[Math] the probability that a coin is fair

bayesianprobability

You paly a game with your friend Alice where you bet on the outcome of a coin toss. The coin has been provided by Alice. You think there is a 50% chance that she would have provided an unfair coin. If the coin is unfair then you believe that the probability that it will turn up heads is uniform in [0, 1].

The question is that,

1: You toss the coin and it comes up head. What is the probability that the coin is fair?

2: You toss the coin for the second time and it comes up head again. Now, what is the probability that the coin is fair?

For me, I solve this problem through this way,

P(fair|data) = $\frac{P(data|fair)P(fair)}{P(data|fair)P(fair)+P(data|unfair)P(unfair)}$

Where I know that P(fair)=0.5, P(data|fair)=$p^1(1-p)^0$=$p=0.5$ (as fair means P(head)=0.5=p), P(unfair)=0.5,

So, the previous equation can be substitued as,

P(fair|data) = $\frac{0.5*0.5}{0.5*0.5+P(data|unfair)*0.5}$

My question is how to express the term of P(data|unfair)?

Thanks.

Best Answer

Part 1: if the coin is fair, it comes up $H$ with probability $\frac 12$. If it is unfair then it comes up $H$ with probability $$\int_0^1 pdp=\frac 12$$. Thus the two cases are symmetric and the answer is $\boxed {\frac 12}$.

Part II. Now the probability that the unfair coin comes up $HH$ is $$\int_0^1 p^2dp=\frac 13$$ While the fair coin of course comes up $HH$ with probability $\frac 14$. Thus we apply Bayes to get $$\frac {\frac 12 \times \frac 14}{\frac 12 \times \frac 13+\frac 12 \times \frac 14}=\boxed {\frac 37}$$

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[Math] what is the probability of BOB winning the game

Hint: Since the two consecutive coin tosses are independent, $\Pr[HT] = \Pr[H]\cdot\Pr[T]$, and $\Pr[TH] = \Pr[T]\cdot\Pr[H]$. Does the winner depend on the bias $p$ at all?

(Note that the probability that someone wins in the first round does depend on the bias; but since the game is repeated until either HT or TH happens, it does not. For more, you may want to read about von Neumann's trick; essentially, Alice and Bob have equal chances of winning, but the (expected) duration of the game will depend on the bias $p=0.72$))

[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)

Best Answer

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