Download plumbstone lesson - Indigenous Perspectives in Mathematics

Plum Stone Game Math Lesson
Created by Karen Gagnon and Len Bonafacio - Edmonton Catholic Schools
History
This game belongs to the second and non-ceremonial class of the games of hazard and is
generally played by women. The Omaha type is here given, but it is similar to the game as
played by kindred tribes.
Properties.
Needed: Five plum stones, a basket or wooden bowl, one hundred counters. The Omaha used
stalks of the blue joint grass as counters, but small twigs or sticks will serve. Two of the stones
are burned black on both sides with a hot iron; on one side of each of these stones a crescent is
marked, and between the lines of the figure the black is carefully scraped so as to leave a clear
design of a new moon on a background of black. On the other side of these two stones a star,
four or five pointed, is drawn and all the black within the lines is scraped off, leaving a brown
star on a background of black. The other three stones are each burned black all over on one side;
the other side is left the natural color of the stones.
Front
Reverse Side
Moon(M) Star(S)
Front
Black(B)
Reverse Side
White(W)
There are 2 Moon/Star disks and 3 Black/White disks in the basket.
Directions
Two players to one basket or bowl. The game is generally to a total of one hundred points.
The two players sit opposite and have the basket or bowl between them, with the five plum
stones lying in the bottom. The one hundred counters are within reach at one side. As points are
made, the winner takes a corresponding number of counters from the general pile and lays them
beside her on the side opposite to the general pile; when this is exhausted, then the winner takes
her counters from the winnings of her opponent. Whoever wins all of the one hundred points has
the game.
Scoring
Two moons and three whites (natural color) = 10 points.
Two stars and three blacks = 10 points.
One moon, one star and three whites (natural color) = 1 point.
One moon, one star and three blacks = 1 point.
No other combinations count for any points in the game.
The probabilities and expected values associated with the scoring outcomes above are fairly
complicated because of 5 disks being involved, so I suggest we build up to that level from
simpler levels.
Level 1
Using only 2 disks, the Moon/Star disks, let’s abbreviate these to symbols M and S.
1
, since there are two sides and
2
1
only one side is M. Likewise, the probability of that disk showing S is also .
2
1. The basic probability of one disk showing M when tossed is
Show the sample space of all possible outcomes of tossing the two M/S disks.
Answer:
MM
MS
SM
SS
or M
S
M
S
With either model we can see that there are 4 possible outcomes.
2. Expected Value of a game is the average result of any one trial of the game for either player.
Expected Value is defined as:
EV = (probability of 1st outcome)(payoff of 1st outcome) + (probability of 2nd outcome)(payoff of 2nd outcome)
If there are more than 2 outcomes, then we continue adding the products for each outcome.
A game is considered to be fair if the Expected Value is zero. Any other result will favor one of
the players.
Let’s say that we set up the scoring at this level so that if player A tosses the 2 disks that both
show the same result(two M’s or two S’s) he gets 3 counters, but if the disks are different(one M
and one S), his opponent, player B gets 3 counters.
Show the expected value for player A of one trial of this game. Is this a fair game?
Answer:
2
2
EV   3    3
4
4
6 6
 
4
4
0
So, this is a fair game.
Level 2
1. Using the 3 Black/White disks, let’s abbreviate these to symbols B and W.
1
Again, the probability of one disk showing B when tossed is , and the probability of it showing
2
1
W is also .
2
Show the sample space of all possible outcomes when the three B/W disks are tossed.
Answer:
BBB
BBW
BWB
WBB
BWW
WBW
WWB
WWW
or B
B
W
B
W
B
W
W
B
W
B
W
B
W
Either way, we see that there are 8 possible outcomes in the sample space.
2. Let’s set up the scoring so that if player A tosses all 3 disks showing the same result he gets 5
counters, but if he tosses 2 disks of one color and 1 disk of the other color, then his opponent,
player B gets 2 counters.
a) Find the expected value for player A of one toss of the three disks. Is this a fair game?
Answer:
2
6
EV   5    2
8
8
10  12
 
8
8
2

8
1

4
No, the game is not fair. It favors player B, since on an average turn of the game, player A
will lose 1/4 of a counter to player B.
b) Can you suggest a way to change the scoring system so that the game becomes fair?
Answer:
If player A got 6 counters instead of 5 for all 3 disks being the same, then the expected value
would be zero, making the game fair.
In fact, as long as the ratio of scoring was a 3:1 ratio (because of the probabilities being 2/8
versus 6/8) the game will be fair.
Level 3
This is the toughest level; we will toss all 5 disks and use the original scoring system.
Front
Reverse Side
Moon(M) Star(S)
Front
Black(B)
Reverse Side
White(W)
There are 2 Moon/Star disks and 3 Black/White disks in the basket.
Scoring:
Two moons and three whites (natural color) = 10 points.
Two stars and three blacks = 10 points.
One moon, one star and three whites (natural color) = 1 point.
One moon, one star and three blacks = 1 point.
No other combinations count for any points in the game.
1. a) Determine the expected value for a player on one toss of the 5 disks.
Answer: Since the probabilities are more complicated, let’s calculate them individually first:
1 1 1 1 1 1
P(twoM , threeW )          
 2 2   2 2 2  32
1 1 1 1 1 1
P(twoS , threeB)          
 2 2   2 2 2  32
2 1 1 1 2
P(oneM , oneS , threeW )         
 4   2 2 2  32
2 1 1 1 2
P(oneM , oneS , threeB)         
 4   2 2 2  32
All other results count for zero points, so we can ignore them.
1
1
2
2
10  10  1  1
32
32
32
32
10 10 2
2

  
32 32 32 32
24

 0.75
32
 EV 
b) Explain the meaning of this result.
Answer: This means that on each turn of the game, one player theoretically will earn 0.75 of a
counter. But it does not mean that the game is unfair since player B does not earn counters on
player A’s turn. It does mean that on one turn it is not too difficult to earn 1 counter, but it is
quite difficult to earn 10 counters. This indicates that the game will generally take a fair amount
of time for one player to win.
Program of Studies Correlation
Math 7
Strand: Statistics and Probability (Chance and Uncertainty)
Specific Outcomes:
4. Express probabilities as ratios, fractions, and percents.
6. Conduct a probability experiment to compare the theoretical probability (determined
using a tree diagram, table, or other graphic organizer) and experimental probability of
two independent events.
Math8
Strand: Statistics and Probability (Chance and Uncertainty)
Specific Outcome:
2. Solve problems involving the probability of independent events.
Math 9
Strand: Statistics and Probability (Chance and Uncertainty)
Specific Outcome:
4. Demonstrate an understanding of the role of probability in society.
Math 30-2
Strand: Probability
Specific Outcome:
3. Solve problems that involve probability of two events.
Math 30-1
Strand: Permutations, Combinations, and Binomial Theorem
Specific Outcome:
2. Determine the number of permutations of n elements taken r at a time to solve
problems.