A fair coin and an unfair coin

[BMAD]

Consider the following process. We have two coins, one of which is fair, and the other of which has heads on both sides. We give these two coins to our friend, who chooses one of them at random (each with probability 1/2). During the

rest of the process, she uses only the coin that she chose. She now proceeds to toss the coin many times, reporting the results. We consider this process to consist solely of what she reports to us.

Given that she reports a head on the nth toss, what is the probability that a head is thrown on the (n + 1)st toss?

Now assume that the process is in state "heads" on both the (n - 1)st and the nth toss. Find the probability that a head comes up on the (n + 1)st toss.

[bushindo]

Calling Dr. Bayes ... Where is Bushindo when you need him?

being heads biases the probability above 1/2 that she is using the double-headed coin. And more so in the second case of two heads. So the asked probability is greater, in both cases, and more in the second case, than the unbiased probability of 3/4.

I won't, on principle, plug numbers into B's formula, so I ask myself this question: if she flips it once and says it's heads, what is the likelihood she flipped the double-headed coin? Seems it would be 2/3. Similarly if it was heads twice, it would be 5/6.

So, if I'm thinking right,

the first answer would be 1/3 x 1/2 + 2/3 x 1 = 5/6

and the second answer would be 1/6 x 1/2 + 5/6 x 1 = 11/12.

On the contrary, I will, on principle, plug the number in Bayes' formula =)

Applying Bayes' formula, if the first N flips are heads, then the chance of the N+1 flip being head is

[ (1/2)^N+1 + 1]/[ (1/2)^N + 1 ]

So the answer to the first question is 5/6, and the answer to the second is 9/10.

[bonanova]

Calling Dr. Bayes ... Where is Bushindo when you need him?

being heads biases the probability above 1/2 that she is using the double-headed coin. And more so in the second case of two heads. So the asked probability is greater, in both cases, and more in the second case, than the unbiased probability of 3/4.

I won't, on principle, plug numbers into B's formula, so I ask myself this question: if she flips it once and says it's heads, what is the likelihood she flipped the double-headed coin? Seems it would be 2/3. Similarly if it was heads twice, it would be 5/6.

So, if I'm thinking right,

the first answer would be 1/3 x 1/2 + 2/3 x 1 = 5/6

and the second answer would be 1/6 x 1/2 + 5/6 x 1 = 11/12.

[BobbyGo]

Calling Dr. Bayes ... Where is Bushindo when you need him?

being heads biases the probability above 1/2 that she is using the double-headed coin. And more so in the second case of two heads. So the asked probability is greater, in both cases, and more in the second case, than the unbiased probability of 3/4.

I won't, on principle, plug numbers into B's formula, so I ask myself this question: if she flips it once and says it's heads, what is the likelihood she flipped the double-headed coin? Seems it would be 2/3. Similarly if it was heads twice, it would be 5/6.

So, if I'm thinking right,

the first answer would be 1/3 x 1/2 + 2/3 x 1 = 5/6

and the second answer would be 1/6 x 1/2 + 5/6 x 1 = 11/12.

I agree with your first answer, but not with your second...

If the biased coin was flipped once, it is 100% likely that the result was heads. If the balanced coin was flipped once, it is only 50% likely that the result is heads. This gives us 2:1 odds (2/3 chance, just like you said) that the biased coin was flipped in the event that heads was reported.

1/3 x 1/2 + 2/3 x 1 = 5/6 probability of next toss being heads

If the biased coin was flipped twice, it is 100% likely that both resulted in heads. If the balanced coin was flipped twice, it is only 25% likely that the results were both heads. This gives us 4:1 odds (4/5 chance, instead of 5/6) that the biased coin was flipped in the event that both reported flips landed as heads.

1/5 x 1/2 + 5/6 x 1 = 9/10 probability of next toss being heads

Of course, I have a knack for being terribly wrong about these types of things.

[bonanova]

Yes. I think the three of us agree that I made a mistake in the second case.

I should say the four of us.

[witzar]

If you don't like Mr Bayes, you can draw a picture.

Now all you have to do is divide green by (green+red).

Edited April 12, 2013 by witzar

[bonanova]

If you don't like Mr Bayes, you can draw a picture.

Now all you have to do is divide green by (green+red).

Nice diagram.

I actually get Bayes, and conditional probability (I think of it as "conditioned" probability.) And I had the right thinking and I had visualized the right numbers. And then I just wrote 5/6 when I should have written 4/5. My Bayesian aversion comes when the numbers are too large or the steps too complicated to visualize. And you just plug, plug, plug, and turn the crank. As an engineer I should love that. That's when the tool really shines. But it also robs me of the opportunity for the puzzle to shape my intuition. and that's part of my love for probability puzzles.