Very Tough Problem

chickenhead · Jan 28, 2002

Let's see if anybody can solve this problem.

You play a game as follows: You toss a coin. If it comes up tails, you toss it again. You keep tossing until you get heads. Once you get heads, you stop tossing.

Under these rules, what is the expected number of tosses that the average person can expect to make on average?

Passions · Jan 28, 2002

1 toss.

chickenhead · Jan 28, 2002

You can't just take random guesses. You have to show your work.

And no, it's not 1 toss. How could it be?

OutHouse · Jan 28, 2002

hummm we did this one in my 10th grade physics class. sorry cant remember

BrunoPuntzJones · Jan 28, 2002

<< Once you get heads >>

tee hee

EpsiIon · Jan 28, 2002

<< Let's see if anybody can solve this problem.

You play a game as follows: You toss a coin. If it comes up tails, you toss it again. You keep tossing until you get heads. Once you get heads, you stop tossing.

Under these rules, what is the expected number of tosses that the average person can expect to make on average? >>

I'd say the average would be about two. Since you have a 50/50 chance of it coming up heads, with a chance that it might not come up heads until 3 or 4, 2 sounds like a reasonable average.

WhiteWonder · Jan 28, 2002

<<

<< Once you get heads >>

tee hee >>

tee hee hee

chickenhead · Jan 28, 2002

<< I'd say the average would be about two. Since you have a 50/50 chance of it coming up heads, with a chance that it might not come up heads until 3 or 4, 2 sounds like a reasonable average. >>

If engineers built bridges the way you do math, millions of people would die every day under collapsed bridges.

AreaCode707 · Jan 28, 2002

Hey chickenhead, I have a question to ask. Were you ever part of these forums under the name greenfirs?

rgwalt · Jan 28, 2002

Define average (mean, median, mode)?

Ryan

lesch2k · Jan 28, 2002

1.9999999999999999 which rounds to 2.0 if you have infinite people participating

figuring 50/50 ratio of heads tails

half the time you stop after 1 toss
half the remaining time you stop after 2 tosses
half the remaining time you stip after 3 tosses etc.

this comes down to a sequence of added fractions which sum to 2

see below for number of tosses

.5*1+.25*2 +.125*3 etc.

gopunk · Jan 28, 2002

<<

<< I'd say the average would be about two. Since you have a 50/50 chance of it coming up heads, with a chance that it might not come up heads until 3 or 4, 2 sounds like a reasonable average. >>

If engineers built bridges the way you do math, millions of people would die every day under collapsed bridges. >>

and if everybody was like you, i'd kill myself.

chickenhead · Jan 28, 2002

<< Define average (mean, median, mode)?

Ryan >>

Expected value is defined by the set of possible values multiplied by their respective probabilities.

Here is a simple example.

You pick a card out of a deck. If it's the king of spades, I give you $52. If it's anything else, I give you nothing. What is your average expected earnings?

The probability of picking the king of spades is 1/52. The probability of picking anything else is 51/52. The value of the king of spades is $52. The value of any other card is $0.

So the average expected value is (1/52 x $52) + (51/52 x $0) = $1

chickenhead · Jan 28, 2002

<< .5*1+.25*2 +.125*3 etc. >>

This adds up to 2? How do you figure?

Show your work please.

EpsiIon · Jan 28, 2002

<<

<< I'd say the average would be about two. Since you have a 50/50 chance of it coming up heads, with a chance that it might not come up heads until 3 or 4, 2 sounds like a reasonable average. >>

If engineers built bridges the way you do math, millions of people would die every day under collapsed bridges. >>

Well, forgive me for doing a quick approximation in my head, which, BTW, engineers and physicists do all the time. If I didn't understand the question, it's likely that you didn't explain it adequately.

chickenhead · Jan 28, 2002

<< it's likely that you didn't explain it adequately. >>

Which part of the question did you have trouble understanding?

lesch2k · Jan 28, 2002

it is a sequence

50% of time you stop after 1 toss
25% you stop after two tosses
12.5% you stop after three tosses

using your expected payout from above

you get
.5 *1+.25*2+.125*3.0625*4+.03125*5+.015625*6 = 1.875

.5 *1+.25*2+.125*3.0625*4+.03125*5+.015625*6 ... + .001015625*10=1.988

any math guy can see what i am saying here

the sum approaches two

gopunk · Jan 28, 2002

lesch2k - don't worry, i can see what you're saying

EpsiIon · Jan 28, 2002

<<

<< it's likely that you didn't explain it adequately. >>

Which part of the question did you have trouble understanding? >>

Well, I thought I understood it. It seemed that you were asking the average (mean) number of times a given person would have to flip a coin to get heads. Intuitively, I'd say the answer is two. Let me know if I'm incorrect.

chickenhead · Jan 28, 2002

The answer is not 2.

It may look like it's approaching 2, but it's approaching another number.

If you properly do the math, you will get the correct answer.

chickenhead · Jan 28, 2002

<< Well, I thought I understood it. It seemed that you were asking the average (mean) number of times a given person would have to flip a coin to get heads. Intuitively, I'd say the answer is two. Let me know if I'm incorrect. >>

Yes, you understood the question correctly. There really isn't much ambiguity.

The answer is not 2.

Ryan · Jan 28, 2002

is it 1.999999999999999999999999999999999999999999999999999999
99999999999999999999999999999999999999999999999999999999999
99999999999999999999999999999999999999999999999999999999999
99999999999999999999999999999999999999999999999999999999999
99999999999999999999999999999999999999999999999999999999999
99999999999999999999999999999999999999999999999999999999999
99999999999999999999999999999999999999999999999999999999999
99999999999999999999999999999999999999999999999999999999999
99999999999999999999999999999999999999999999999999999999999
99999999999999999999999999999999999999999999999999999999999
99999999999999999999999999999999999999999999999999999999999
99999999999999999999999999999999999999999999999999999999999
99999999999999999999999999999999999999999999999999999999999
99999999999999999999999999999999999999999999999999999999999
99999999999999999999999999999999999999999999999999999999999
99999999999999999999999999999999999999999999999999999999999
99999999999999999999999999999999999999999999999999999999999
99999999999999999999999999999999999999999999999999999999999
99999999999999999999999999999999999999999999999999999999998?

Which is REALLY two

EpsiIon · Jan 28, 2002

<< The answer is not 2.

It may look like it's approaching 2, but it's approaching another number.

If you properly do the math, you will get the correct answer. >>

Hmmm, interesting. I'd like to see what it is. Because I just did a tiny, tiny sample with a quarter, and I seemed to average about two flips...

chickenhead · Jan 28, 2002

1.9 bar (an infinite number of 9's) is equal to 2.

Here is a neat way to show that.

Let x = 1.9 bar
10x = 19.9 bar

10x = 19.9 bar
x = 1.9 bar
-------------------
9x = 18
x = 2

chickenhead · Jan 28, 2002

<< Hmmm, interesting. I'd like to see what it is. Because I just did a tiny, tiny sample with a quarter, and I seemed to average about two flips... >>

I took a tiny sample with women, and it seemed like they were all bisexual.

(I live in a college town.)

The lesson to learn from this is to never trust tiny samples.

Very Tough Problem

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