stated probability of 1/101.
I wouldn't say so; it's only off by a multiplicative factor of 3. If
the MF were 100 or so, then I'd say it's "far off".
> But it is not necessarily true that the probability that a
> blank will be drawn by a given player is 1/2, because
> over the course of the games, the player who places more
> tiles on the board during his moves will draw more tiles
> from the bag than the other player will.
And the player who draws more tiles will likely get more points; so
getting few blanks ought to correspond roughly to the number of games
won, which makes "I lost every game" plausible. OTOH, if he drew 3
blanks over the course of 9 games and won all 9, *that* would look
suspicious.
> So we need to find p such that
> C(18,0)*1*q^18 + C(18,1)*p*q^17 + C(18,2)*p^2*q^16
> + C(18,3)*p^3*q^15 = 1/101, that is,
> (1-p)^18 + 18*p*(1-p)^17 + 153*p^2*(1-p)^16 + 816*p^3*(1-p)^15 = 1/101.
>
> I don't see a way to solve this short of siccing a search program on
> this for finding the general root of f(x)=0. [...]
Using fsolve on Maple, I get an answer of about p = 0.4587483900.
(There is a negative real root, and no other roots are real.)
--- Christopher Heckman
jacques
> In the game Scrabble (see http://en.wikipedia.org/wiki/Scrabble),
> you start with a bag of 100 square tiles, on each of which a letter
> of the alphabet is printed. Among the 100 tiles, E appears on 12
> tiles; A, I on 9 tiles each; O on 8 tiles; N, R, T on 6 tiles each;
> D, L, S, U on 4 tiles each; G on 3 tiles; B, C, F, H, M, P, V, W, Y, ?
> on 2 tiles each; and J, K, Q, X, Z on 1 tile each. (The letter "?"
> represents a blank tile, which a player can use for any letter.)
>
> Let's calculate the probability of drawing a given set of tiles
> when randomly drawing 7 tiles from the full bag at the beginning
> of the game.
>
> As an example, let's calculate P{MINIMAL}, which is defined as
> the probability, upon randomly drawing 7 tiles from the 100-tile
> bag, of drawing, in any order, tiles that can be rearranged to
> spell out the word MINIMAL.
> There is C(2,2)=1 subset of 2 M's from the 2 M's in the bag.
> There are C(9,2)=36 subsets of 2 I's from the 9 I's in the bag.
> There are C(6,1)=6 subsets of 1 N from the 6 N's in the bag.
> There are C(9,1)=9 subsets of 1 A from the 9 A's in the bag.
> There are C(4,1)=4 subsets of 1 L from the 4 L's in the bag.
>
> Thus there are 1*36*6*9*4=7776 distinct subsets of
> 2 M's, 2 I's, and 1 each N, A, L. The 7 distinct tiles can
> be arranged in 7! ways. Thus there are 7776*7! distinct
> permutations of tiles that can be rearranged to spell MINIMAL.
> The probability of drawing such a set of tiles from the bag is
> this number of permutations, divided by the number of all
> permutations of 7 tiles drawn (without repetition) from the
> bagful of 100 tiles, which is P(100,7) = 100*99*98*97*96*95*94.
>
> So P{MINIMAL} = 7776*7!/P(100,7) = 7776/(P(100,7)/7!)
> = 7776/C(100,7).
> C(100,7) = 2^5*3*5^2*7*11*19*47*97 = 16,007,560,800.
> P{MINIMAL} works out to 81/166,745,425.
Just reading this from a couple of months ago ...
Does this 7776 figure take the blanks into account?
It seems you may have taken the blanks into account when talking about
the full bag (100 tiles) but not for the word MINIMAL itself.
So, what is the probability of a blank replacing each one of those 7
letters? For instance, with the M's you say ....
> There is C(2,2)=1 subset of 2 M's from the 2 M's in the bag.
But what about the additional probability of one of both of the two
blanks replacing part of the word MINIMAL? If I pull 7 tiles from the
bag, there is a chance that any one (or two) of those tiles I extract can
be a blank.
There are 7776 ways I can pull MINIMAL from the bag, but how do I go
about calculating an additional probability of blanks?
Can you demonstrate with 2 words, let's say ...
INTONES 9*15*6*8*12*4 = 311040 (.0000194)
ACETINS 9*2*12*6*9*6*4 = 279936 (.0000175)
In most scrabble software ACETINS is ranked above INTONES, despite having
a lesser probability using your blanks.
Thanks