1. Case Western Reserve University
Weatherhead School of Management
STAT 538 - Stochastic Models: Diffusive Phenomena and
Stochastic Differential Equations
Coin toss problem
Yanbing Sun (yxs492)
September 10th
, 2015

2. Introduction
Coin toss is always recognized as a typical random event. For example, we use coin toss in
soccer game to decide which team will select side. Because we do not know the accurate initial
state of the motion of the coin, it is hard to know which side is going to face up after the coin
settles down. And it appears that events of heads or tails facing up at the end are equally likely to
occur. But we already know the rules the coin will follow in its motion. So if we can know the
actual initial state, we can know exactly which side of the coin will face upward. It seems there is
no randomness in coin toss process.
Now we try to find out under small uncertainty of initial state, is coin toss an random event
and is possibility of heads up happening equal to tails up happening.
Hypotheses
To simplify this problem, we make such assumptions.
1). The coin is a perfect circle. And we do not consider the thickness of the coin.
2). We do not consider the air resistance.
3). At initial state, we hold the coin perfectly level.
4). We toss the coin upwards vertically, which means we exert a force to the coin so it have an
initial velocity upwards vertically and an angular velocity.
5). When the coin touches the ground, it will not rebound, so the side which faces up will not
change
Symbol description
𝑣: The initial velocity of the coin.
𝜔: The angular velocity at the edge of the coin.
𝑕: The initial distance from the coin to the ground.
𝑔: The gravity of earth.
𝑟: The radius of the coin.
Figure 1 describes the initial state of coin toss.
Figure 1

3. Methodology
The whole motion of the coin is a uniformly accelerated motion. It is easy to see that if the
distance from ground to the center of the coin is no longer than coin’s radius, the coin would not
have enough room to spin half circle. So the side which will face up should be decided at the time
when the distance from ground to the center of the coin is its radius. We can calculate the time 𝑇
it takes from when the coin is tossed till when the side will face upwards is decided.
𝑇 = √
𝑣2
𝑔2
+
2(𝑕 − 𝑟)
𝑔
+
𝑣
𝑔
Assume that the side which faced up at the initial state is 0, and the other side is 1. So when
the coin spin
𝜋
2
to
3𝜋
2
, the side facing up would be 1. Otherwise, it would be 0. It can be
described as follows.
𝑓(𝑥) = {
0, 𝜔𝑇 ∈ (2𝑛𝜋, 2𝑛𝜋 +
𝜋
2
] ∪ (2𝑛𝜋 +
3𝜋
2
, 2𝑛𝜋 + 2𝜋]
1, 𝜔𝑇 ∈ (2𝑛𝜋 +
𝜋
2
, 2𝑛𝜋 +
3𝜋
2
]
𝑛 = 0,1,2, ⋯
If 𝑕 and 𝑟 are fixed, we can draw the function out in a coordinate system.
Figure 2 (initial position of 𝑕 = 0.01 , 𝑟 = 0.01)

4. Figure 3 (initial position of 𝑕 = 5 , 𝑟 = 0.01)
In figure 2 and figure 3, dark areas represent the side faces up when the coin touch the ground
will be the side faces up at initial state. The light areas represent the side faces up will be the side
faces down at initial state. It is obvious that if the velocity upwards is very close to 0, if initial
height is very low, the coin will have no room to spin, the side faces up will always be the side
which faced up at the beginning. If the angular velocity at the edge of the coin is very close to 0,
the coin will not spin during the whole process. So the side faces up will always be the side which
faced up at the beginning too. As the angular velocity increase, the coin would spin faster. The
areas would be thinner and thinner because the it uses less time to change side which face upwards.
When 𝑣 approaches to infinity, the relationship between time of the coin toss 𝑇 and 𝑣 will
approach to linear relationship, because
lim
𝑣→∞
(
𝑇
𝑣
) =
√
𝑣2
𝑔2 +
2(𝑕 − 𝑟)
𝑔 +
𝑣
𝑔
𝑣
=
2
𝑔
If 𝜔 is fixed, the areas which represent heads or tails face up will appear alternately on the
line of time which the coin spend in air.
Figure 4
Because when 𝑣 approaches to infinity 𝑣 and 𝑇 will have linear relationship, we can use 𝑣
to replace 𝑇, and the area will appear in same way.

5. When 𝑣 and 𝜔 approach to infinity, we pick up a rectangle area. If the dark area and light
area are equal, then we can say event of heads up happens and event of tails up happens have the
same possibility. First we cut this rectangle into slices which have same width 𝛿. We already know
that areas which represent heads or tails face up will appear alternately on the line of velocity. So
when 𝛿 approaches to infinitely small, the dark area to light area ratio is the same as dark length
to light length ratio at the left edge of the slice.
Figure 5
It is easy to notice that ratio 𝑅 ∈ [
𝑛
𝑛+1
,
𝑛+1
𝑛
], where 𝑛 = ⌊
𝑇𝜔
2𝜋
⌋ = ⌊
𝑣𝜔
𝑔𝜋
⌋. As 𝜔 approaches to
infinity, 𝑛 also approaches to infinity. So according to Squeeze Theorem, 𝑅 will approach to 1.
Every single slice in the rectangle area have the same ratio = 1 . After we add up all the slices,
the ratio has to be 1. That means as 𝜔 and 𝑣 approach to infinity, in a small rectangle area, the
possibility of heads up events is equal to possibility of tails up events.
We can test it empirically to see whether the area of dark or light zone in rectangle is equal to
half of the area of rectangle zone. We pick a rectangle zone of 𝑣 ∈ [106
, 106
+ 0.001] and
∈ [106
, 106
+ 0.001] . The ratio of light area to the rectangle area is 0.500174510188944
which is very close to theoretical value of 0.5 .
Conclusion
As initial velocity 𝑣 and angular velocity 𝜔 approach to infinity, within a small area of
uncertainty of 𝑣 and 𝜔, the possibility of heads up is equal to the possibility of tails up.

6. Appendix: Matlab code
function [r] = coin ( vb,ve,wb,we,r,h,d )
n=(we-wb)/d+1;
m=(ve-vb)/d+1;
x=[wb];
for i=1:(n-1)
x(i+1)=x(i)+d;
end
y=[vb];
for j=1:(m-1)
y(j+1)=y(j)+d;
end
for i=1:m
t(i)=(2*(h-r)/9.8+y(i)^2/9.8^2)^0.5+y(i)/9.8;
end
z=[0];
r=[];
for i=1:n
for j=1:m
z(i,j)=floor(mod(2*x(i)*t(j)/pi,4));
if z(i,j)==0
r(i,j)=0;
elseif z(i,j)==1
r(i,j)=1;
elseif z(i,j)==2
r(i,j)=1;
else
r(i,j)=0;
end
end
end
end
[x1,y1]=meshgrid(0:0.01:20);
r1=coin(0,20,0,20,0.01,0.01,0.01);
mesh(x1,y1,r1)
[x2,y2]=meshgrid(0:0.01:20);
r2=coin(0,20,0,20,0.01,5,0.01);
mesh(x2,y2,r2)
r3=coin(10^6,10^6+0.001,10^6,10^6+0.001,0.01,5,0.00001);
rto=sum(sum(r3))/(0.001/0.00001+1)^2