Puzzling Prob Stats / Bayes problem

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The discussion revolves around calculating the probability of flipping heads on a fourth toss after obtaining heads in the first three tosses with a fair coin and a double-headed coin. Initially, an incorrect assumption led to a probability of 1/2, but further analysis revealed that the probability of having the double-headed coin is significantly higher after observing three heads. The correct calculation shows that the probability of getting heads on the fourth toss is 17/18, considering the likelihood of having either coin after the first three results. The participants clarified their reasoning and corrected their calculations, ultimately arriving at the accurate probability. The final consensus emphasizes the importance of Bayesian reasoning in probability assessments.
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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.
 
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Quantumduck said:
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.
 
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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.
 
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Quantumduck said:
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.
 
bala.l said:
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)
 
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!
 
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