Three Fun Games of Probability, and the Big Bang Lottery
or
The Fatal Flaw of Arguing for Intelligent Design from Probability
Dr. Bob L. Sturm, Assistant Professor
Department of Architecture, Design and Media Technology
Aalborg University Copenhagen
Lautrupvang 15, 2750 Ballerup, Denmark
August 2, 2011
Abstract
I present some basic results of probability theory that reveal human intuition can be completely wrong even when our assumptions appear reasonable. Game 1 shows a brilliant
example of this using birthdays. Game 2 shows how events with zero probability can still
occur, and how unreasonable results can come from apparently reasonable models. Game 3
shows that we cannot reasonably talk about aspects of the real random world using probability until we define a space of possible outcomes, and build a model underlying those. In
section 2, I present the “Big Bang Lottery (BBL),” which shows the futility of arguing that
because I won a lottery, and because the probability of me entering the lottery is assumed to
be nearly zero, then I was meant to win the lottery, and the lottery was designed such that I
and only I would win it. This extremely intuitive argument, called the argument for design
from probability, is very persuasive as a proof that the universe was intelligently designed;
but it is fatally flawed, and elementary probability theory reveals why.
1
1.1
Three Games
Game Number One: Birthdays
Given a room of 25 independently and randomly selected people, I will give you $50 if no one
in the room shares the same birthday. Otherwise you give me $100. Are you willing to play?
Figuring out the probability of you winning is elementary. Consider first one person. What
is the probability this person does not share the same birthday with anyone else in the room?
Obviously there are 365 possibilities; and since there is no more than one person in the room,
365 of these are possibilities of the event that no one share the same birthday. This gives a
probability of 365/365 = 1, which is a “certain event.” Extending this for two people is a simple
application of counting. For the first person we have 365 possible dates from which to choose.
For the second person to not share the same birthday with the first person, there are only 364
possible dates. So the total number of possible pairs of dates that are not the same is 365 ∗ 364;
and the total number of possible dates is 365 ∗ 365. This gives a probability that they do not
share the same birthday as:
365 364
= 0.9973.
365 365
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Three Fun Games, Bob L. Sturm
In other words, the probability of these two people sharing the same birthday is 1 − 0.9973 =
0.002739. That is less than three to a thousand against me.
We repeat this procedure for three people. There are 365 possibilities for the first; 364 for
the second; and 363 for the third. Thus, the probability that none of them share a birthday is
365 ∗ 364 ∗ 363
= 0.9918.
3653
Thus the probability of at least two of them sharing the same birthday, which includes the
possibility that they all were born on the same day of the year, is 1 − 0.9918 = 0.0082. In this
case, the odds are 8 to 1000 against me, which is not very good. Continuing in this way for
25 independently and randomly selected people, we find that the probability that none of them
share the same birthday is
365 ∗ 364 ∗ . . . ∗ 342 ∗ 341
= 0.4313.
36525
Thus, the probability of at least two of them sharing the same birthday is 1 − 0.4313 = 0.5687,
i.e., greater than 50%. This means I should expect to win this game more than half the time!
In fact, on average, I should expect to win 0.5687 ∗ 100 − 0.4313 ∗ 50 = $35.03 each time I play.
If we play three times, it is likely I will be up by at least $100.
Of course, if the people selected to be in the room were selected from a convention of fraternal
twins, for instance, the odds will obviously be horribly skewed toward me. And if we happened
upon a group of people with members that avoid people of their same zodiac sign, the odds
will be horribly skewed toward you. Thus, the above analysis is true given the assumption that
our group of independently and randomly selected people have birthdays that are distributed
uniformly over all 365 possible days of the year.
1.2
Game Number Two: The Exponential Bus
Consider waiting for the special Exponential Bus at one of its scheduled stop. The Exponential
Bus is never early, but often arrives within a few minutes of its scheduled time, nearly always
arrives within ten minutes of the scheduled time, and very seldom arrives after ten minutes
past the scheduled time. Furthermore, our wait time for the Exponential Bus is a continuous
exponential probability distribution function, e.g.,
fT (t) =

 1 e−t/5 ,
t≥0
0,
t<0
5
(1)
where t is time in minutes and is continuous, beginning from the scheduled arrival time of the
bus at the stop in consideration. Also assume that this model is not time-varying, i.e., the
arrival delay of the bus does not change with time, e.g., later in the day, or on another date.
Figure 1 shows this function.
Assuming that we arrive at the bus stop at the scheduled time, our expected wait time for
the bus is five minutes:
Z ∞
E[T ] =
−∞
tfT (t)dt =
Z ∞
t −t/5
e
dt = 5.
0
2
5
(2)
August 2, 2011
Three Fun Games, Bob L. Sturm
0.2
0.18
0.16
0.14
T
f (t)
0.12
0.1
0.08
0.06
0.04
0.02
0
0
2
4
6
8
10
12
t (minutes)
14
16
18
20
Figure 1: Probability density function modeling the wait time for the Exponential Bus.
And the probability of waiting longer than 10 minutes is only 0.13:
P {T > 10} =
Z ∞
fT (t)dt =
Z ∞
1 −t/5
e
dt = 0.13.
10
10
5
(3)
Thus the probability of having to wait no longer than 10 minutes is 1 − 0.13 = 0.87. Thus,
eighty-seven percent of the time we will wait no longer than 10 minutes; and half of our waits
will likely be less than 5 minutes. This model sounds reasonable and realistic to apply to all
buses that never arrive early.
Now, what is the probability of waiting exactly five minutes? Integrating the probability
density function above we see that:
P {T = 5} =
Z 5
Z 5
1 −t/5
fT (t)dt =
e
dt = 0.
5
5
5
(4)
Even though it is the expected wait-time, the probability we will wait exactly 5 minutes is
zero. In fact, for any particular time between minus infinity and infinity, the probability of
waiting that amount of time is zero. With any probability distribution function defined over a
set of continuous regions, there exists no probability mass at single points, or even collections
of single points. This means that the probability of waiting for any countably infinite subset of
[0, ∞)1 , say the natural numbers, is zero. The bus does arrive at some time, however, and the
probability of it arriving at that precise time is zero. This shows that just because an event has
zero probability, it is not precluded from occurring.
Now suppose that we have been waiting for this bus for ten minutes and it still has not
come. I will bet you $100 we will be waiting at least another four minutes. If the bus comes
before then I will pay you $50. As calculated above, the probability that this bus will arrive
after ten minutes of waiting is 0.13; and for it to arrive between 10 to 14 minutes of waiting,
the probability is 0.0149. Will you take the bet?
Now that we have new information, i.e., that we have been waiting for ten minutes, we can
use this information to update the model describing our new wait time of the bus. Thus, given
1
A countably infinite set is one for which we can pair each member with a unique integer.
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Three Fun Games, Bob L. Sturm
our assumption that the model is not time-varying,
fT ≤t|T >10 (t) =

1.478e−t/5 ,
t > 10
0,
t ≤ 10.
(5)
We know we have waited 10 minutes already, so the probability of its arrival delay to be less
must be zero. Now, what is the probability that the bus will be here in the next four minutes,
given we have already waited 10 minutes? This is simply
P {10 < T ≤ 14|t > 10} =
Z 14
10
fT ≤t|T >10 (t)dt = 0.449.
(6)
Surprisingly, I will win this bet with a probability of 0.551.
What if the bus does not arrive after sixty minutes, but we know for certain that the bus is
still coming. The probability of that occurring is very slim; in fact it is 0.2e−60/5 = 0.0000012.
But given that it has not arrived after sixty minutes, I will bet you $100 we will have to wait at
least four minutes more. If not, I will give you $50. Now, will you play? Who wouldn’t, given
that the bus is certainly still coming our way?
Similar to the above, we can update the probability density function with this new information:

32, 550e−t/5 , t > 60
(7)
fT ≤t|T >60 (t) =
0,
t ≤ 60
The probability that the bus will arrive in the next four minutes then is:
P {60 < T ≤ 64|t > 60} =
Z 64
60
fT ≤t|T >60 (t)dt = 0.449.
(8)
Yet again I will win with the same probability of 0.551.
In fact, the wait time for the Exponential Bus given that we have been waiting any amount
of time, even 100 years, will always be five minutes more. This curious and unintuitive result
comes from conditioning the exponential model with the information that we have been waiting
for some time; but of course, most of the time, we will be waiting under five minutes after the
scheduled arrival time.
1.3
Game Number Three: Guess my Number
Me: I am thinking of a number. I will give you $1,000 if you correctly guess it, but it costs a
dime to play. Would you like to play?
You: Who wouldn’t? But how can I be sure you wont change the answer if I guess correctly?
Me: I will write each number down before you guess, and show it to you after your guess.
You: Sounds fair. Here’s ten dollars; I will play one hundred times! I’ll guess five.
Me: Nope. Seven.
You: How about three?
Me: Nope. Twenty-two.
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Three Fun Games, Bob L. Sturm
You: Eleven?
Me: Negative four.
You: Wait a second. Isnt it a number between one and a hundred?
Me: I said I was thinking of a number. Negative four is a number.
You: Oh. Can we make it a number between one and ten? ... to be more fair?
Me: Alright. I am thinking of a number between one and ten.
You: Four?
Me: Nope, four and one tenth.
You: Six point three?
Me: So close! Six point three three.
You: Oh come on! That was close enough.
Me: Not for this game.
You: Can we make it an integer between one and ten?
Me: That is not very fair to me. You would have at least a one in ten chance of guessing
correctly. Would you be happy selecting an integer between one and fifty thousand?
You: The chances of winning that are only one to fifty thousand.
Me: Not always. The odds could be much better.
You: What do you mean? There are fifty thousand possibilities and I choose one.
Me: Consider I choose an integer between one and ten, ninety percent of the time. And ten
percent of the time I choose an integer between eleven and fifty thousand.
You: Ok. Five.
Me: Nope. One hundred seventy two.
You: Five.
Me: One hundred seventy three.
You: One hundred seventy four?
Me: Eighty.
You: Eighty?
Me: Three thousand two hundred and twenty.
You: Five?
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Three Fun Games, Bob L. Sturm
Me: One.
You: Ok, this is frustrating! Tell me how you are picking the numbers.
Me: I can’t tell you. But there is a way to find out.
You: How?
Me: By making hundreds, maybe thousands, of observations and learning the statistics of the
process that way. Then, you might be able to make more informed choices. But, if I am
picking numbers uniformly and am able to pick the same number numerous times, this
will not help you no matter how many times you play. Your odds of winning will be one
in fifty thousand every time; but take comfort in the fact that that is your worst odds
possible. And what is the probability of that being the case??
You: Give me back my money.
Me: Here’s $9.00. Pleasure doing business with you.
1.4
Discussion
These three games reveal very important and basic, yet surprising, properties of probability, and
the real world. Games #1 and #2 show us that probability can be contrary to our intutition,
which we often believe to be logical, or rational. Who would believe that with twenty-five people
in a room, and 365 days in a year, that the probability of at least two of them sharing the same
birthday is better than one to one odds? Often, when two people meet and discover they share
the same birthday, it is treated as a miracle, or a sign that they share a cosmic fate. With each
group of 23 new people you randomly meet, you should expect about half of the time to meet
one that also has your birthday.
Game #2 shows us, again contrary to “good” sense, no matter how long we have been
waiting for a bus with a wait time modeled by an exponential probability distribution (which
as demonstrated is not an unreasonable model given our experience), we should still expect to
wait five minutes more. We also see that the probability of any particular wait time is zero, even
when we consider a countably infinite number of possible delay times; but that does not mean
any one wait time is impossible. More deeply, this game shows that additional information —
that the bus has not shown up in some amount of time — gives no (as in none, and not even a
little) new information that improves our expected wait time. The slogan for the Exponential
Bus is, “Never more, never less — always five minutes more!”
Finally, game #3 demonstrates the importance of properly defining a model in order to reasonably talk in terms of odds, i.e., the space of possible outcomes, and a measure of probability
on that space. For instance, the model in game #1 is the set of 365 days with the probability of
each day being a birthday of a randomly selected person is 1/365. In game #2, the wait time
for the Exponential Bus is modeled by a continuous exponential random variable having a mean
of five minutes defined on the sample space [0, ∞). In the worst case, prior information comes
only as uninformed assumptions; but with many observations of the outcomes of some event or
process, you can improve your assumptions and model, and increase your odds of winning. In
game #3, the bettor made a poor assumption that the numbers being picked are only positive,
and in 1-100. Then another poor assumption was that the number had to be an integer. You
can learn about the distribution of the selected number only by repeatedly guessing; but if
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the number selection follows a discrete uniform distribution over the integers in [1, 50000], the
bettor cannot improve his odds better than one in 50,000. Without numerous observations —
where “numerous” is the operative word — and without a model, talking meaningfully about
statistics is not possible.
2
The Big Bang Lottery (BBL)
As you walk late to work, taking a path you do not normally take, and listening to a radio station
to which you normally do not listen, you hear an advertisement about an exciting lottery. It is
one dollar per ticket, and you can enter as many tickets as you want. One ticket will be selected
and if it is yours you win one trillion dollars. The only condition is that you tell no one you are
or were playing; for if you do, you forfeit your chance to play, or all your winnings if you win.
Will you play?
Hopefully, having learned from the games above, and being careful with your money, you
will want to determine how likely it is you will win before you play. But you can do no such
thing for a number of reasons. First, being sworn to secrecy with all the others, you do not
know how many other tickets are being played. If you are one of two people playing, you might
figure your odds are pretty good. In that case, though, you still don’t know how many tickets
the other person bought. It could also be that your tickets are included with other tickets for
which there are no winners. So even if you are the only person playing, your ticket might still
not be selected. One trillion dollars is an attractive amount though.
Second, in this mysterious lottery, you know that a ticket will be selected, but you do not
know how the winning ticket will be selected. You might be one of two players, but the person
selecting the winning ticket might have a grudge against you and will try to make sure you
don’t win. On the other hand, the ticket selector might be a friend. Or the ticket selector could
be a fair and impartial computer. One trillion dollars is too attractive of an amount.
Third, you can’t determine your chances of winning from previous lotteries because you do
not know if this type of lottery has ever happened before; and you don’t know whether it will
ever happen again. These three reasons make it impossible for you to determine any odds of
winning or losing. All you know is that you could seriously benefit by winning because, Dude!
its one trillion dollars!
So you decide to take the plunge and go on faith — after all there are no numbers you
can associate with the lottery. You borrow $1000 dollars from family and friends, only telling
them of a wonderful business opportunity, and commit to paying them back whether it comes
through or not. The big day comes, and to your surprise and disbelief you win, and the check
actually comes in the mail and does not bounce when you deposit it.
One day, you reflect on that famous day, and wonder what the chances were. Since you and
other players were sworn to secrecy, you do not know how many people were playing, or if there
even were any. You do not know how the lottery worked, or who or what selected the winning
ticket. You cannot compare it to any lotteries like it. So you begin to think that the lottery
could have been designed such that you would win. in other words, someone really likes you.
What evidence in this scenario supports the conclusion that someone or something rigged the
lottery so that only you would win? The only evidence we have that supports that conclusion
is the fact that you won. Seeing as how that is the only evidence you have, and what terrible
evidence it is — for it leads to circular reasoning — let us make due and try to compute the
probability of the lottery being rigged in your favor given the fact that you won.
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Let H be the event that the lottery was rigged in just your favor, and H c the event that
it wasn’t rigged in just your favor, i.e., all other possible ways for the lottery to operate such
that you winning is not a certain event. You don’t know which of the above two scenarios is
accurate, but those are the only two possibilities. Let W be the event that you won the lottery.
You want to determine the probability that the lottery was rigged only in your favor given that
you win, i.e., P (H|W ). Using the fundamental law of conditional probabilities, you know
P (H|W )P (W ) = P (W |H)P (H)
(9)
which reads, the probability of the lottery being rigged only in your favor given that you win
times the total probability that you win, is equal to the probability of you winning given the
lottery was rigged only in your favor times the total probability the lottery was rigged only in
your favor.
Rearranging this expression gives the familiar Bayes’ rule, and allows you to judge the
probability the lottery was rigged only in your favor given the outcome that you win the lottery:
P (H|W ) =
P (W |H)P (H)
P (W |H)P (H)
=
P (W )
P (W |H)P (H) + P (W |H c )P (H c )
(10)
using the law of total probability to evaluate P (W ). Since you winning given the lottery was
rigged for just you to win is certain, P (W |H) = 1. Thus, the above boils down to
P (H|W ) =
P (H)
P (H) + P (W |H c )[1 − P (H)]
(11)
which leaves you to evaluate the probability that the lottery was rigged such that just you win,
P (H), and the probability that you win the lottery given it operated any other way, P (W |H c ).
You might be tempted to assume that P (W |H c ) is so small that P (H|W ) = P (H)/[P (H) +
] ≈ 1, with > 0 very small, but to do so conflates winning and rigging since P (W ) =
P (W |H)P (H) + P (W |H c )P (H c ) = P (H) + ≈ P (H) in that case. Saying you win is not the
same as saying you were meant to win; that there leads to circular reasoning. Thus, you see
that you can say something sensible about the probability of your special status given that you
won only if you know P (H) and P (W |H c ). In short, winning provides no evidence that you
were meant to win, or that the lottery was rigged in just your favor, unless you know about all
the other possibilities and can associate them with measures of probability.
The question of your specialness continues to plague you. You recollect that fateful day
you learned about the lottery. You didn’t take the usual route to work. It took an extra two
minutes, during which time you learned about the lottery from a radio station to which you do
not normally listen. The only reason you remembered the phone number to call was that the
digits were your own birthday, curiously. Perhaps out of the thousand tickets you purchased, the
thousandth one was the winner. If you didnt have such trusting friends and family you might
not have won. You begin to realize the multitude of events that had to happen in order for you
to have even entered the lottery, going all the way back to your career path after college, your
parents “getting it on,” their parents, ahem, procreating, and so on to the beginning of time.
The coincidences continue to pile into a collection suggesting that the probability of you even
entering the lottery in the first place has got to be nearly zero! Then you think of the person
who or thing that selected the winning ticket. If anything was different in their past, even the
oil on their fingers, or the speed of the spinning drum containing all the tickets, or the flow of
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the air tossing around those numbered balls, then your ticket might not have been picked. You
are convinced that nothing leading up to you winning could be different, otherwise you wouldn’t
have won. Therefore, you reason, “Since the series of events leading to my incredibly fortunate
blessing had to have been so finely tuned and arranged, I was meant to enter that lottery, that
lottery was meant for me, and I and only I was meant to win it.” With all these coincidences
now, and the assumption that the probability of you entering this lottery is nearly zero, what
is the probability that the lottery must have been fixed such that you would win?
Let S be the event of you entering the lottery. By the above reasoning you assume that
P (S) must be vanquishingly small, and almost zero. If one event of the above sequence of events
were different, you may have never entered the lottery, and thus never have won. Using the law
of conditional probabilities, you can then say:
P (H|W ∩ S)P (W ∩ S) = P (W ∩ S|H)P (H)
(12)
where the event W ∩ S signifies entering the lottery and winning it. Rearranging these terms
produces again the familiar Bayes’ rule:
P (H|W ∩ S) =
P (W ∩ S|H)P (H)
.
P (W ∩ S)
(13)
By the axioms of probability, you know 0 ≤ P (H|W ∩ S) ≤ 1, but you might think you can
find a better lower bound. You can claim with certainty that P (W ∩ S) ≤ P (S), which simply
states that the probability of you entering and winning the lottery (whether it is rigged or not)
is less than or equal to the probability of you entering the lottery, with equality if the lottery
was in fact rigged so that only you would win if you entered it. With this, you can bound from
below P (H|W ∩ S)
P (W ∩ S|H)P (H)
P (W ∩ S|H)P (H)
=
≤ P (H|W ∩ S) ≤ 1
P (S)
P (S|H)P (H) + P (S|H c )P (H c )
(14)
using the law of total probability to expand P (S). Now you have a lower bound on the probability that the lottery was rigged just for you, given that you entered it and won; but the number
of terms you have to evaluate has increased, i.e., P (W ∩ S|H), P (H), P (S|H) and P (S|H c ).
You know from above that you are missing P (H); but what about all the others?
Is it possible to evaluate P (W ∩ S|H)? This is the probability you entered and won the
lottery given that it was designed for you to win. It is conceivable that part of the design of the
lottery was to make sure you heard about it and entered it, which says that you were meant
to enter the lottery, and that the lottery was rigged in only your favor. In this case, since H
implies S is a certain event, P (S|H) = 1, and obviously P (W ∩ S|H) = 1. Thus, you will
certainly enter and win a lottery rigged such that you will enter and win. You gain nothing but
more circular reasoning by making such an assumption.
So you instead say that H does not imply S, or that the lottery was rigged only in your favor
if you entered, and that the probability of you entering in the first place is very small. But here
you still have a problem because you have to evaluate P (S) = P (S|H)P (H) + P (S|H c )P (H c ).
So, you say, “I will assume P (S|H) and P (S|H c ) are very small to make P (S) very small, since,
you know, all those events leading up to my entering the lottery compounded are extremely
unlikely, whether or not the lottery was rigged for me to win if I entered.” Then, to make
P (H|W ∩ S) have a large lower bound, you just want P (W ∩ S|H)P (H) to be large. But
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here you have another problem: by the axioms of probability, P (W ∩ S|H)P (H) ≤ P (S) since
P (H|W ∩ S) ≤ 1. The largest P (W ∩ S|H) can be is one, and in this case, 0 ≤ P (H)/P (S) ≤ 1.
Assuming P (S) is very small, we see from this that P (H), the probability that the lottery was
designed such that only you would win, can only be as big as P (S), which we assume is very
small. The smaller you make the chances of you entering the lottery, the smaller you make the
chances of the lottery being rigged only in your favor. Obviously something is not right. In
fact, when you assume P (W ∩ S|H) = 1, you are again assuming H implies W and S.
To avoid such ridiculousness, you are then forced to assume P (W ∩ S|H) < 1, which says
that if the lottery is rigged such that you win if you entered, the lottery is not rigged for you.
Then the real problem becomes the other term, P (H) — the probability the lottery is designed
such that only you win. As you saw above using the fact that you won, adding the assumption
that you entering the lottery in the first place has a very small probability, you are still unable
to evaluate the probability of the lottery being rigged such that only you will win because you
simply do not know P (W ∩ S|H) or P (H). Thus, you are left where you were at the beginning,
when you were trying to determine P (H|W ). The fact that you won, and the assumption that
the probability of you entering the lottery is nearly zero, provide no evidence that the lottery was
designed such that you and only you would win!
3
Conclusion
Intelligent design (ID), the idea that our universe must have been designed by an intelligent
agent because the chances of our existence otherwise are too slim, is a persuasive argument for
the existence of at least an intelligent agent some time before “the beginning.” The numerous
“statistics” that ID proponents quote, the multitude of “coincidences” they tender, and the
“testimonies” of scientists who “see design in everything,” seems to leave little doubt that our
existence in the universe is so improbable that it just cannot be an accident, and we must have
been planned for all along. How could it be any different with probabilities like one hundred
billion trillion trillion million to one?
Advocates of ID claim that our expanding scientific knowledge is finally shedding indisputable scientific proof on the existence of an intelligent designer; and many even believe it has
proven certain religious texts to be true, e.g., the Bible, and the Koran. The intricate balance of
the speed of light, Newton’s Gravitational constant, and the mass and charge of the electron, is
well-quoted. ID says that if any one of these universal constants were one-hundred-thousandth
of a percent different, the universe would be void of all the elements necessary to human life
because either stars wouldn’t be able to form, or their lifespan would be so short that the time
and conditions necessary for life to form would be missing. Thus, if these constants were arrived at randomly, the probability of our existence is so close to zero that we can consider it
impossible. It is not surprising how influential the idea of ID continues to be. “Statistics” and
“probabilities” can be extremely convincing for people who do not understand them, and this
can make then dangerously misleading.
However, the argument of design from probability is very poor. One doesn’t even have
to argue against it by pointing out its inexhaustible naivety of a cosmos that is indisputibly
inhospitable to human life, or its human-centeredness assuming that human life is the goal of
the universe, and not a mouse named Jimmy. It is fundamentally flawed because of its circular
reasoning, and its misuse of probability. In short, ID claims H, but considers only one particular
H c (“pure chance”) without justification. The two choices thus form a false dichotomy until
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one justifiably defines the sample space from which the universe was selected. Furthermore,
ID never provides a probability measure on this sample space, and thus leaves undefined the
critical components P (H) and P (W |H c ). What shows that H is even a possible event? Have
we seen such examples before? Currently, we only have one sample from the sample space: our
own universe. We are as stuck as the player in Game #3. And because of this, essentially,
ID and its argument from probability are as fatally flawed as proving the lottery was rigged
for you to win, and you were meant to win it, given that you win, and that you entering to
win is extremely unlikely. No matter how many coincidences are given by ID proponents, no
matter how many “intricately balanced” factors of life are found, no matter how many scientists
“see design everywhere,” any attempt to use probability to prove that our existence must have
been planned for because it is so unlikely otherwise, is fatally flawed until one can produce a
reasonable model for the random variable mapping the set of events that includes our universe
to a measure of probability.
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