# quick coin flip question

• Feb 6th 2008, 05:00 PM
xfyz
quick coin flip question
A professor flips two balanced coins. Both fall to the floor and roll under his desk. A student informs the professor that he can see only one coin and it shows tails. What is the probability that the other coin is also tails?

Is it 1/2? I said it was 1/2 because they are independent events? If the question asked what is the probability of getting a combination of T-T , it would be 1/4.
is this right?
• Feb 6th 2008, 05:39 PM
colby2152
Quote:

Originally Posted by xfyz
A professor flips two balanced coins. Both fall to the floor and roll under his desk. A student informs the professor that he can see only one coin and it shows tails. What is the probability that the other coin is also tails?

Is it 1/2? I said it was 1/2 because they are independent events? If the question asked what is the probability of getting a combination of T-T , it would be 1/4.
is this right?

Probability of a combination is 25%, but the probability of a combination (T-T) given a T is 50% because as you say, the events are independent.(Clapping)
• Feb 6th 2008, 05:47 PM
Soroban
Hello, xfyz!

This is a classic trick question . . .

Quote:

A professor flips two balanced coins.
Both fall to the floor and roll under his desk.
A student informs the professor that he can see only one coin and it shows tails.
What is the probability that the other coin is also tails?

When two coins are flipped, there are four possible outcomes: .$\displaystyle HH,\:HT,\:TH,\:TT$

Since the student saw a Tail, there are only three possible situations: .$\displaystyle HT,\:TH,\:TT$

Among them, only one of them has both Tails.

The probabitlity is .$\displaystyle {\color{blue}\frac{1}{3}}$

~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

This is the basis for a "Sucker Bet".

I have four playing cards, one of each suit, face down on the table.

You pick any two of them.

You will bet that the cards are of the same color.
I will bet that they have opposite colors.
. . And we bet "even money".

Are you being hustled?

Let's reason it out . . .

There are only two outcomes: the colors match or they do not match.
. . Since these outcomes are equally likely, the bet is "fair".

Okay, there are four outcomes: (Red, Red), (Red, Black), (Black, Red), (Black, Black)
. . Since we each win half the time, the bet is fair.

The above explanations are comforting and reasonable . . . but wrong!

. . There are six outcomes: .$\displaystyle (\heartsuit\,\diamondsuit),\;(\heartsuit\,\spadesu it),\;(\heartsuit\,\clubsuit),\;(\diamondsuit\,\sp adesuit),\;(\diamondsuit\,\clubsuit),\;(\spadesuit \,\clubsuit)$

. . And in only two of them, $\displaystyle (\heartsuit\,\diamondsuit),\;(\spadesuit\,\clubsui t)$, the colors match.

Your probability of winning is: .$\displaystyle \frac{2}{6}\:=\:\frac{1}{3}$

• Feb 6th 2008, 05:48 PM
Jhevon
Quote:

Originally Posted by Soroban

This is a classic trick question . . .

you know a lot of those, don't you? :D
• Feb 6th 2008, 05:53 PM
Jhevon
Quote:

Originally Posted by Soroban
This is the basis for a "Sucker Bet".

I have four playing cards, one of each suit, face down on the table.

You pick any two of them.

You will bet that the cards are of the same color.
I will bet that they have opposite colors.
. . And we bet "even money".

Are you being hustled?

Let's reason it out . . .

There are only two outcomes: the colors match or they do not match.
. . Since these outcomes are equally likely, the bet is "fair".

Okay, there are four outcomes: (Red, Red), (Red, Black), (Black, Red), (Black, Black)
. . Since we each win half the time, the bet is fair.

The above explanations are comforting and reasonable . . . but wrong!

. . There are six outcomes: .$\displaystyle (\heartsuit\,\diamondsuit),\;(\heartsuit\,\spadesu it),\;(\heartsuit\,\clubsuit),\;(\diamondsuit\,\sp adesuit),\;(\diamondsuit\,\clubsuit),\;(\spadesuit \,\clubsuit)$

. . And in only two of them, $\displaystyle (\heartsuit\,\diamondsuit),\;(\spadesuit\,\clubsui t)$, the colors match.

Your probability of winning is: .$\displaystyle \frac{2}{6}\:=\:\frac{1}{3}$

[/size]

i'll keep that in mind next time i go gambling, thanks :D
• Feb 10th 2008, 07:35 AM
Soroban
Hi, Jhevon!

Quote:

You know a lot of those, don't you?

Yes, I do . . .

Over the years, I've been surprised (or been had) by dozens of these "sucker" bets.
. . The first was probably the "Birthday Paradox".

A similar bet can be made with a friend while standing on a street corner.

Bet your friend that, among the next 15 license plates that go by,
. . some two will end in the same two-digit number.

[You might agree to disregard those ending in letters.]

Argument: "Hey, there are a hundred different two-digit numbers.
. . . . . . . . What are the chances?"

Super Hustle: The same bet with 18 cars.
. . . . . . . . . . The odds are about 4-to-1 in your favor.

• Feb 12th 2008, 10:31 AM
connick
Quote:

Originally Posted by Soroban
Hello, xfyz!

This is a classic trick question . . .

[size=3]
When two coins are flipped, there are four possible outcomes: .$\displaystyle HH,\:HT,\:TH,\:TT$

Since the student saw a Tail, there are only three possible situations: .$\displaystyle HT,\:TH,\:TT$

Among them, only one of them has both Tails.

The probabitlity is .$\displaystyle {\color{blue}\frac{1}{3}}$

I was just posed this problem today, and a quick google search led me to this thread. I still can't seem to get my head around the explanation presented to me, both by the guy who asked the question and by Soroban. Can someody explain where my train of throught is going wrong here?

According to Soroban's explanation, if the student can see one coin, which is tails, that eliminates the HH possiblity, leaving HT, TH, and TT. Out of those possilbilities, there is a 2/3 chance of the other coin being heads, and only 1/3 being tails. HOWEVER, to me this implies that you are choosing which coin you want to see AFTER they've been flipped. In other words, you will flip a HT/TH combo 50% of the time, and when this happens, the coin you see WILL ALWAY BE THE ONE SHOWING TAILS. This simply does not make sense to me. You will flip a HT/TH combo 50% of the time, but when that happens, you will only see the tails coin half the time, and the other half you will see heads, so that situation doesn't count in this riddle.

The way I see it, 50% of the time the student will see heads, thus removing those situations from consideration (this would be the HH and HT flips). The other 50% of the time, the student would see tails, and this would be the TT and TH flips. Of those where the student sees tails, the other coin being tails would have a 50/50 chance. Can someone clearly explain to me how i'm misinterpreting this?
• Feb 13th 2008, 09:47 AM
colby2152
I was going with the fact that a known coin showed up as Tails... hmmm! (Headbang)