CareerCup

8/9 is the (conditional) probability that (after 5 tosses) the coin was the coin that has heads on each side.
The probability of having heads "the next time" is independent of the previous 5 tosses, but conditioned on the pull from bag. The answer of Anonymous below, i.e., 3/5 is the correct answer in this case.

This is not correct.
If what you are saying is correct, then the probability of two heads in two tosses is 3/5 * 3/5 = 9/25, right?
But the probability of two heads in two tosses should be calculated as (4/5 * 1/2 * 1/2) + (1/5 * 1 * 1) = 2/5.

Gursharan's comment explains the answer neatly.

i think 3/5 is the right answer as the result of a toss doesnt depend on the previous tosses:)
@riderchap: in your answer 9/25, there is a mistake, you arrive at the result when u choose a different coin for each toss. which is not the case here...

I was exposing that your answer is indeed based on the assumption that each time you are pulling a coin from the bag. But that is not the case here.

The result of a toss doesn't depend on previous tosses if you know what you are tossing. If I am tossing a fair coin the probability of next head after 5 (or any number of) heads is 1/2. If that is a double headed the probability is 1. Here you don't know what you are tossing.

For a fair coin the probability of a head after 5 heads is 1/2 as we all know, and we can reach that solution by the same solution proposed by Gursharan.

P(6 heads in a row ) =  1/64
P(5 heads in a row ) =  1/32
P( Next head ) = P( Next head | 5 heads in a row)
                ----------------------------------
                       P(5 heads in a row )
But P( Next head | 5 heads in a row) = P(6 heads in a row )
P( Next head ) = P(6 heads in a row )
                 ---------------------
                 P(5 heads in a row )

P( Next head ) = 1/64
                 ----  = 1/2
                  1/32

For the problem given.

P(6 heads in a row ) =  17/80
P(5 heads in a row ) =  9/40
P( Next head ) = P( Next head | 5 head in a row)
                ----------------------------------
                       P(5 heads in a row )
But P( Next head | 5 head in a row) = P(6 heads in a row )
P( Next head ) = P(6 heads in a row )
                 ---------------------
                 P(5 heads in a row )

P( Next head ) = 17/80
                 ----  = 17/18
                  9/40

Imagine this, I have a fair coin in my hand and I got say 1000000000000000000 heads in a row, since its a fair coin I know the probability of next head is 1/2.

Now in the given scenario I got say 100 heads in a row, I don't know its fair coin or a double headed coin in my hand. At first (when we pulled the coin from the bag) there was a 4/5 chance that its a fair coin and 1/5 chance that its a double headed coin in my hand. But now since we saw 100 heads in a row there is 2^n/(4+2^n) = 2^100/(4 + 2^100 ) = 0.99999999999999999999999999999684 chance that its a double headed coin (see http://math.arizona.edu/~jwatkins/f-condition.pdf). Still do you think that the probability of next head is 3/5?

My answer comes as 39/32

Explanation.

First time when you pick coin out of bag. you have
4 conins with head and tail and one coin with head only.
Probability of getting head five time can be count as follows.
with first coin with head and tail five times head = 1/32. second coin and 5times head 1/32 and upto 4th coin 1/32
for the 5th coin probability of getting head is 1 its surely returns 1.
(1/32 + 1/32 + 1/32 +1/32 + 1) = 36/32 = 9/8.

now when one coin is already removed we have 4 coins left.
two posibilities. one might be fair or might not be. we have to consider both case.

I) when one fair coin is there
probability = (1/32 + 1/32 + 1/32 +1) = 35/32

II) when one fair coin is not there
probability = (1/32 + 1/32 + 1/32 + 1/32 ) = 4/32

adding two posibilities 35/32 + 4/32 = 39/32.

how can probability be greater than 1..:O..:S

I think the answer, 3/5, is correct. That is:
(4/5 * 1/2) + (1/5 * 1) = 3/5.
The comment above from RiderChap, which shows 3/5 * 3/5 for a two-heads toss would, expanded out, be:
[(4/5 * 1/2) + (1/5 * 1)]*[(4/5 * 1/2) + (1/5 * 1)]
But this brings in all kinds of cross terms, which don't seem to make much sense.

Also, Gursharan's comment below is taking into account the previous 5 tosses, which have nothing to do with the outcome being asked for. Do you really think that the probability of getting a head is that close to 1, when there is a 4/5 probability that the coin is a fair one? I don't.

>>"The comment above from RiderChap, which shows 3/5 * 3/5 for a two-heads toss would, expanded out, be:
[(4/5 * 1/2) + (1/5 * 1)]*[(4/5 * 1/2) + (1/5 * 1)]
But this brings in all kinds of cross terms, which don't seem to make much sense."

Yes, it does not make any sense. I put that to explain that 3/5 is not correct. I am sorry if my comment was misleading.


If I am getting all heads for 100000 tosses in a row, there is a greater chance that I pulled a two headed one from the bag, don't you think? So if the previous outcomes are given you need to consider that, it could give you (in the given question it will) a more accurate probability for what kind of coin is in your hand.

See http://math.arizona.edu/~jwatkins/f-condition.pdf

It is not P(H/5H) we need to find. We need to find P(6H/5H).

Find P(6H/5H) and you will get 17/18.

Let A be the event that we tossed a fair coin, B be the event that we tossed a "unfair" coin, and C be the event that all 5 times head turned up.

P(A)=4/5, P(B)=1/5, P(C|A)=(1/2)^5,P(C|B)=1

Apply the Bayes' Rule,

P(A|C)= {P(C|A)P(A)}/{P(C|A)P(A)+P(C|B)P(B)}=1/9

P(B|C)= {P(C|B)P(B)}/{P(C|A)P(A)+P(C|B)P(B)}=8/9

The probability that next toss turns out to be Head should be

P(A|C)*(1/2)+ P(B|C)*1=17/18.

No.

He seems to be barking up the wrong tree. Way to give him a ruff time, dawg.

Hi. This answer is correct only if the coin is redrawn after each toss. Because you are using the same coin, you must assume that with each successive head, it is that much more likely that you are using the trick coin, therefore it will carry more weight.