Binomial distribution problem / biased coin toss

binomial distributionprobability

I've got a text where the following statement appears:

You have a biased coin that has a 51% chance of coming up heads and 49% chance of coming up tails. Tossing it 1,000 times, you will generally obtain more or less 510 heads and 490 tails, majority of heads. If you do the math, you will find that the probability of obtaining a majority of heads after 1,000 tosses is close to 75%. The more you toss the coin, the higher the probability (e.g., with 10,000 tosses, the probability climbs over 97%).

This is a Bernoulli experiment executed 1000 times so we are dealing with a binomial distribution.

So doing the math (with python)

from scipy.stats import binom

n = 1000  # tosses
p = 0.51 # probability of head (coin is biased)
x = 501  # to get majority of heads out of 1000 tosses I need just 501
binom.pmf(x, n, p)  # scipy library

0.021451693818843456

And for 10000 tosses

n = 10000  # tosses
p = 0.51 # probability of head (coin is biased)
x = 5001  # to get majority of heads out of 10000 tosses I need just 5001
binom.pmf(x, n, p)  # scipy library

0.0011231912772331355

Assuming I'm wrong, where is my error?

Thanks

Best Answer

You are using the probability mass function when you should be using the cumulative distribution function or its complement

1 - binom.cdf(x-1, n, p) might give you what you are looking for. Or you could take a sum

Related Solutions

[Math] Is the coin fair

With a sample as large as $n = 1000$ and $p$ in the vicinity of 1/2, a normal approximation should work well.

To test the null hypothesis $H_0: p = .5$ against the alternative $H_a: p > .5$ you need to compute the test statistic $$Z = \frac{\hat p - p_0}{\sqrt{p_0(1-p_0)/n}}.$$

In your example, you have $\hat p = 0.51,\,p_0 = 0.5,$ and $n = 1000.$ You would reject the null hypothesis $H_0,$ saying that the data from the $n = 1000$ flips of the coin are inconsistent with the behavior of a fair coin, at the 5% level if $Z > 1.645.$ (The 'critical value' cutting 5% of the probability from the upper tail of a standard normal distribution, can be found from printed normal CDF tables or from software.)

I will leave it to you to finish this. Please compare this with what is in your textbook, and leave a Comment if you still need help.

Notes: It would not be unusual to get 510 or more Heads in 1000 tosses of a fair coin. That would happen over 25% of the time. A truly unusual outcome would be to get exactly 500 Heads in 1000 tosses (probability about 0.025). Results from R statistical software:

1 - pbinom(509, 1000, .5)
## 0.2739864                # P(510 Heads or more)
dbinom(500, 1000, .5)
## 0.02522502               $ P(EXACTLY 500 heads)

Probability to pick up biased coin

Using Bayes rule.

$F$: denote the event you picked the fair coin

$B$: denote the event you picked the biased coin

$D$: Data collected, i.e., observed 3 heads in 3 tosses

We want to calculate: $P(F|D)$ and $P(B|D)$

Assume that the trials are independent. This is equivalent to saying: $P(D|F)=1/8$

$P(D|B)=1$

Let us also assume that $P(F)=P(B)=0.5$, implying that "you randomly pick coin" with equal chances of picking either one.

Apply Bayes theorem, which basically starts from:

$P(B|D)P(D)=P(D|B)P(B)$

Bring over the second term on the left hand side to the right hand side and expand by considering total probability as:

$P(D)=P(D|F)P(F)+P(D|B)P(B) = 1/8*1/2+1*1/2$

$P(B|D) = [P(D|B)P(B)]/[(P(D|F)P(F)+P(D|B)P(B))]$

I have got $\dfrac{8}{9}$ for biased coin.

Related Question

Coin Toss Probability Using Bayes Theorem