Probability Theory – How to Generalize Probability of ‘Head’ Occurring on a Coin

Question

Supposedly I toss a coin 1000 times a day for a period of 10 days. Each day the probability of head occurring on the coin changes. For instance – 670 times getting a head on day 1, 567 times getting a head on day 2 and so on. So each time, the probability of a head occuring isn't 0.5, so can someone please explain why we generalise that the probability of a head occuring on tossing of a coin is 0.5??

Answer 1

In a straightforward 10 day experiment with 1000 tosses each day with a fair coin you might see the following numbers of Heads on the 10 days. [Sampling in R.]

set.seed(2022)
x = rbinom(10, 1000, .5)
x
[1] 488 488 502 493 495 490 524 487 500 506

If someone doesn't know you used a fair coin, then they might use the procedure prop.test to see whether the Heads probability each day might be $0.5.$

prop.test(x, rep(1000,10), p=rep(.5,10))

        10-sample test for given proportions 
        without continuity correction

data:  x out of rep(1000, 10), null probabilities rep(0.5, 10)
X-squared = 4.988, df = 10, p-value = 0.892
alternative hypothesis: two.sided
null values:
 prop 1  prop 2  prop 3  prop 4  prop 5  prop 6  prop 7  prop 8  prop 9 prop 10 
    0.5     0.5     0.5     0.5     0.5     0.5     0.5     0.5     0.5     0.5 
sample estimates:
 prop 1  prop 2  prop 3  prop 4  prop 5  prop 6  prop 7  prop 8  prop 9 prop 10 
  0.488   0.488   0.502   0.493   0.495   0.490   0.524   0.487   0.500   0.506 

The null hypothesis is that $p=P(\mathrm{Heads}) = 0.5$ each day, against the alternative hypothesis that $p \ne 0.5$ on one or more of the days. $H_0$ cannot be rejected because the P-value $0.892 > 0.05 = 5\%.$

Clearly, not all daily proportions of Heads are exactly $0.5,$ but the random variation (among observed proportions, $0.487$ to $0.524)$ is not more than would be expected by chance.

By contrast, if you got results as in the vector y below, then someone might wonder whether you used a fair coin each day.

set.seed(228)
y = rbinom(10, 1000, seq(.2, .8, len=10))
y
[1] 182 250 332 408 456 508 611 668 772 809

Then prop.test can be used to see if Heads probabilities are all the same (the default null hypothesis, in case nothing else is specified in the input).s prop.test(y, rep(1000,10))

    10-sample test for equality of proportions 
    without continuity correction

data:  y out of rep(1000, 10)
X-squared = 1649.3, df = 9, p-value < 2.2e-16
alternative hypothesis: two.sided
sample estimates:
 prop 1  prop 2  prop 3  prop 4  prop 5  prop 6  prop 7  prop 8  prop 9 prop 10 
  0.182   0.250   0.332   0.408   0.456   0.508   0.611   0.668   0.772   0.809 

For this test, the P-value is very near $0$ so $H_0$ is rejected.

Notes: (1) Both versions of prop.test are roughly equivalent to chi-squared tests. You could use chi-squared tests if you like, but I find the output from prop.test to be more informative for current purposes.

(2) This Answer is based on one possible interpretation of your Question, which I found to be somewhat vague. If you have something else in mind, please edit your question to be more specific.