Puzzling Prob Stats / Bayes problem

[Quantumduck]

Hello All,
I have a problem that goes like this:I have one fair coin and one double heads coin. I pick one at random and flip the same coin three times. All three times it comes up heads. What is the probability that this same coin will come up heads a fourth time?

I said that the prob on the fourth throw is 1/2. Why?

The probability for the fair coin to throw 4 heads is .5 x .5^4 = 1/32. (the first .5 because of the random choice between the two coins.)

The probability of the double heads throwing 4 heads is .5 x 1^4 = .5

Therefore, the probability of throwing a 4th heads is .5, because it is much more likely that I have the double heads coin in my hand than the fair coin.

My question is, did I make an error in figuring this? It seems too easy, but I can not find any flaws in my reasoning (because I am too close to it)

Thanks in advance.

 

[Rogerio]

Hello All,
I have a problem that goes like this:I have one fair coin and one double heads coin. I pick one at random and flip the same coin three times. All three times it comes up heads. What is the probability that this same coin will come up heads a fourth time?

Your universe: 1/32 + 1/2
Probability of having the double heads coin: (1/2) / ( 1/32 + 1/2 ) = 32/33
Probability of having the fair coin: (1/32) / ( 1/32 + 1/2 ) = 1/33
( of course you could get the last result by "1 - 32/33" )

Then, the probability of a 4th head is:
1.0 * 32/33 , if you have the double heads coin
0.5 * 1/33 , if you have the fair coin

So, the answer is 65/66.

 

[bala.l]

If the fair coin is chosen, the probability of 4th head would be (1/2)^4.

If the biased coin is chosen, the probability of 4th head would be 1.

So, the probability of 4th head could be
Prob(Fair Coin)*Prob(4th head on Fair Coin) OR Prob(Biased Coin)*Prob(4th head on Biased Coin)
= 1/2 * (1/2)^4 + 1/2 * 1
= 17/32

Is there a mistake somewhere?

I didn't follow Rogerio's reply, and the divisions are wrong in that.

EDIT--The mistake is that I haven't taken into account that the first three tosses result in head.

 

[CRGreathouse]

Hello All,
I have a problem that goes like this:I have one fair coin and one double heads coin. I pick one at random and flip the same coin three times. All three times it comes up heads. What is the probability that this same coin will come up heads a fourth time?

1/8 chance to get all heads with the normal coin, 8/8 with the double-head coin. The chance that you have the normal coin is thus 1/9 since you were equally likely to have picked up either coin. The chance of getting a tails on the normal coin is 1/2, for an overall chance of 1/18; otherwise, you have four heads. 17/18 is thus the chance of getting four heads, given the first three are heads.

 

[Rogerio]

I didn't follow Rogerio's reply, and the divisions are wrong in that.

My mistake.
Correcting the values:

Your universe: 1/16 + 1/2
Probability of having the double heads coin: (1/2) / ( 1/16 + 1/2 ) = 8/9
Probability of having the fair coin: (1/16) / ( 1/16 + 1/2 ) = 1/9
( of course you could get the last result by "1 - 8/9" )

Then, the probability of a 4th head is:
1.0 * 8/9 , if you have the double heads coin
0.5 * 1/9 , if you have the fair coin

So, the answer is 17/18.
(8/9 + 1/18)

 

[Quantumduck]

Thank you!
I noticed that the probability in the first reply were for the 5th throw coming up heads, but I understood what he was doing so I did it for the 4th throw and got ... 17/18! That rocks.
Thank you all for your help!