592
A Aspects of Combinatorial Analysis
probabilities have changed with the number of experiments performed. This tells
us that we have not performed enough experiments for our results to be converged.
Let us then use all the data that we have acquired. We would perhaps find that
N H = 245, N T = 255, (N = 500), so that p H = 0.49 and p T = 0.51. We are still
seeing a change in the values of the probabilities with the number of experiments!
Let us now examine the sequence of results that we have obtained.
N
p H
p T
100
0.42
0.58
200
0.53
0.47
500
0.49
0.51
·
·
·
·
·
·
·
·
·
∞
0.50
0.50
The answer to the original question that we posed is therefore: ‘All that I can tell
you is that the probability of obtaining H is
1
2 , i.e., a 50% chance.’
In a sense we have made a guess, but it was a reasoned guess, based upon the
assumptions that the coin is an ‘honest’ coin (i.e., that the centre-of-mass of the
coin coincides with the geometrical centre), and that there are only two possible
outcomes for our coin toss (H or T). We have seen that in the long run the numbers
N H and N T will tend to be equal, i.e., N H = N T =
1
2 N
p H =
N H
N
−→ in the long run
1
2 N
N
=
1
2
(A.1.4)
We can speak of probabilities only when the number of experiments or trials
is extremely large—equivalently, we can only make proper statistical analyses on
large quantities of data. In the case just discussed in detail we have the normalization
condition p H +p T = 1, since we know for certainty that the outcome of any trial will
be either H or T. More generally, for experiments leading to M mutually exclusive
outcomes, we have seen that the normalization condition is
M
r=1
p r = 1 .
(A.1.5)
1. Let us now ask about the probability of occurrence of either outcome r or
outcome s.
• N r systems in the ensemble exhibit outcome r
A Aspects of Combinatorial Analysis
probabilities have changed with the number of experiments performed. This tells
us that we have not performed enough experiments for our results to be converged.
Let us then use all the data that we have acquired. We would perhaps find that
N H = 245, N T = 255, (N = 500), so that p H = 0.49 and p T = 0.51. We are still
seeing a change in the values of the probabilities with the number of experiments!
Let us now examine the sequence of results that we have obtained.
N
p H
p T
100
0.42
0.58
200
0.53
0.47
500
0.49
0.51
·
·
·
·
·
·
·
·
·
∞
0.50
0.50
The answer to the original question that we posed is therefore: ‘All that I can tell
you is that the probability of obtaining H is
1
2 , i.e., a 50% chance.’
In a sense we have made a guess, but it was a reasoned guess, based upon the
assumptions that the coin is an ‘honest’ coin (i.e., that the centre-of-mass of the
coin coincides with the geometrical centre), and that there are only two possible
outcomes for our coin toss (H or T). We have seen that in the long run the numbers
N H and N T will tend to be equal, i.e., N H = N T =
1
2 N
p H =
N H
N
−→ in the long run
1
2 N
N
=
1
2
(A.1.4)
We can speak of probabilities only when the number of experiments or trials
is extremely large—equivalently, we can only make proper statistical analyses on
large quantities of data. In the case just discussed in detail we have the normalization
condition p H +p T = 1, since we know for certainty that the outcome of any trial will
be either H or T. More generally, for experiments leading to M mutually exclusive
outcomes, we have seen that the normalization condition is
M
r=1
p r = 1 .
(A.1.5)
1. Let us now ask about the probability of occurrence of either outcome r or
outcome s.
• N r systems in the ensemble exhibit outcome r