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(Hope this hasn't been done before)

1. There are 10 coins on a table. 9 are normal but 1 of them has 2 heads on it (i.e. will always land on heads). Without looking you pick one at random and flip it 8 times. It lands on heads each time. What is the probability of it landing on heads the next time?

Bonus question: (UNRELATED TO THE FIRST ONE) You flip a coin 20 times and it lands on tails each time. What is more likely for it to land on next, heads or tails?

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dunno about the first one, but 4 the second:

if the first few tests were to see if the coin would land heads or tails, the chances it will be tails is 100% according to those studies.

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I flipped a coin once (a normal coin) and it landed on its edge. I've never repeated it since, but maybe you should factor that into your calcs? I struggle to count to 10, so please don't ask me to do it!

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it doesn't say if you picked the normal coin or the 2 headed coin. if you picked the normal coin your chance will be 50/50, if you piked the 2 headed coin your chance will be 100%

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wow so i just realized y you all were sounding like idiots to me. your talking about a different part of the puzzle my b. ADD gets the better of me again i saw the first problem and went off didnt even see the second

anyway second part 50-50 unless your really counting landing on the edge or the small increased probability that coins land on tails since the head ways more

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[spoiler='answer

']

55%

We don't know which coin it is, even though it landed on heads 8 times.

So, there is a 90% chance that it is a normal coin, which have a 50% chance of landing on heads.

But, there is a 10% chance that it is the rigged coin, which has a 100% chance of landing on heads.

So,

.9 x .5 + .1 x 1 = .45 + .1

= .55 or 55% B))

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OK. First of all I wasnt insulting anybody. Sorry if it sounded like I was :( .

Well a few of you seem to disagree with the answer to this bonus question.

There are 2 possibilities:

1. It was a normal coin and you witnessed something extraordinary (over 1 in a million chance)

2. You were using a fake coin.

Think about it. Which is more likely? I'm sure there are plenty of fake coins out there making the second option the more likely one. I never said that that it was a fair coin.

I, for one, was monumentally insulted.

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theres a real easy way to do these problems you find the probability of flipping 8 heads

find the probability of flipping 9 heads and divide like so

P(8 heads)=.5^8*9/10 +1^8*1/10

P(9 heads)=.5^9*9/10+1^9*1/10

P(9heads given 8heads)=P(9heads)/P(8 heads)= .1017578125/.103515625= .9830188679

and done

Wow...I like this one better. Nice.

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Bonus question: (UNRELATED TO THE FIRST ONE) You flip a coin 20 times and it lands on tails each time. What is more likely for it to land on next, heads or tails?

Wording is awkward and inadequate. Yet, great questions none-the-less.

1. It was a fair coin which would make the chance of the next tails 50/50 or 0.5.

2. It is a biased coin making the probability of tails the next time greater then 0.5 (lets call it x, where 0.5<x<1)

Using the same logic that was used to answer the first question in the OP we can find the chance of the next tails. The probability that number 2 is correct is unkown. Also we wouln't know how biased the coin was. But we don't need to know it. If P is the chance of the next tails then all we know is that:

0.5<P<x

so P is greater than 0.5 and hence tails is more likely.

Finally, what if you toss a coin a million times and it landed tails every time. How can you say that that coin isnt biased? It is so unlikely that I would be more than willing to bet my life (yes really) that there is something not quite right about that coin. So clearly, the next land, we would assume, would be tails.

Now I don't mind debating this with someone, but I'm sure you'll have a hard time suggesting that the second point I made is an inaccurate assumption.

I went back and read all the posts. I was skimming through. I will apologize for not reading your restatements of the question.

The way I read your original post it was is the likely hood of it being heads or tails and the only assumption we can make w/o more info is that it is a 50/50; however, I re-read your restatements and realized what you were trying to say. I was ignorant from lack of reading some of the posts which is completely my fault. Yes... it is more likely that this is a bias coin and the probability is .5<P<x; however great/minute that difference may be.

As far as flipping the coin a million times, that is what I am getting at with your sample size. It is a very small sample size. I was only giving examples where the odds seem astronomically but can happen. Twenty flips all landing on one side has an extremely low probability (one in a million - I'll trust your math) but not enough to make huge judgment calls with such a small sample size. Again, that is why posting the last 20-40 outcomes on the roulette tables have brought such great profits.

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I flipped a coin once (a normal coin) and it landed on its edge. I've never repeated it since, but maybe you should factor that into your calcs? I struggle to count to 10, so please don't ask me to do it!

Although it does play a part, it becomes infinitely small:

If you base the chances of landing on a part of the coin on its surface area, it is a very small amount of area on the edge.

When compared to the area of the sides of the coin, the side is much smaller. it is connected to the surface of the table, I assume, by not quite a tangent, but very close, it would have to rest on two ridges on the side of the coin (which are very close together).

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No I wouldn't be dumb enough to jump the first time :lol:.

I say the second part belongs in the riddle forum, because you are expecting us to "assume" certain situations. I get where you were going with it, but this forum is for logical/math problems. The information given is what we as posters go off. So if you want assumtions, then I say everyone is wrong with the answer to the first part, if you flip heads that many times in a row, I assume you have a trick coin, or you have the double-headed coin. So its obviously 100%. :P

No offense with my earlier post. ;)

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(Hope this hasn't been done before)

1. There are 10 coins on a table. 9 are normal but 1 of them has 2 heads on it (i.e. will always land on heads). Without looking you pick one at random and flip it 8 times. It lands on heads each time. What is the probability of it landing on heads the next time?

Bonus question: (UNRELATED TO THE FIRST ONE) You flip a coin 20 times and it lands on tails each time. What is more likely for it to land on next, heads or tails?

for the coin probability problem answer will be 521/5120 :):)

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(Hope this hasn't been done before)

1. There are 10 coins on a table. 9 are normal but 1 of them has 2 heads on it (i.e. will always land on heads). Without looking you pick one at random and flip it 8 times. It lands on heads each time. What is the probability of it landing on heads the next time?

Bonus question: (UNRELATED TO THE FIRST ONE) You flip a coin 20 times and it lands on tails each time. What is more likely for it to land on next, heads or tails?

This is an interesting thread, and as some people pointed out, many of you are suffering from the "gambler's fallacy". In short, the prior history does not change the current probablity.

The bonus question is clearly 50% with allowances for the edge landing and slight variations in weight.

However, the first question is deceptively clever, but the gambler's fallacy still is in play. The prior history of 8 heads changes nothing! It only gives us a probablity that the coin is the fake coin, but it does not provide certainty. IE: while the coin is "highly likely" to be the double heads coin, it is in not "certain" that the coin is the double heads coin.

As such, we can not change the initial probablity: we have 10 coins, therefore we have 20 sides. 9 sides are tails and 11 sides are heads. You have an 11 out of 20 chance to flip heads. No matter how many times you flipped heads previously. 11/20 = 55%

PS you should all watch/read Rosencrantz and Guildenstern are Dead.

PSS sorry, but the spoiler button is not working for me. :huh:

Edited by theharangue

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Ok, the odds of flipping heads again on the 9th try is equal to:

the probability that you picked a regular coin and flip 9 heads in a row,

PLUS

the probability that you picked the two headed coin and flip 9 heads in a row,

this is because these are the only two ways to achieve nine straight heads according to the parameters of the ridle.

probability of picking normal coin = .9

probability of flipping 9 heads in a row with normal coin = .00195

probability of picking normal coin and then flipping nine straight heads = .9 *.00195 = .00176

PLUS

probability of picking two headed coin = .1

probability of flipping nine heads on two headed coin = 1

probability of picking two headed coin and then flipping 9 straight heads = .1*1 = .1

so, overall probability that the next flip is heads = .100176

as for the bonus, each flip is independent and odds are 50/50

small calculation mistake....

.1+.00176=.10176 :):)

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im sorry if i didnt explain this well enough before but ill try again because i hate when people belittle others when they don't understand or know where they are. If the test was pick a coin randomly and flip it once then the chance of heads is 11/20 correct. if all the coins are equal and were told 8 heads whats next 50-50. but this isnt the case. There are 10 coins which u pick ranomly from think of it this way you have two guns. one always fires when the trigger is pulled, when one fires on average one out of 2 trigger pulls but randomly. you pick one of them not knowing which it is and pull the trigger. it fires. you think this is probably the gun that fires (66% chance your right) but u think let me try again. pull the trigger it fires. well im almost positive this is the gun that always fires but u know what i need to make sure. pull the trigger one more time. it fires. then someone walks up and bets you the gun wont fire another time in a row.

would you take the bet?

yes

now granted i wouldnt bet my life on it you got something like a 10% chance that its the wrong gun but still 90% win and if thats a gamblers fallacy then i think the casinos should find a new day job.

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by the way almost positive answers are

part 1 .9830188679

part 2 .5

if you want explanations chuck posted one, i posted one, and maybe someone else in the million posts (half mine sorry)

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(Hope this hasn't been done before)

1. There are 10 coins on a table. 9 are normal but 1 of them has 2 heads on it (i.e. will always land on heads). Without looking you pick one at random and flip it 8 times. It lands on heads each time. What is the probability of it landing on heads the next time?

Bonus question: (UNRELATED TO THE FIRST ONE) You flip a coin 20 times and it lands on tails each time. What is more likely for it to land on next, heads or tails?

The first one could be very tricky mathematically, if you try and calculate the probability that the coin in your hand has two heads given that it hs landed on heads 8 times. On the other hand, if you look at the overall picture, you can argue for a simpler answer.

The second question is a very old one and is related to the gamblr's fallacy. See spoiler.

The gambler thinks, "Because it has landed on tails so many times, if it is a fair coin, it must eventually land on heads, so it's more likely to land on heads this time." However, the rules of probability for random events tell us that they are unrelated to each other. Thus, assuming it is a fair coin, the probability of a "head" is, and will remain, a half.

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OK. I’ve had chance to think about this now. I realized that actually the bonus question follows on quite nicely form the main question. In the first question you are given the initial odds of picking a fake coin and have to determine the chance of another head based on the results so far.

In the bonus question this time you don't know the initial odds. But in reality it would be greater than 0. For the sake of argument we could say that there was only 1 coin in the world with 2 sides the same so the chance that you just picked it up was 0.00000000000000000000000001 or whatever.

Now someone (I forget who sorry) posted this simple calculation.

P(8 heads)=.5^8*9/10 +1^8*1/10

P(9 heads)=.5^9*9/10+1^9*1/10

P(9heads given 8heads)=P(9heads)/P(8 heads)= .1017578125/.103515625= .9830188679

However in this case we threw it 20 times and lets call the probability that were picked up a coin with tails on both sides, x, since we have no idea what it is. So the chance of picking a normal coin is (1-x).

P(20 tails)=.5^20*(1-x) + 1^20*x

P(21 tails)=.5^21*(1-x) + 1^21*x

P(21 tails given 20 tails)=P(21 tails)/P(20 tails)

= 0.5^21*(1-x) + x / 0.5^20*(1-x) + x

Now I'm not sure how to find the range of vales for this where 0<x<1. But my attempt is:

Multiply the denominator by 0.5 (effectively doubling the probability)

0.5^21*(1-x) + x / 0.5^21*(1-x) + 0.5x

Now by inspection you can see this is greater than 1 so probability must have been greater than a half (i.e. tails is more likely next throw)

But lets put x = 0.000001 ( I tried smaller values but my calculator didn’t like it :) )

P(21 tails given 20 tails) = 0.7559281452

I hope this makes sense to everyone. It is really just a more general case of the first problem.

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Regarding the bonus: Already too many explanations and re-explanations, but I'll toss one more point of view that hasn't been said yet. (I read through all of them.)

Imagine I said to you, "I will now toss a coin 21 times, and the first 20 will be heads, but the 21st will be tails." The odds that I would actually be right and get those 21 flips perfect would be 1 in 2,097,152.

Imagine I said to you, "I will now toss a coin 21 times, and the first 20 will be heads, AND the 21st will ALSO be heads." The odds that I would actually be right and get those 21 flips perfect would be 1 in 2,097,152.

Both series of 21 flips have an equal chance of happening.

(I know this doesn't address all the stuff about the chances the coin is biased because we saw that it flipped 20 heads already, but I just thought I'd toss out this other perspective. Also, I know it's been said in other ways...just not quite like this.)

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Another look at the bonus question. Prior to flipping the coin, the chance of scoring 20 straight heads is the same as scoring any unique series of heads and tails, roughly 1 in a million. Long odds you say? If you did this experiment once a day, odds are you would see your chosen unique series--in this case 20 straight heads--once every approximately 2740 years. BUT IT COULD HAPPEN NEXT TUESDAY. Now, for the gamblers in the group--give me a dollar and specify a unique series of flips. If your series comes up I'll pay you $500,000. Real sucker's bet, right? Millions of folks per day take relatively equal variations of that bet. That's why many of us in Florida call our lottery a tax on the challenged.

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Asking, "what is the chance of the next toss" vs. "what is the chance of 20 tosses of heads and 1 toss of tails" are two very different problems. You should try to check your wording next time. The "next toss" implies, it's independant of the other tosses.

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To all who are interested, this is a good read on using information to change probabilities. The moral of the read is that it is not a truely proper gauge of the actual occurence, but gives a good indicator on how you can use information to predict further chance. It relies on subjective evidence to provide an answer, rather than objective evidence. I believe that this is the exact same argument we are all trying to roll over in our heads.

Either way, good mind puzzle.

http://en.wikipedia.org/wiki/Bayesian_inference

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ok without using any math or logic im going to go with 60% heads 50/50 for a normal add 1/10 or 10% for the double headed coin and that comes to 60%... ok so i used simple minded logic.

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