# Conditional Probability brainteaser

• Nov 15th 2011, 10:24 AM
sfspitfire23
Conditional Probability brainteaser
Suppose I flip two coins without letting you see the outcome, and I tell you that at least one of the coins came up heads. What is the probability that the other coin is also heads?

Through cond prob. I think the answer is 1/4, but others are saying its 1/3. There are 4 possible combinations, and knowing that at least one coin turned up heads doesnt change the probability of getting HH in flipping 2 coins.

we cant just eliminate TT from the equation bc before you flipped the coins, TT has 1/4 chance of turning up.

But if you said right from the start that the coins are somehow biased that at least 1 coin will always turned up heads, then i would agree that the probability of getting HH is 1/3.

otherwise if getting HH, HT, TH or TT have equal chances, then it should be 1/4.

Thoughts?
• Nov 15th 2011, 10:54 AM
wnvl
Re: Conditional Probability brainteaser
For me the answer is 1/3

HH HT TH TT have equal chance in general.

One of the coins is head, this means that only

HH HT TH

are possible, all with equal chance.

-> 1/3
• Nov 15th 2011, 11:03 AM
sfspitfire23
Re: Conditional Probability brainteaser
Quote:

Originally Posted by wnvl
For me the answer is 1/3

HH HT TH TT have equal chance in general.

One of the coins is head, this means that only

HH HT TH

are possible, all with equal chance.

-> 1/3

I still think that you can't eliminate the proability of TT in calculations just because one is heads after the fact.
• Nov 15th 2011, 11:08 AM
wnvl
Re: Conditional Probability brainteaser
Yes you can, because the chance for TT is zero after telling that at least one of the coins came up H.
• Nov 15th 2011, 01:16 PM
takatok
Re: Conditional Probability brainteaser
Another way of looking at this problem: I have a bag with 1 red, 1 blue, 1 green and 1 black marble. I pull a marble out. At this point the probability are:
\$\displaystyle P(red)=0.25\$
\$\displaystyle P(blue)=0.25\$
\$\displaystyle P(green)=0.25\$
\$\displaystyle P(black)=0.25\$
\$\displaystyle P(red)+P(blue)+P(green)+P(black) = 1\$
Which makes sense. I have 100% chance of having 1 marble if I pull 1 marble.

Now I look at it and tell you its NOT black. The chances are now
\$\displaystyle P(red)=0.3333\$
\$\displaystyle P(blue)=0.3333\$
\$\displaystyle P(green)=0.3333\$
\$\displaystyle P(black)=0.0\$
\$\displaystyle P(red)+P(blue)+P(green)+P(black) = 1\$
Again i have 100% of having at least 1 marble in my hand.

If I tried to insist the probabilites remained .25 for the colors I'd have this:
\$\displaystyle P(red)=0.25\$
\$\displaystyle P(blue)=0.25\$
\$\displaystyle P(green)=0.25\$
\$\displaystyle P(black)=0.0\$
\$\displaystyle P(red)+P(blue)+P(green)+P(black) = .75\$
Which is saying after pulling 1 marble out of the bag, looking at it and saying its not black. There is now a 75% chance of me having a marble in my hand. Which is obviously absurd :) I definitely have a marble in my hand.

Now you can equate this:
red = HH
green = HT
blue = TH
black = TT.

Now telling you that at least one coin is Heads is the same as saying its not black. So the odds of double heads coming up (red) is 1/3.