If you flip a coin 100 times how many times will it land on heads?

[nmanlamb]

Homework Statement

If you flip 1 quarter 100 times what are the odds all 100 flips would land heads?
My friend argues that it would be a 50% chance since each individual flip has a 50% chance of being heads.

Homework Equations

The Attempt at a Solution

I tried saying that its 50% chance per flip but it decreases by (0.50) ^ x each flip. x being the number of flips. Is there anyone out there who can help me explain this to him? His platform is that flipping 1 coin 100 times is different from flipping 100 coins 1 time. The probability of each should be the same though right?

 

[LCKurtz]

You are correct. It is $\frac 1 {2^{100}}$, very very small. Have your friend try it for just 5 in a row and he will soon see it isn't 1/2 even for that.

 

[Ray Vickson]

Your are right: the probability of getting 100 heads in 100 tosses is about 0.79x10^(-30), or 0.000 ... 00079 (with 30 0s after the decimal point). If your friend still wants to argue the point, try to explain it with the easier case of tossing 2 times instead of 100. Now there are four possible outcomes, and it is easy enough to list them all. Only one of those four has two 'heads'. If your friend STILL wants to argue the point, ask him what he thinks are the probabilities of each of the four outcomes.

RGV

 

[genericusrnme]

you are correct

You could try getting him to write our every possible configuration you could end up with after 100 coin tosses, then asking him (or his ancestors via a message passed down since both of you would be dead by the time it's dnoe) how many out of the 2^100 configurations has all tails

 

[teknicalissue]

I don't mean to revive a super old thread but this is exactly what I was looking for.

I'm doing a personal project and I'm curious to know if there is an equation for this? something along the lines of, The more times you flip the less likely it will be (in percentage) to have the coin land in all heads/tails consecutively.

I'm not a math nor physics major so if this is something relatively easy.. Feel free to attack me with some constructive mathematical verbiage.

 

[LCKurtz]

As pointed out earlier, the probability of getting all heads in $n$ tosses is $\left (\frac 1 2\right)^n$. So, for example, to get all heads in 3 tosses the probability would be 1/8. For four tosses it would be 1/16. It multiplies by 1/2 for each additional toss.

 

[teknicalissue]

I'm assuming that for an evenly 3 sided dice, the formula would be (1/3)^n ?

 

[HallsofIvy]

Yes, there are 3 equally likely outcomes, then the probability of exactly one of those outcomes happening n times in a row is (1/3)^n.

 

[teknicalissue]

wow, I swear, If I didn't major in CS, Physics would have most likely been my next choice :P

How about this scenario (and last one I promise)

Lets say we have an 8 sided dice. Let's say one of the sides of the dice is not a number but a color.

As I land on the numbers continuously what is the probability of the dice landing on the color?

IE.

Let's assume I'll be rolling the dice 20 times. I've rolled all numbers up to my 15th roll. What are the chances my next roll would land on a color in contrast to a number?

 

[LCKurtz]

I'm assuming that for an evenly 3 sided dice, the formula would be (1/3)^n ?

A previously undiscovered 3 sided platonic solid.

 

[teknicalissue]

A previously undiscovered 3 sided platonic solid.

hahaha XD

 

[teknicalissue]

wow, I swear, If I didn't major in CS, Physics would have most likely been my next choice :P

How about this scenario (and last one I promise)

Lets say we have an 8 sided dice. Let's say one of the sides of the dice is not a number but a color.

As I land on the numbers continuously what is the probability of the dice landing on the color?

IE.

Let's assume I'll be rolling the dice 20 times. I've rolled all numbers up to my 15th roll. What are the chances my next roll would land on a color in contrast to a number?

(sorry for double post)

Now that I think about it, Let's make it easier with my undiscovered 3 sided Dice.

I'm rolling for 10 times, and up to my 4th roll I've landed 1's and 2's. What are the chances I would land on 3 the next roll?

I think what I'm getting at is that the more times I roll the dice and NOT hit a number, the chances of it hitting that number (in this case 3) increases.

 

[Ray Vickson]

(sorry for double post)

Now that I think about it, Let's make it easier with my undiscovered 3 sided Dice.

I'm rolling for 10 times, and up to my 4th roll I've landed 1's and 2's. What are the chances I would land on 3 the next roll?

I think what I'm getting at is that the more times I roll the dice and NOT hit a number, the chances of it hitting that number (in this case 3) increases.

Absolutely false. If you got 1s and 2s on the first 9 tosses, your probability of getting a '3' on toss #10 is still 1/3. (All this assumes a 'fair' die and an unskilled 'tosser' who cannot deliberately influence the outcomes.) You don't believer it? Well, what does the die do? Does it think to itself "Hmmm... I have come up 1s and 2s for the last 9 tosses, so I had better up my chances of coming up '3' on the next toss." I don't think a die thinks that way.

BTW: the word you want is "die", not "dice" = more than one "die".

Probably you are thinking of something like the "Law of Averages", so that if you have had an unusually small number of 3s, the chances of getting a '3' needs to increase in order to maintain an average. There are several versions of Laws of Averages, and not one of them works like I just indicated.

 

[LCKurtz]

To add to Ray's comments. Let's just take a regular 6-sided die to keep it simple. If you roll it 100 times and it always comes up with an even number, since you know how unlikely that is, it might lead you to suspect that the die is not a fair die. This is a statistical concept -- the idea that what you have just witnessed is so very unlikely that it calls into question your assumption that the die is fair. This is the statistical idea of hypothesis testing. If you test the hypothesis that the die is fair (P(even) = 1/2) such an outcome would lead you to reject that hypothesis. So, in the "real world", if you witnessed such an outcome, your natural conclusion would to be to suspect the die is loaded. And hence, you might also suspect that the next outcome will be even also.

But that doesn't counter anything explained above. If the die is fair, the probability that the next roll will be even is still only 1/2. And if you experienced 100 even tosses in a row, you have witnessed an extremely unlikely event. In theory it might happen. But if you try it, it won't.

 

[teknicalissue]

Hmm.. interesting. Then I guess My question to this would be, Then why does (1/2)^n work? Wouldn't it follow the same principle?

you flip a coin heads 10 times, the more the coin lands on heads there's a less likely hood it will land on heads again?
(1/2)^2 vs (1/2)^8

As you keep flipping the percentage drops right?

 

[HallsofIvy]

wow, I swear, If I didn't major in CS, Physics would have most likely been my next choice :P

How about this scenario (and last one I promise)

Lets say we have an 8 sided dice. Let's say one of the sides of the dice is not a number but a color.

As I land on the numbers continuously what is the probability of the dice landing on the color?

IE.

Let's assume I'll be rolling the dice 20 times. I've rolled all numbers up to my 15th roll. What are the chances my next roll would land on a color in contrast to a number?

If, as is normally assumed, all sides are equally likely, all rolls are independent. There are 8 sides, 1 of which is a color. The probability of rolling a color on any roll is 1/8. the fact that "I've rolled all numbers up to my 15th roll" is irrelevant.

 

[Ray Vickson]

Hmm.. interesting. Then I guess My question to this would be, Then why does (1/2)^n work? Wouldn't it follow the same principle?

you flip a coin heads 10 times, the more the coin lands on heads there's a less likely hood it will land on heads again?
(1/2)^2 vs (1/2)^8

As you keep flipping the percentage drops right?

Wrong, wrong, wrong! I don't know why you think a coin has "memory" and can adjust its own 'heads' probability as time goes on. No: if it is a fair coin, the probability that (before doing any tosses) your first 9 tosses will be all heads is (1/2)^9. However, if your first 9 tosses did happen to be all heads, that would not affect the probability of getting heads again; that probability remains 1/2. As LCKurtz has indicated, for a coin that you suspect might not be fair (so for which you do not actually *know* the head/tail probabilities), seeing 9 heads in a row may cause you to change your mind somewhat about P{head}. The coin itself is completely unaffected; what IS affected would be your "state of knowledge" about the coin.

If you just reach into your pocket and pull out a standard coin, its head/tail probabilities are going to be essentially 1/2. A "rigged" coin might have head/tail probabilities different from 1/2, but even then the probabilities would not be affected by the outcomes of some initial tosses. For example, even if I have a weighted coin with P{Head} = 0.6 and P{Tail} = 0.4, the probability of my first 9 tosses being all heads is (0.6)^9; the probability that the 10th toss is heads is still 0.6, no matter what happened in tosses 1--9.

So, given all this, why do we say that our state of knowledge could ever be toss-dependent? Well, if you are a 'Bayesian' (which I, personally, am...more-or-less), you would start by saying that the coin has some fixed, but unknown probability P{head} = p, and where p itself is subject to a probability distribution f(p) (such as uniformly distributed between 0 and 1). For a uniform prior, our a priori probability of heads on one toss is 1/2. However, after getting our first n tosses as all heads, out new expected probability of getting a 'head' is now 1/(n+1). So, I my first 9 tosses were all heads, my new head probability estimate would be 1/10.

I was somewhat reluctant to mention any of the Bayesian stuff because I am not convinced you fully grasp yet how probabilities actually work. I can only hope that I have not confused you.

 

[Ray Vickson]

Wrong, wrong, wrong! I don't know why you think a coin has "memory" and can adjust its own 'heads' probability as time goes on. No: if it is a fair coin, the probability that (before doing any tosses) your first 9 tosses will be all heads is (1/2)^9. However, if your first 9 tosses did happen to be all heads, that would not affect the probability of getting heads again; that probability remains 1/2. As LCKurtz has indicated, for a coin that you suspect might not be fair (so for which you do not actually *know* the head/tail probabilities), seeing 9 heads in a row may cause you to change your mind somewhat about P{head}. The coin itself is completely unaffected; what IS affected would be your "state of knowledge" about the coin.

If you just reach into your pocket and pull out a standard coin, its head/tail probabilities are going to be essentially 1/2. A "rigged" coin might have head/tail probabilities different from 1/2, but even then the probabilities would not be affected by the outcomes of some initial tosses. For example, even if I have a weighted coin with P{Head} = 0.6 and P{Tail} = 0.4, the probability of my first 9 tosses being all heads is (0.6)^9; the probability that the 10th toss is heads is still 0.6, no matter what happened in tosses 1--9.

So, given all this, why do we say that our state of knowledge could ever be toss-dependent? Well, if you are a 'Bayesian' (which I, personally, am...more-or-less), you would start by saying that the coin has some fixed, but unknown probability P{head} = p, and where p itself is subject to a probability distribution f(p) (such as uniformly distributed between 0 and 1). For a uniform prior, our a priori probability of heads on one toss is 1/2. However, after getting our first n tosses as all heads, out new expected probability of getting a 'head' is now 1/(n+1). So, I my first 9 tosses were all heads, my new head probability estimate would be 1/10.

I was somewhat reluctant to mention any of the Bayesian stuff because I am not convinced you fully grasp yet how probabilities actually work. I can only hope that I have not confused you.

Sorry: there is an error above, and the system will not let me edit the message now. I should have said "However, after getting our first n tosses as all heads, out new expected probability of getting a 'head' is now (n+1)/(n+2). So, I my first 9 tosses were all heads, my new head probability estimate would be 10/11."

 

[1MileCrash]

Sounds like the (new) poster is committing a textbook Gambler's Fallacy.

http://en.m.wikipedia.org/wiki/Gambler's_fallacy

(1/2)^n doesn't imply that the chance of rolling another heads gets smaller. It only implies that the entire sequence previously rolled (HHHHHH.. n times) is part of a larger and larger sample space of possible sequence of results from the coin flip. Thus, the ratio of this result to all results gets smaller.

In other words, if you roll heads 200 times, the probability of the next flip being heads is 50%, but the probability of flipping heads 201 times in a row is completely different because it considers the probability of you gettingto that initial 200 in a row in the first place.