Heads - 2 in a row

Question

I have two coins, Fair coin - gives H and T with 0.5 probability each, and a Biased coin, always gives H.

I pick one of the two coins randomly and toss it. It comes up as heads. I toss it again. What is the probability that it comes up as heads again?

Answer 1

F: fair coin

B: biased coin

P(F)=P(B)=0.5

Fair coin: P(H|F)=0.5

Biased coin: P(H|B)=1

P(B|H)=P(H|B)P(B)​/P(H)

P(H)=P(H|F)P(F)+P(H|B)P(B)=0.5⋅0.5+1⋅0.5=0.25+0.5=0.75

P(B|H)=1⋅0.5/​0.75=2​/3

P(F/H)=1/3

P(H on 2nd)=P(H|B)P(B|H)+P(H|F)P(F|H) =1⋅2/3+0.5⋅1/3=2/3+1/6=5/6=0.83

Answer 2

Hi professor,

Let F denote the fair coin and B denote the biased coin.

P(F) = 1/2

P(B) = 1/2

We observe that the first toss results in heads.

P(H on first toss | F) = 1/2

P(H on first toss | B) = 1

Using conditional probability, we update the likelihood of having chosen each coin:

P(B | first toss = H)

= [P(H | B) P(B)] / P(H)

= [1 × (1/2)] / [(1/2 × 1/2) + (1 × 1/2)]

= (1/2) / (3/4)

= 2/3

Therefore, P(F | first toss = H) = 1/3

Now we calculate the probability that the second toss also results in heads:

P(second toss = H | first toss = H)

= P(H | B)P(B | first toss = H) + P(H | F)P(F | first toss = H)

= (1 × 2/3) + (1/2 × 1/3)

= 2/3 + 1/6

= 5/6

Answer 3

Hello Professor,

Let:

F = picked the fair coin

B = picked the biased coin

H₁ = got heads on the first toss

H₂ = got heads on the second toss

Initially:

P(F) = 1/2

P(B) = 1/2

The chance of getting heads on the first toss is:

P(H₁ | F) = 1/2

P(H₁ | B) = 1

So the total probability of getting heads on the first toss is:

P(H₁) = (1/2 × 1/2) + (1/2 × 1)= 1/4 + 1/2 = 3/4

Now update the probabilities using Bayes’ theorem:

P(F | H₁) = (1/2 × 1/2) / (3/4) = 1/4 ÷ 3/4 = 1/3

P(B | H₁) = (1/2 × 1) / (3/4)= 1/2 ÷ 3/4= 2/3

Now find the probability that the second toss is heads:

P(H₂ | F) = 1/2

P(H₂ | B) = 1

So:

P(H₂ | H₁) = (1/2 × 1/3) + (1 × 2/3)
= 1/6 + 2/3 = 1/6 + 4/6 =  5/6

Final Answer:

P(H₂ | H₁) = 5/6

Answer 4

Two coins:
Fair coin: P(H) = 1/2, P(T) = 1/2
Biased coin: P(H) = 1, P(T) = 0

Choose a coin at random:
P(Fair) = 1/2
P(Biased) = 1/2

First toss is Heads.

Total probability of observing Heads:
P(H) = P(H | Biased) * P(Biased) + P(H | Fair) * P(Fair)
 = (1)(1/2) + (1/2)(1/2)
 = 1/2 + 1/4
 = 3/4

Posterior probabilities after observing Heads:
P(Biased | H) = (P(H | Biased) * P(Biased)) / P(H)
 = (1 * 1/2) / (3/4)
 = 2/3

P(Fair | H) =  1/3

Probability that the second toss is Heads:
P(H_2 | H_1) = P(H_2 | Biased) * P(Biased | H_1) + P(H_2 | Fair) * P(Fair | H_1)
 = (1)(2/3) + (1/2)(1/3)
 = 2/3 + 1/6
 = 5/6

Final answer:
5/6

Answer 5

Salam

Fair coin (F): H = 0.5, T = 0.5 

Biased coin (B): always H = 1

You choose a coin randomly, so: P(F) = 1/2 P(B) = 1/2

Firstly we got Heads on the first toss

Now we update our belief: which coin is more likely?
If it was Fair: P(H) = 0.5

If it was Biased: P(H) = 1

So:  P(F and H) = 1/2 × 1/2 = 1/4 and  P(B and H) = 1/2 × 1 = 1/2

Total probability of getting H:  1/4 + 1/2 = 3/4

Now update:  P(F | H) = (1/4) / (3/4) = 1/3  and  P(B | H) = (1/2) / (3/4) = 2/3

So after seeing a head, it’s more likely we picked the biased coin.

Then Second toss

Now we compute the chance of heads again: 

If Fair coin: 0.5 chance 

If Biased coin: 1 chance

So
P(H2​∣H1​)=(1/3)⋅(1/2)+(2/3)⋅(1)=1/6+2/3=1/6+4/6=5/6

Answer 6

Hello Professor,

Let F be the fair coin and B be the biased coin.

At the start:

P(F) = 1/2

P(B) = 1/2

The fair coin gives heads with probability 1/2, while the biased coin always gives heads.

First, we find the probability of getting heads on the first toss:

P(H1) = P(H1 | F)P(F) + P(H1 | B)P(B)

P(H1) = (1/2)(1/2) + (1)(1/2)

P(H1) = 1/4 + 1/2 = 3/4

Now, since we already saw heads on the first toss, we update the probability that the coin was biased:

P(B | H1) = [(1)(1/2)] / (3/4)

P(B | H1) = 2/3

So the probability that it was the fair coin is:

P(F | H1) = 1/3

Now we calculate the probability that the second toss is also heads:

P(H2 | H1) = P(H2 | B)P(B | H1) + P(H2 | F)P(F | H1)

P(H2 | H1) = (1)(2/3) + (1/2)(1/3)

P(H2 | H1) = 2/3 + 1/6

P(H2 | H1) = 5/6

Final answer:

5/6, or about 83.33%

So after seeing one head, it becomes more likely that the coin is the biased coin, which makes the probability of getting heads again higher than 1/2