Old Maid is
a card game that consists of pairs of matching cards and one unmatched card
called the “old maid”. Although
specialized Old Maid decks of cards can be used, a standard deck of playing
cards can work just as well by using cards with the same type and color for the
matching cards (for example, the ace of spades would match with the ace of
clubs, the king of hearts would match with the king of diamonds, etc.), but
discarding one card, traditionally the queen of clubs, to leave the queen of
spades for the old maid.
The game
begins with the dealer dealing out all the cards to each player. Each player looks at his or her hand and
discards any matches that were dealt to them.
Then starting with the player to the left of the dealer and going in a
clockwise direction, the player picks a card from the hand of the player to the
right of him or her. If that card makes
a match with one of the cards in the player’s hand, that match can be
discarded, otherwise that card is added to the player’s hand. Then the next player takes a turn by picking
a card and either discarding the match or adding the card to his or her hand. Gameplay continues in this fashion until all
the matches our discarded, leaving the old maid. Whoever has the old maid at the end of the
game is the loser.
This game is
popular among children because it is so simple.
All you have to do is pick a random card and make matches. There’s no strategy involved, which means the
outcome of every game is based solely on chance. But a game based solely on chance can be
analyzed with the laws of probabilities, which means we should be able to predict
which player has an advantage of winning, whether it be the dealer or the non-dealer,
and whether it be the player who starts with the old maid or the player who doesn’t
start with the old maid.
Old Maid: 2 Players, 1 Card
Before
answering these questions, we should first analyze the simplest version of Old
Maid and work our way through versions of increasing complexity, and hope to
find some general rules and patterns.
The simplest version of Old Maid is to play with 2 players and 1
card. It’s actually not much of a game
at all, but at least gives us a good starting point. The dealer deals out the one and only card to
the other player, and the other player is “left” with the old maid and
automatically loses. In this version,
the dealer wins 100% of the time, and the non-dealer wins 0% of the time. Note also that the player who starts out with
the Old Maid wins 0% of the time.
Old Maid: 2 Players, 3 Cards
The next
simplest version of Old Maid is with 2 players and 3 cards, where one card is
the old maid and the other two cards are a match. Let us define the old maid as the queen of
spades (Q♠), and the match
as the king of hearts (K♥) and the king of diamonds (K♦). Now when the
dealer deals out the cards, the non-dealer will receive two cards and the
dealer will receive one card. Since the
dealer receives only one card, and there are three cards to choose from (Q♠, K♥, K♦), there are only three scenarios to consider:
1) {Q♠ | K♥, K♦} – The dealer receives the old maid (Q♠), and the non-dealer receives
the other two matching cards (K♥, K♦). In this case, the
non-dealer discards his or her matching pair, and the dealer is left with the
old maid (Q♠). The dealer wins 0% of the time, and the
non-dealer wins 100% of the time. The
player who starts with the old maid (the dealer) wins 0% of the time.
2) {K♥ | Q♠, K♦} – The dealer receives one card of the matching pair (K♥), and the non-dealer receives the old maid and the other match
(Q♠, K♦). In this case, no
one starts with the matching pair, so the non-dealer takes his turn picking the
dealer’s card (K♥), makes a match (K♥, K♦), and is left with the old maid (Q♠). The
dealer wins 100% of the time, and the non-dealer wins 0% of the time. The player who starts with the old maid (the non-dealer)
wins 0% of the time.
3) {K♦ | Q♠, K♥} – The dealer receives the other card of the matching pair
(K♦), and the non-dealer receives the old maid and the other
match (Q♠, K♥). In this case, no
one starts with the matching pair, so the non-dealer takes his turn picking the
dealer’s card (K♦), makes a match (K♥, K♦), and is left with the old maid (Q♠). The
dealer wins 100% of the time, and the non-dealer wins 0% of the time. The player who starts with the old maid (the non-dealer)
wins 0% of the time.
In this
version of Old Maid, the dealer wins 0% of the time in 1/3
of the scenarios (when he or she is dealt the old maid), but wins 100% of the
time in 2/3 of the scenarios (when he or she is not dealt
the old maid), and therefore overall wins 0(1/3) + 1(2/3)
= 2/3 = 67% of the time, which means that the non-dealer
wins 100 – 67 = 33% of the time. The
player who starts out with the old maid wins 0% of the time in all three
scenarios.
Old Maid: 2 Players, 5 Cards
Things start
getting a little more interesting when Old Maid is played with 2 players and 5
cards, because now the outcome is not solely based upon the hand that is dealt
to the players but also upon the selections the players make during gameplay. Let us once again define the old maid as the
queen of spades (Q♠), but
one match as the king of hearts (K♥) and the king of
diamonds (K♦) and another match as the ace of hearts (A♥) and the ace of diamonds (A♦). Here are the possible hands that can be
dealt:
#
|
Hand
|
Discard
Matches
|
Leftover
|
1
|
{K♥, K♦ | Q♠,
A♥, A♦}
|
{
|
1 card
|
2
|
{A♥, A♦ | Q♠,
K♥, K♦}
|
{
|
1 card
|
3
|
{Q♠,
K♥ | K♦, A♥, A♦}
|
{Q♠,
K♥ | K♦,
|
3 cards
|
4
|
{Q♠,
K♦ | K♥, A♥, A♦}
|
{Q♠,
K♦ | K♥,
|
3 cards
|
5
|
{Q♠,
A♥ | K♥, K♦, A♦}
|
{Q♠,
A♥ |
|
3 cards
|
6
|
{Q♠,
A♦ | K♥, K♦, A♥}
|
{Q♠,
A♦ |
|
3 cards
|
7
|
{K♥, A♦ | Q♠,
K♦, A♥}
|
{K♥, A♦ | Q♠,
K♦, A♥}
|
5 cards
|
8
|
{K♥, A♥ | Q♠,
K♦, A♦}
|
{K♥, A♥ | Q♠,
K♦, A♦}
|
5 cards
|
9
|
{K♦, A♦ | Q♠,
K♥, A♥}
|
{K♦, A♦ | Q♠,
K♥, A♥}
|
5 cards
|
10
|
{K♦, A♥ | Q♠,
K♥, A♦}
|
{K♦, A♥ | Q♠,
K♥, A♦}
|
5 cards
|
Because the
non-dealer receives 3 cards out of the 5 possible, we can use mathematical
combinations to and observe that there are 5C3 = 10 total
hands. To have 1 card leftover, the
non-dealer must receive 1 match out of the 2 possible, which happens 2C1
= 2 times. To have 3 cards leftover, the
non-dealer must receive 1 match out of the 2 matches possible, and then either
of the 2 parts of the 1 match out of the 1 leftover matches possible, which
happens 2C1∙1C1∙21
= 4 times. To have 5 cards leftover, the
non-dealer must receive either 2 parts of the 2 matches out of the 2 matches possible,
which happens 2C2∙22 = 4 times.
When there
is only one card leftover after the matches have been discarded, which happens 2/10
= 1/5 of the time, the probabilities are the same as the
2 player 1 card game: the dealer wins 100% of the time, the non-dealer wins 0%
of the time, and whoever starts with the Old Maid wins 0% of the time.
When there
are only three cards leftover after the matches have been discarded, which
happens 4/10 = 2/5 of the time, the
non-dealer starts out by discarding one match and then selects one of two cards
from the dealer. To analyze this
situation better, let us pick one of its specific hands and follow the
resulting gameplay. Let’s say that the
dealer was dealt the queen of spades (Q♠)
and the king of hearts (K♥), and the non-dealer was dealt the king
of diamonds (K♦), the ace of hearts (A♥), and the ace of
diamonds (A♦) (hand {Q♠,
K♥ | K♦, A♥, A♦}). The non-dealer
starts out by discarding the matching aces (A♥, A♦) leaving the king of diamonds (K♦) leftover. Now the non-dealer must select one of the two
cards from the dealer (Q♠, K♥). If the non-dealer
picks the king of hearts (K♥), which is a ½
probability, he or she will have a match of kings (K♥, K♦) and the dealer is left with the queen of spades (Q♠) and loses. However, if the non-dealer picks the queen of
spades (Q♠), which is also
a ½ probability, he or she will have the queen of spades and the king of
diamonds (Q♠, K♦). Now the dealer
would have a turn, and would also have a ½ probability of picking the king of
diamonds (K♦), making a match and winning, or a ½ probability of picking
the queen of spades (Q♠),
not making a match and continuing the game.
The game would continue on forever until one player finally does not
pick the queen of spades (Q♠)
and wins the game.
Perhaps it
would be easier to represent the probability of this Old Maid scenario with a
simple coin game. In this game, you and
your friend take turns flipping a coin, and whoever flips a head first
wins. (Flipping a tail would be the equivalent
of choosing the queen of spades (Q♠)
in the Old Maid equivalent.) The first
player has a ½ probability of winning on the first turn by flipping a head. However, if the first player flips a tail,
then the second player takes a turn and now has a ½ probability of winning by
flipping a head, for an overall ½ ∙ ½ = ¼ probability of winning. If the second player flips a tail, then the
first player takes another turn and now has a ½ probability of winning by
flipping a head, for an overall ½ ∙ ½ ∙ ½ = ⅛
probability of winning. From the first
player’s perspective, the overall probability of winning is ½, then ⅛, then 1/32, and so on, which is an
infinite sum where t1 = ½ and r = ¼, making the overall probability
of winning S = t1/1 – r = ½/1 – ¼ =
2/3. From the
second player’s perspective, the overall probability of winning is ¼, then 1/16, then 1/64,
and so on, which is an infinite sum where t1 = ¼ and r = ¼, making
the overall probability of winning S = t1/1 – r = ¼/1
– ¼ = 1/3.
These are
the same probabilities in the 3-card-leftover scenario of the 2 player 5 card
Old Maid game: the dealer (who starts with the old maid) wins 33% of the time, and
the non-dealer (who doesn’t start with the old maid) wins 67% of the time.
When no
matches can be discarded, leaving all five cards leftover, which happens 4/10
= 2/5 of the time, the non-dealer starts out by selecting
one of two cards from the dealer and discarding a match. To analyze this situation better, let us once
again pick one of its specific hands and follow the resulting gameplay. Let’s say that the dealer was dealt the king
of hearts (K♥) and the ace of diamonds (A♦), and the non-dealer
was dealt the queen of spades (Q♠),
the king of diamonds (K♦), and the ace of hearts (A♥) (hand {K♥, A♦ | Q♠, K♦, A♥}), and begins by selecting the king of
hearts (K♥) from the dealer and discarding the matching kings (K♥, K♦), leaving the dealer with the ace of
diamonds (A♦) and the non-dealer with the queen of spades and the ace of
hearts (Q♠, A♥). Now the
probabilities are the same as the 3-card-leftover scenario of the 2 player 5
card Old Maid game, except now the dealer has the advantage: the dealer (who
doesn’t start with the old maid) wins
67% of the time, and the non-dealer (who does start with the old maid) wins 33%
of the time.
In this 5
card version of Old Maid, the dealer wins 100% of the time in 1/5
of the scenarios (when only one card is leftover), 33% of the time in 2/5
of the scenarios (when three cards are leftover), and 67% of the time in 2/5
of the scenarios (when all five cards are leftover), and therefore overall wins
1(1/5) + 1/3(2/5)
+ 2/3(2/5) = 3/5 =
60% of the time, which means the non-dealer wins 100 – 60 = 40% of the
time. The player who starts out with the
Old Maid wins 0% of the time in 1/5 of the scenarios
(when only one card is leftover), 33% of the time in 2/5 of
the scenarios (when three cards are leftover), and 33% of the time in 2/5
of the scenarios (when all five cards are leftover), and therefore overall wins
0(1/5) + 1/3(2/5)
+ 1/3(2/5) = 4/15 =
27% of the time.
Old Maid: 2 Players, More Cards
The more
cards there are, the more complex the probabilities get. For the 7 card version of Old Maid, the
dealer wins 0(3/35) + 1(12/35) + 1/4(12/35)
+ 3/4(8/35) = 3/5 =
60% of the time, and the player who starts with the Old Maid wins 0(3/35)
+ 0(12/35) + 1/4(12/35)
+ 1/4(8/35) = 1/7 =
14% of the time. For the 9 card version
of Old Maid, the dealer wins 1(6/126) + 1/3(24/126)
+ 2/3(48/126) + 2/5(32/126)
+ 3/5(16/126) = 19/35
= 54% of the time, and the player who starts with the Old Maid wins 0(6/126)
+ 1/3(24/126) + 1/3(48/126)
+ 2/5(32/126) + 2
/5(16/126) = 12/35
= 34% of the time. Following the
pattern, we can create a piecewise function with sum notation to generalize
these probabilities. If p is the number
of pairs in the deck, the probability of the dealer winning is:
and the
probability of the player winning who starts with the Old Maid is:
Computer Simulation
Another way to test these conjectures and equations is to
create a computer program that simulates millions of Old Maid games and keeps
track of all the desired data. The following
program, written in Python, simulates a 2 Player Old Maid game for a defined
number of games with a defined number of cards, and keeps track of the average
amount of times the dealer wins, the average amount of times the player who
starts with the old maid wins, and the average amount of times the old maid is
picked per game:
01
|
# Old Maid Simulator
|
02
|
# Python 2.7.3
|
03
|
# After running, input the number cards in the
deck and the number of games
|
04
|
# you want
to simulate.
|
05
|
|
06
|
# import python libraries
|
07
|
import random
|
08
|
import math
|
09
|
|
10
|
# the combination function
|
11
|
def nCr(n,r):
|
12
|
f = math.factorial
|
13
|
return f(n) / f(r) / f(n-r)
|
14
|
|
15
|
# function that calculates the expected
probability of the dealer winning
|
16
|
# using
the equation
|
17
|
def DealerWinProbability(p):
|
18
|
probdealerwin = 0
|
19
|
if
p % 2 == 0:
|
20
|
for k in range(p / 2 + 1):
|
21
|
probdealerwin = probdealerwin +
1.0 * (k + 1) / (2 * k + 1) * nCr(
|
22
|
p, p / 2 - k) * nCr(p / 2 +
k, 2 * k) * 2 ** (2 * k) / nCr(
|
23
|
2 * p + 1, p + 1)
|
24
|
for k in range(p / 2):
|
25
|
probdealerwin = probdealerwin +
1.0 * (k + 1) / (2 * k + 3) * nCr(
|
26
|
p, p / 2 - k) * nCr(p / 2 + k,
2 * k + 1) * 2 ** (
|
27
|
2 * k + 1) / nCr(2 * p + 1, p
+ 1)
|
28
|
else:
|
29
|
for k in range((p - 1) / 2 + 1):
|
30
|
probdealerwin = probdealerwin +
1.0 * k / (2 * k + 2) * nCr(p, (
|
31
|
p - 1) / 2 - k + 1) * nCr((p
- 1) / 2 + k, 2 * k) * 2 ** (
|
32
|
2 * k) / nCr(2 * p + 1, p +
1)
|
33
|
for k in range((p - 1) / 2 + 1):
|
34
|
probdealerwin = probdealerwin +
1.0 * (k + 2) / (2 * k + 2) * nCr(
|
35
|
p, (p - 1) / 2 - k) * nCr((p
- 1) / 2 + k + 1, 2 * k + 1) * (
|
36
|
2 ** (2 * k + 1)) / nCr(2 * p
+ 1, p + 1)
|
37
|
return probdealerwin
|
38
|
|
39
|
# function that calculates the expected
probability of a the player who
|
40
|
# starts
with the old maid winning using the equation
|
41
|
def OldMaidWinProbability(p):
|
42
|
proboldmaidwin = 0
|
43
|
if p % 2 == 0:
|
44
|
for k in range(p / 2 + 1):
|
45
|
proboldmaidwin = proboldmaidwin +
1.0 * k / (2 * k + 1) * nCr(
|
46
|
p, p / 2 - k) * nCr(p / 2 +
k, 2 * k) * 2 ** (2 * k) / nCr(
|
47
|
2 * p + 1, p + 1)
|
48
|
for k in range(p / 2):
|
49
|
proboldmaidwin = proboldmaidwin +
1.0 * (k + 1) / (2 * k + 3) * (
|
50
|
nCr(p, p / 2 - k)) * nCr(p / 2 + k, 2 * k
+ 1) * 2 ** (
|
51
|
2 * k + 1) / nCr(2 * p + 1, p
+ 1)
|
52
|
else:
|
53
|
for k in range((p - 1) / 2 + 1):
|
54
|
proboldmaidwin = proboldmaidwin +
1.0 * k / (2 * k + 2) * nCr(p, (
|
55
|
p - 1) / 2 - k + 1) * nCr((p
- 1) / 2 + k, 2 * k) * 2 ** (
|
56
|
2 * k) / nCr(2 * p + 1, p +
1)
|
57
|
for k in range((p - 1) / 2 + 1):
|
58
|
proboldmaidwin = proboldmaidwin +
1.0 * k / (2 * k + 2) * nCr(p, (
|
59
|
p - 1) / 2 - k) * nCr((p - 1)
/ 2 + k + 1, 2 * k + 1) * 2 ** (
|
60
|
2 * k + 1) / nCr(2 * p + 1, p
+ 1)
|
61
|
return proboldmaidwin
|
62
|
|
63
|
# function that simulates Old Maid games
|
64
|
def OldMaidSimulator(cards,
games):
|
65
|
#
assign variables
|
66
|
players = 2
|
67
|
pairs = (cards - 1) / 2
|
68
|
|
69
|
#
create a virtual deck, where each card is labeled with a whole number
|
70
|
# the
old maid is 0, and a match is two cards that add up to the
|
71
|
# total number of cards
|
72
|
deck = []
|
73
|
for i in range(cards):
|
74
|
deck.append(i)
|
75
|
|
76
|
#
counting variables for stats
|
77
|
count_dealerwin = 0
|
78
|
count_dealerstartoldmaid = 0
|
79
|
count_dealerwinstartoldmaid = 0
|
80
|
count_otherplayerwinstartoldmaid = 0
|
81
|
count_oldmaidpicked = 0
|
82
|
|
83
|
# loop
through by the defined number of games
|
84
|
for i in range(games):
|
85
|
#
new hand
|
86
|
hand = []
|
87
|
for j in range(players):
|
88
|
hand.append([])
|
89
|
|
90
|
#
shuffle deck
|
91
|
random.shuffle(deck)
|
92
|
|
93
|
#
deal cards
|
94
|
for j in range(cards):
|
95
|
hand[j % players].append(deck[j])
|
96
|
|
97
|
#
match pairs and remove them
|
98
|
for j in range(players):
|
99
|
# find matches
|
100
|
removecard = []
|
101
|
for k in range(len(hand[j])):
|
102
|
if (cards - hand[j][k]) in
hand[j]:
|
103
|
removecard.append(hand[j][k])
|
104
|
|
105
|
# remove matches
|
106
|
for k in range(len(removecard)):
|
107
|
hand[j].remove(removecard[k])
|
108
|
|
109
|
#
set up variables
|
110
|
playerpicker = 0
|
111
|
currentcount_oldmaidpicked =
count_oldmaidpicked
|
112
|
currentcount_dealerstartoldmaid =
count_dealerstartoldmaid
|
113
|
if 0 in hand[1]:
|
114
|
count_dealerstartoldmaid += 1
|
115
|
|
116
|
#
loop game play until only the old maid is left
|
117
|
while len(hand[0]) + len(hand[1])
> 1:
|
118
|
# define whose turn it is
|
119
|
playerpicked = (playerpicker + 1)
% players
|
120
|
|
121
|
# pick a random card
|
122
|
pickedcard =
random.choice(hand[playerpicked])
|
123
|
hand[playerpicked].remove(pickedcard)
|
124
|
|
125
|
# if the card is the old maid (0), then keep it
|
126
|
if pickedcard == 0:
|
127
|
count_oldmaidpicked += 1
|
128
|
hand[playerpicker].append(pickedcard)
|
129
|
|
130
|
# if the card is not the maid (0), then
discard it and its match
|
131
|
else:
|
132
|
hand[playerpicker].remove(cards - pickedcard)
|
133
|
|
134
|
# switch players
|
135
|
playerpicker = (playerpicker + 1)
% players
|
136
|
|
137
|
#
update stats
|
138
|
if len(hand[1]) == 0:
|
139
|
count_dealerwin += 1
|
140
|
if
currentcount_dealerstartoldmaid != count_dealerstartoldmaid:
|
141
|
count_dealerwinstartoldmaid
+= 1
|
142
|
else:
|
143
|
if
currentcount_dealerstartoldmaid == count_dealerstartoldmaid:
|
144
|
count_otherplayerwinstartoldmaid += 1
|
145
|
|
146
|
# get
expected probabilities based on the equation
|
147
|
probdealerwin = DealerWinProbability(pairs)
|
148
|
proboldmaidwin =
OldMaidWinProbability(pairs)
|
149
|
|
150
|
# print
final stats
|
151
|
print "For a 2 Player, " + str(
|
152
|
cards) + " card game of Old
Maid, simulated " + str(games) + " times:"
|
153
|
print " The dealer won " + str(round(
|
154
|
100.0 * count_dealerwin / games, 1))
+ "% of the games played. " + str(
|
155
|
"(Exp: ") + str(round(100.0
* probdealerwin, 1)) + "%)"
|
156
|
print " The player who started with the old maid
won " + str(round(
|
157
|
count_otherplayerwinstartoldmaid)) /
(games), 1)) + (
|
158
|
"% of the time. (Exp: ") +
str(round(100.0 * proboldmaidwin, 1)) + "%)"
|
159
|
print " The old maid was picked an average of
" + str(round(
|
160
|
1.0 * count_oldmaidpicked / games,
2)) + " times per game."
|
161
|
|
162
|
# Get user input values of the number of cards in
the deck and the number of
|
163
|
while True:
|
164
|
# get
the user input value of the number of cards in the deck
|
165
|
while True:
|
166
|
try:
|
167
|
cards = int(input("Please enter the number of cards in the "
+ (
|
168
|
"deck (must be an odd
number): ")))
|
169
|
except NameError:
|
170
|
print "You must enter an odd
number."
|
171
|
continue
|
172
|
except SyntaxError:
|
173
|
print "You must enter an odd
number."
|
174
|
continue
|
175
|
else:
|
176
|
if cards % 2 == 1:
|
177
|
break
|
178
|
else:
|
179
|
print "You must enter an
odd number."
|
180
|
|
181
|
# get
the user input value of the number of games to simulate
|
182
|
while True:
|
183
|
try:
|
184
|
games = int(input("Please
enter the number of games " + (
|
185
|
"to simulate: ")))
|
186
|
except NameError:
|
187
|
print "You must enter a
number."
|
188
|
continue
|
189
|
except SyntaxError:
|
190
|
print "You must enter a
number."
|
191
|
continue
|
192
|
else:
|
193
|
break
|
194
|
|
195
|
# run
the simulation
|
196
|
print ""
|
197
|
OldMaidSimulator(cards, games)
|
198
|
print ""
|
199
|
print ""
|
For example,
simulating a 5 card game 1,000,000 times results in the following:
>>
Please enter the number of
cards in the deck (must be an odd number): 5
Please enter the number of
games to simulate: 1000000
For a 2 Player, 5 card game of
Old Maid, simulated 1000000 times:
The dealer won 60.0% of the games played.
(Exp: 60.0%)
The player who started with the old maid
won 26.8% of the time. (Exp: 26.7%)
The old maid was picked an average of 0.8
times per game.
|
The higher
the number of games that are simulated, the closer the simulated probabilities
are to the expected probabilities, which shows that the complex probability
equations are indeed accurate.
Summary
The
following is a summary of the expected probabilities for the dealer winning (and
for the player who starts with the old maid winning) a 2 Player Old Maid game for
a deck of 1 card to a deck of 51 cards (which is the number of cards for Old
Maid on a standard deck of playing cards):
cards
|
pairs
|
dealer
wins
|
old
maid wins
|
cards
|
pairs
|
dealer
wins
|
old
maid wins
|
|
1
|
0
|
100.0%
|
0.0%
|
3
|
1
|
66.7%
|
0.0%
|
|
5
|
2
|
60.0%
|
26.7%
|
7
|
3
|
60.0%
|
14.3%
|
|
9
|
4
|
54.3%
|
34.3%
|
11
|
5
|
55.8%
|
23.2%
|
|
13
|
6
|
52.4%
|
38.1%
|
15
|
7
|
53.7%
|
28.8%
|
|
17
|
8
|
51.5%
|
40.4%
|
19
|
9
|
52.5%
|
32.5%
|
|
21
|
10
|
51.0%
|
42.0%
|
23
|
11
|
51.8%
|
35.1%
|
|
25
|
12
|
50.8%
|
43.1%
|
27
|
13
|
51.3%
|
37.1%
|
|
29
|
14
|
50.6%
|
43.9%
|
31
|
15
|
51.0%
|
38.5%
|
|
33
|
16
|
50.5%
|
44.6%
|
35
|
17
|
50.8%
|
39.7%
|
|
37
|
18
|
50.4%
|
45.1%
|
39
|
19
|
50.7%
|
40.7%
|
|
41
|
20
|
50.3%
|
45.5%
|
43
|
21
|
50.6%
|
41.5%
|
|
45
|
22
|
50.3%
|
45.9%
|
47
|
23
|
50.5%
|
42.1%
|
|
49
|
24
|
50.2%
|
46.2%
|
51
|
25
|
50.4%
|
42.7%
|
Analyzing
the table we can see several patterns. First
of all, the dealer is always more likely to win than the non-dealer. Secondly, the player who starts with the old
maid is always more likely to lose than the player who does not start with the
old maid. Third, as the number of cards
increases, the likelihood of the dealer winning (mostly) decreases, and the likelihood
of the player who starts with the old maid winning (mostly) increases, and both
percentages approach a limit of 50%.
The most
unexpected part of all this is that these probabilities are only continuous if the
data is split between even and odd matching pairs (as presented in the table). If you consider a deck with 4, 5, and 6 matching
pairs (9, 11, and 13 cards), we might expect the dealer percentages to
continually decrease, but in fact they do not, as the respective percentages are
54.3%, 55.8%, and 52.4%. The
probabilities only continuous decrease if you separate the data between even
and odd matching pairs, for example a deck with 4, 6, and 8 matching pairs (9,
13, and 17 cards) do have decreasing percentages 54.3%, 52.4%, and 51.5%. This means that the dealer has a slight
advantage if there is an odd number of matching pairs as opposed to an even
number.
So if you’re
playing Old Maid with a friend and you want to stack the probabilities in your
favor, make sure you’re the dealer, play with as little the number of cards as
you dare, and try to make sure you have an odd number of matching pairs. But most importantly, don’t forget to have
fun – it is a game, after all.
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