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

Hello Professor,

Let:

- F = picked the fair coin

- B = picked the biased coin

- H = got heads on the first toss

Initially:

- P(F) = 1/2

- P(B) = 1/2

The chance of getting heads on the first toss is:

- P(H | F) = 1/2

- P(H | B) = 1

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

P(H) = (1/2 * 1/2) + (1/2 * 1)
= 1/4 + 1/2
= 3/4

Now update the probability of which coin was chosen after seeing heads.

Probability it was the fair coin:

P(F | H) = (1/2 * 1/2) / (3/4)
= (1/4) / (3/4)
= 1/3

Probability it was the biased coin:

P(B | H) = (1/2 * 1) / (3/4)
= (1/2) / (3/4)
= 2/3

Now find the probability of getting heads on the second toss:

P(second head | first head)
= P(F | H) * P(H on second toss | F) + P(B | H) * P(H on second toss | B)
= (1/3 * 1/2) + (2/3 * 1)
= 1/6 + 2/3
= 1/6 + 4/6
= 5/6

So, the probability that the second toss is also heads is 5/6.

Answer 2

Salam!
 

Let F = fair coin

Let B = biased coin

P(F) = 1/2

P(B) = 1/2

First toss is heads.

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

P(H on first toss | B) = 1

So:

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

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

Now we can compute the probability the second toss is 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

Final answer: 5/6 = 83.33%

Answer 3

Hello everyone,

In this case, we have 

1. Fair coin P(H) = 0.5

2. Biased coin P(B) = 1

3. Choice of each coin = 1/2

Now, we want to calculate the probability of Biased coin given Head:

P(B∣H)=P(H∣B)P(B) / P(H∣B)P(B) + P(H∣F)P(F) = (1 * 0.5) / 1 * 0.5 + 0.5 * 0.5 = 2 / 3

Which means the probability that it is biased = 2/3, then the probability that it is fair = 1/3​

Probability that second toss is head = P(Hs)=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.

It means the second toss is most probably going to be Head again. 

Answer 4

Explanation:
Let the coins be Fair (F) and Double-headed (D), each chosen with probability 1/2. After observing the first toss is Heads:

P(H) = 1/2 * 1 + 1/2 * 1/2 = 0,75 

P(D | H) = (1 * 1/2)/3/4 = 2/3

P(​F | H) = 1/3

Then the probability second toss is Heads: 

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

So, after seeing one head, it is more likely the coin is double-headed, making the second Heads highly probable. 

Answer 5

H1 - first is heads

H2 - second is heads

P(rigged) = 0.5

P(fair) = 0.5

P(H1|rigged) = 1

P(H2|rigged) = 1

P(H1|fair) = 0.5

P(H2|fair) = 0.5

P(H2) = 0.25

P(H1) = P(fair)*0.5 + P(rigged)*1 = 0.5*0.5 + 0.5*1 = 0.25 + 0.5 = 0.75

Solution:

P(H2|H1) = P(H2|rigged)*P(rigged|H1) + P(H2|fair)*P(fair|H1) = 1*(1*0.5/0.75) + 0.5*(0.5*0.5/0.75) = 1*2/3 + 0.5*1/3 = 2.5/3 = 0.83333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333333

Answer 6

Let:

• P(F)=0.5, P(B)=0.5
• P(H|F)=0.5, P(H|B)=1

After first head:

P(F|H₁) = (0.50*0.5) / (0.5 + 1*0.5) = 0.25/0.75 = 1/3
P(B|H₁) = 2/3

Second toss:

P(H₂|H₁) = P(F|H₁)*0.5 + P(B|H₁)*1
= (1/3)*0.5 + (2/3)*1
= 1/6 + 2/3
= 5/6

Answer: P(H₂|H₁) = 5/6

Answer 7

So, for sake of simplicity, let's suppose F is the fair coin, while B is the biased coin. Then from the statement "I pick one of the two coins randomly", we get below probabilities:

• P(F)=1/2​
• P(B)=1/2

But we also have this condition, which says that fair coin will give H with 0.5 probability each, while biased coin will always (i.e., with 1.0 probability) give H. Thus, we have:

• P(H | F) = 1/2
• P(H | B) = 1

So, using total probability formula, we get that the probability of getting heads on the first toss is:
                      P(H) = P(H | F) * P(F) + P(H | B) * P(B) = 1/2 * 1/2 + 1 * 1/2 = 3/4

Now to answer to the question "What is the probability that it comes up as heads again?", we will need below formula:

                     P(second toss = H | first toss = H) = P(B | H) * 1 + P(F | H) * 1/2​          (1)
We can find unknown values above using using the Bayes' rule:

• P(B | H) = (P(H) * P(H | B) ) / P(B) = (1 * 1/2) / 3/4 = 2/3
• ​P(F | H) = (P(H) * P(H | F) ) / P(F) = (1/2 * 1/2) / 3/4 = 1/3

Thus, above equation (1) becomes:
                               P(second toss = H | first toss = H) = 2/3 + 1/3*1/2 = 5/6

The answer is 5/6 or ~83.3%

Answer 8

Hello,

According to the given instructions the solution will be like the following:

We have 2 coins:

Fair coin:
P(H) = 0.5, P(T) = 0.5

Biased coin:
P(H) = 1

We choose one coin randomly, so:

P(Fair) = 1/2

P(Biased) = 1/2

First toss comes up Heads.

We want:

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

So first update which coin is more likely after seeing a Head.

Probability of getting Head on first toss:

P(H1) = P(H1|Fair)P(Fair) + P(H1|Biased)P(Biased)

P(H1) = (0.5)(1/2) + (1)(1/2)

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

Now find posterior probabilities.

Probability that the chosen coin was Fair given first Head:

P(Fair|H1) = [(0.5)(1/2)] / (3/4)

P(Fair|H1) = (1/4) / (3/4) = 1/3

Probability that the chosen coin was Biased given first Head:

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

P(Biased|H1) = (1/2) / (3/4) = 2/3

Now compute probability of Head on second toss:

P(H2|H1) = P(H2|Fair)P(Fair|H1) + P(H2|Biased)P(Biased|H1)

P(H2|H1) = (0.5)(1/3) + (1)(2/3)

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

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

Final answer:

Probability that it comes up Heads again = 5/6 ~= 0.8333

Answer 9

Hii, everyone, 

Use conditional probability (Bayes reasoning):

• Two coins: Fair (P(H)=0.5), Biased (P(H)=1)

• Pick randomly → P(Fair)=P(Biased)=0.5

• First toss = Heads

Update belief:

• P(Biased | H1) = (1 × 0.5) / (1×0.5 + 0.5×0.5) = 2/3

• P(Fair | H1) = 1/3

Second toss:

• If biased → always H

• If fair → H with probability 0.5

Final probability:

P(H2 | H1) = (2/3)×1 + (1/3)×0.5 = 5/6

Answer: 5/6 ≈ 0.833

Answer 10

Prior: P(Fair)=P(Biased)=0.5

P(H)=0.5*0.5+1*0.5=0.75

P(Biased∣H)=0.5/0.75=2/3​

P(Fair∣H)=1/3​

Second toss: P(H2​|H1​)=1*2/3+0.5*1/3=5/6

Answer: 5/6