BrainDen.com - Brain Teasers
• 0 ## Question (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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• 0 The odds that it is a fair coin causing the 8 heads is: 9/10 * (1/2)^8 = 1 in 284.444...

The odds that it is a double-headed coin is: 1/10 * 1 = 1 in 10

In other words, it is 28.444... times more likely that it is a double-headed coin. The combined odds results in a 3.515% chance of a 50/50 shot and a 96.485% chance of a guaranteed heads. The final odds are: 98.242% chance of heads.

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• 0 For those of you who said both or 50/50: are you seriously telling me that if a coin lands on tails 20 times in a row with a probability of 1/1048576 that this coin is not fake or you are throwing it properly. Clearly you would have to be stupid to think that heads would be likely to come up again.

So the answer is tails is the most likely.

It was really a common sense problem more than a maths problem. Definitely innapropriate to suggest your readers are "Clearly...stupid". The nature of the first part of the teaser certainly suggests a more mathmatically tailored problem so it doesn't seem to be much of a stretch to me that we'd lean that way.

Having said that, I'll agree that I fell into the trap of assuming certain conditions about the 2nd problem without any evidence. I will agree that, in a different context, I would most definitely be suspicious of a coin that consistently came up heads (or tails) 10,15, 20 times in a row and would have to assume that the results of the next 'toss' would not be constrained by pure statistics.

BUT, that, by no means, should imply that I or any of the other readers are stupid.

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• 0 Definitely innapropriate to suggest your readers are "Clearly...stupid". The nature of the first part of the teaser certainly suggests a more mathmatically tailored problem so it doesn't seem to be much of a stretch to me that we'd lean that way.

Having said that, I'll agree that I fell into the trap of assuming certain conditions about the 2nd problem without any evidence. I will agree that, in a different context, I would most definitely be suspicious of a coin that consistently came up heads (or tails) 10,15, 20 times in a row and would have to assume that the results of the next 'toss' would not be constrained by pure statistics.

BUT, that, by no means, should imply that I or any of the other readers are stupid.

That was just a phrase meant in a light hearted way. Like when someone says "no one in there right mind..." but they aren't implying that everyone who disagrees with them are clinically insane. .

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• 0 The odds that it is a fair coin causing the 8 heads is: 9/10 * (1/2)^8 = 1 in 284.444...

The odds that it is a double-headed coin is: 1/10 * 1 = 1 in 10

In other words, it is 28.444... times more likely that it is a double-headed coin. The combined odds results in a 3.515% chance of a 50/50 shot and a 96.485% chance of a guaranteed heads. The final odds are: 98.242% chance of heads.

Having reconsidered my prior position, response answered the question "Given these 10 coins, one being double headed, what is the probability of picking one and tossing 9 heads in a row" that would be %10.17.

However, the question states, having picked one of the ten and having tossed 8 heads, what is the probability of another head.

I think pieeater has nailed this one, on the head (sorry, had to)

but the answer is definitely not 55/45

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• 0 9 coins, 1 trick coin = 11 heads, 9 tails.

There for probablility of getting a heads on the next flip is 11/20.

For the bonus, probability is always 1/2 no matter how many times you've flipped it

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• 0 (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?

thee answer i think is: 101/180 ....for the first part

and 1/2 for the second part...

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• 0 How is it that people are under the assumption that the number of iterations will change the outcome? If we have not changed the initial conditions of the problem, then the chance of flipping a coin heads on the 1st or the Nth iteration remain constant. (you do not get decreasing odds from multiple iterations if you do not change the data set) If this was true, then for each iteration, you would say that your odds are constantly changing (55/45, 45/55. 35/65, etc... after each flip, which is not how statistics work)

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• 0 This is the same reason that a single "normal coin" does not have decreasing odds. You would not say if you had 4 flips heads that your odds are 50/50, 25/75, 12.5/87.5, etc...

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• 0 How is it that people are under the assumption that the number of iterations will change the outcome? If we have not changed the initial conditions of the problem, then the chance of flipping a coin heads on the 1st or the Nth iteration remain constant. (you do not get decreasing odds from multiple iterations if you do not change the data set) If this was true, then for each iteration, you would say that your odds are constantly changing (55/45, 45/55. 35/65, etc... after each flip, which is not how statistics work)

You might want to search for Monty Hall Problem on Google .

The probability of initial conditions can change. Lets say you had a lottery ticket. You may think that the odds are fixed of you wining. But if you knew that your first 2 numbers were right then the chance that it is a winning ticket has now changed.

Edited by psychic_mind
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• 0 How is it that people are under the assumption that the number of iterations will change the outcome? If we have not changed the initial conditions of the problem, then the chance of flipping a coin heads on the 1st or the Nth iteration remain constant. (you do not get decreasing odds from multiple iterations if you do not change the data set) If this was true, then for each iteration, you would say that your odds are constantly changing (55/45, 45/55. 35/65, etc... after each flip, which is not how statistics work)

Because the exact coin being flipped does not change between flips. Your 55% chance is based on a new coin being chosen randomly each time. The fact is there is only one coin being flipped over and over. Once the original decision was made, that coin is the only one ever flipped again. That original pick was the only random action. Now we just needed to figure out the odds of the original coin pick being fair vs. being the double-headed coin. The eight flips of heads is an indicator of which of the ten coins was chosen originally. The likelihood that the original pick was the double-headed coin increases the more heads that get flipped in sequence. Does that answer your question?

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• 0 This is the same reason that a single "normal coin" does not have decreasing odds. You would not say if you had 4 flips heads that your odds are 50/50, 25/75, 12.5/87.5, etc...

This is because with a normal coin, each consecutive flip is done with a normal coin. after 20 heads, nothing has changed, and the 21st flip is still a normal coin.

however, once you throw a false coin into the mix, flipping 20 consecutive heads changes the condition of the 21st flip. it is now highly likely that you are holding the false coin, therefore diminishing the odds of a tails. you are right in that probability is calculated for each instance, but when new imformation is gained (ie. the higher probability of holding the fake) it factors in to the instance in question.

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• 0
How is it that people are under the assumption that the number of iterations will change the outcome? If we have not changed the initial conditions of the problem, then the chance of flipping a coin heads on the 1st or the Nth iteration remain constant. (you do not get decreasing odds from multiple iterations if you do not change the data set) If this was true, then for each iteration, you would say that your odds are constantly changing (55/45, 45/55. 35/65, etc... after each flip, which is not how statistics work)

I think there is a fundamental difference in the thinking of the two camps here. The first camp assumes a priori that p, chance of a head, is fixed at 1/2. Consequently, regardless of the sequence of heads, the chance of the next toss is still 1/2.

psychic_mind and others are taking the bayesian point of view, which is that p is unknown. They then estimate p given the data. In that approach, given the 20 heads, the probability of head is very near 0.

Given the original wording, I'd go with the second approach. Perhaps this is a rewording of the original problem that makes it clearer. Assume that psychic_mind approaches you and proposes a game. He'll take a coin out of his pocket and flip it. If you can correctly guess the coin's orientation, he gives you 1 dollar. If you're incorrect, you give him a dollar.

You figure head is equally likely as tail, so you guess tail 20 times. The coin come up head 20 times, and you're down 20 dollars. Would you guess tail on the 21 try?

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• 0 ...or for a more extreme example. What if I had flipped a coin 1000 times and they were all heads. Is the chance of a next head still 55/100 ?

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• 0 I don't think anyone's gotten it yet, so...

First we will figure out what the probability of having the 2-headed coin, GIVEN that we have flipped 8 heads in a row.

P(A|B) = P(A and B)/P(B)

Where P(A|B) is the probability of A, GIVEN that B has already occurred. In English, the formula above states that the probability of event A, GIVEN that event B has already occurred, is the probability of the two events occurring together divided by the probability of event B occurring alone.

P(We have the 2-headed coin | We flipped 8 heads in a row) = P(2-headed coin AND 8 heads in a row) / P(8 heads in a row)

The probability that we get the 2-headed coin AND flip 8 heads in a row = 1/10, since if we pick the 2-headed coin, we will always get heads.

The probability of 8 heads in a row = 1/10 + 9/10*1/256 (probability of getting the 2-headed coin plus the probability of getting a fair coin times the probability of getting 8 heads with a fair coin) = 256/2560 + 9/2560 = 265/2560 = 53/512

Thus, the probability that we have the 2-headed coin given that we flipped 8 heads in a row is (1/10)/(53/512) = 512/530 = 256/265

At this point, the probability that we have the 2-headed coin is 256/265. Thus, the probability that we have a fair coin is 9/265. The probability that our next flip is heads is now P(Heads with the 2-headed coin)*256/265 + P(Heads with fair coin)*9/265, which is

1*256/265 + 1/2*9/265 = 512/530 + 9/530 = 521/530

Part B: Given that it's a fair coin, neither one is more likely. That said, if it occurred to me that the coin might not be fair, I would guess tails.

Edited by Chuck
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• 0 How about this. If you flip just one time, and get heads, have you learned anything from the problem? You do not know for sure what type of coin you are holding still. The monty hall explanation only applies if you actually garner information from the results. Since a heads can be indicative of any of the ten coins on the table, you have not changed the problem. Maybe you should read further into the monty hall problem, as it has good explanations of why the odds do or do not change. Thanks.

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• 0 Congratulations Chuck. I think you are the first to get it right .

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• 0 <("<) (>")> <("<) (>")>

\m/ (^.^) \m/

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• 0
How about this. If you flip just one time, and get heads, have you learned anything from the problem? You do not know for sure what type of coin you are holding still. The monty hall explanation only applies if you actually garner information from the results. Since a heads can be indicative of any of the ten coins on the table, you have not changed the problem. Maybe you should read further into the monty hall problem, as it has good explanations of why the odds do or do not change. Thanks.

Lets restate the problem so we know where we're going. We have 10 coins, 9 of which are fair, and 1 is a 2-headed coin. You randomly pick a coin and toss it. It comes up head. You claim that you gained no information from the toss. Actually, you just gained some information, just that it isn't a lot of information.

After you pick up the coin, but before the toss, you have no information whatsoever, so the probability that your coin is biased is 1/10. After one toss and the coin comes up head, the chance of it being a biased coin just increased a bit. Instead of giving calculations, let me present two examples.

To argue that you gain information from 1 toss, let's look at the opposite side. Assume that you randomly pick a coin and it comes up tail upon the flip. The probability of the coin being 2-headed goes from (1/10) before the toss to 0. You just gained some information from 1 toss.

Suppose you have ten 1000-sided dice. 9 of which have sides numbered from 1-1000. The last die has all sides labelled with 314. You randomly pick a die and toss it once. The number 314 comes up. What die do you think you got?

Edited by bushindo
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• 0 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

and done

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• 0 Chuck beat me to the probability equation; however, people that said 50/50 are, in my opinion, correct. You wanted people to make an assumption without enough information even though you foreshadowed your answer in the 1st question. It is still going to be 50/50 probability. This is called the gambler fallacy and is one of the reasons casino's started seeing 20% increase in profits from the roulette tables when they started posting previous spins.

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• 0 Chuck beat me to the probability equation; however, people that said 50/50 are, in my opinion, correct. You wanted people to make an assumption without enough information even though you foreshadowed your answer in the 1st question. It is still going to be 50/50 probability. This is called the gambler fallacy and is one of the reasons casino's started seeing 20% increase in profits from the roulette tables when they started posting previous spins.

I'll try to explain the bonus question one final time.

There are 2 possibilities:

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.

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• 0 I completely understand what you are "trying" to get to; however, your sample size was very small. For someone to assume it is not a valid or fair coin is unrealistic. Again, you tried implying the direction you wanted us to go but 20 flips is not enough to be suspicious. I have seen 2 royal flushes in one night over the course of 4 texas hold em tournaments; I have seen 7 four of a kinds in one night (3 of which were deuces). I have seen back to back straight flushes. Why did I not questions these. My sample size was large enough. Your margin of error is too great. You would need a sample size of over 100 before you even start to reach 4 std dev.

-royal flush odds 1 in 650,000

-straight flush odds 1 in 72,000 (one time)

-4 of a kind odds 1 in 4,000

These are not exact and you may want to look them up. However, the point is: With the way you worded your question, it is more likely that (with a sample size of 20) that I just got lucky with a fair coin, versus picking up a random coin and having it have heads (tails) on both sides.

The only way I could even concede the point is that you foreshadow it could be an unfair coin.

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• 0 for the bonus the probability of it landing on tails 20 times is equal to 1/ and therefore rolling a head next should be [(2 ^20) - 1] /(2 ^20)

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• 0 I completely understand what you are "trying" to get to; however, your sample size was very small. For someone to assume it is not a valid or fair coin is unrealistic. Again, you tried implying the direction you wanted us to go but 20 flips is not enough to be suspicious. I have seen 2 royal flushes in one night over the course of 4 texas hold em tournaments; I have seen 7 four of a kinds in one night (3 of which were deuces). I have seen back to back straight flushes. Why did I not questions these. My sample size was large enough. Your margin of error is too great. You would need a sample size of over 100 before you even start to reach 4 std dev.

-royal flush odds 1 in 650,000

-straight flush odds 1 in 72,000 (one time)

-4 of a kind odds 1 in 4,000

These are not exact and you may want to look them up. However, the point is: With the way you worded your question, it is more likely that (with a sample size of 20) that I just got lucky with a fair coin, versus picking up a random coin and having it have heads (tails) on both sides.

The only way I could even concede the point is that you foreshadow it could be an unfair coin.

Well the chance you get lucky is 1 in 1048576 and even if there was only a tiny chance that the coin was biased then that would have to be taken into account. It may be 50.00000001% but its still more than half.

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• 0 i dont understand how your saying assuming its the biased coin. no one is assuming, there is a significant probability that it is the biased coin, and even if it is in the small percentage that it is a fair coin. That too is considered in the answer. I am almost positive chuck and psychic are right. Its like saying 2 people are throwing darts (ones really good, one is really bad). one steps up and hits the bullseye 10 times in a row. There is a significant probability of hitting the bullseye again because with the data we have its almost 100% chance that the really good ones throwing.

Anyway if anyone is still looking at this thread. It might help to think of this. you pick a random coin of the ten and flip 4 heads in a row whats the probabilty that you selected the biased coin.

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