Probability again :D

[DarkSider]

You are trying to toss a coin and get heads 10 times in a row. Probabilisticly speaking, in average, how many times you have to start over, and how many times you tossed the coin ?

[Darryl]

Well the probability of flipping a coin ten times and it landing on heads each time is 1/1024, so on average....I'm not sure. Either 1024, or somewhere in the middle, 512?

[Nonny]

I agree with Darryl that the chance of flipping 10 heads in a row is 1 in 1024.

Total number of flips though.....

As an educated guess I'd say up the probability of reaching each milestone on the way?

That sounds too low... Maybe that's the number of restarts? Restarts then I'm saying 2047 times....

Total number of flips....!!!!???

Flipping hell is all I can say right now!!! ;) Over 10,000 though I'll say as a rough guess.

[Darryl]

Total number of flips....!!!!???

Flipping hell is all I can say right now!!! ;) Over 10,000 though I'll say as a rough guess.

It depends on whether you always flip it ten times and then restart over, or whether you give up when you flip a tails.

If it's the first of the two, I'm going to say, on average, 5120.

[DarkSider]

If you would try to get 10 heads in a row .. would you keep going once you get a tail, or start again ?

[Nonny]

2 4 8 16 32 64 128 256 512 1024

1024 - 1 * 10 = 10
512 - 2 * 9 = 18
256 - 4 * 8 = 32
128 - 8 * 7 = 56
64 - 16 * 6 = 96
32 - 32 * 5 = 160
16 - 64 * 4 = 256
8 - 128 * 3 = 384
4 - 256 * 2 = 512
2 - 512 * 1 = 512

Total flips - 2036...

Which messes up my total restarts number... If I go with my new formula restarts will be ....
1023 ..... I must confess these numbers are lower than I'd expect but I'll run with them.

Lol, it just hit me, if probability says you'll flip heads ten times in a row on the 1024th attempt then .... you restart 1023 times!! I'm definately running with this one! :D

[BlackWolf]

There is no set number.
This question is impossible. Theres amount of throws that gives you propability of almost zero throws but you cant ever reach zero so theres no such number that would get you there.
As long as propability to throw each side of coin is 50% you could as well throw 10 heads on first attempt as it could take forever.

[Cheese]

There is no set number.
This question is impossible. Theres amount of throws that gives you propability of almost zero throws but you cant ever reach zero so theres no such number that would get you there.
As long as propability to throw each side of coin is 50% you could as well throw 10 heads on first attempt as it could take forever.

Don't be stupid it's not impossible lol...

[BlackWolf]

Yes it is impossible. Setting of that question makes it impossible.

[Nonny]

There is no set number.
This question is impossible. Theres amount of throws that gives you propability of almost zero throws but you cant ever reach zero so theres no such number that would get you there.
As long as propability to throw each side of coin is 50% you could as well throw 10 heads on first attempt as it could take forever.


I see what you mean, but when talking probability numbers you're essentially talking about the number of times you would have to do something to ensure it happens assuming no such thing as luck.

I.e. the chance of tossing a coin and getting heads is as you say 50%, or one in two. Without luck, if you tossed a coin twice you would get one head and one tail. In practise you could do it first time, or it could take you ten. Probabilistically though if you toss the coin twice you will get heads once. For the equation we have to assume it would be on the second attempt.

So the answer we're looking for is based on the above.

Assuming I haven't fallen completely out of touch with modern day probability mathematics? :)

[BlackWolf]

You have.
What is propability that you will throw 6 with dice?
1/6 what is propability that you will throw 2 times 6 in row?
1/36
But how many times you must throw dice to get 6 for 2 times = impossible to answer.
It can be anything between 2 and unlimited. You might never throw 2 sixes in row but propability for that to happen gets higher and higher after every throw but it will never reach 100% as such this is impossible question to answer.

I hope that clears it out. Propability that you will throw heads for 10 times in row is 1/2048 or something. But how many times you must throw is impossible to calculate.

[Nonny]

You are trying to toss a coin and get heads 10 times in a row. Probabilisticly speaking, in average, how many times you have to start over, and how many times you tossed the coin ?


This line should clear it up.

We're not talking realistically speaking because it's impossible due to luck being an unknown.

The question does state both probabilitically speaking and on average though so we can ditch both reality and luck for the purposes of this question.

[BlackWolf]

No you dont get it.
How many times you must toss a coin to get heads?
Tell me. Cause you cant answer that.

[Nonny]

BW I believe your last post was made whilst I was editing mine. The original draught may not have been clear enough. I'll assume it is now.

[BlackWolf]

No its not cause it still doesnt work that way.
You could ask what is propability for you to toss heads 10 times in row. Ok we can answer that 1/2048 or 1/1024 or something like that.
So time that with 10 and you get number of throws and -10 of right tosses and you get amount of failures.
With large enought of pool of trials and errors you would eventually get closer and closer to line X which says that propability to throw heads is 50% but you wouldnt ever reach that exact amount cause its impossibility.

Thats why this question if stated as it was stated is impossible to answer.

On average you might toss coing 10 times or unlimited amount of times. On average combined amount of tosses would be either 10 or unlimited. So what is answer? On average you have failed to toss heads once to unlimited amount of times.

Still we can answer what is propability that you will throw heads 10 times in a row. But it has nothing to do with amount of tosses as that is something that cant be answered.

[Nonny]

Oh ok, I see where you are coming from here.

With regard to number of throws, although it wasn't stated as a definate rule, I'm assuming that once you toss a tail you start again.
So, leaving luck aside. My numbers are based on tossing 10 in a row once, 9 in a row twice, 8 in a row 4 times etc. Only with this assumption is it possible to work out the probabilistic number of throws. At least in my book anyway. :)
Unless it's assumed you throw ten times every attempt then it's 10,240 tosses.
Anything in between restarting on a tail or flipping the 10 come what may, without clear rules, would indeed make it impossible to answer the question.

I'll stick with restarts being the logical 1023 for the reason given, but I think it's just the number of tosses which were causing the problem?

I will amend my total for flips though as it doesn't allow for the times a tail will spoil the sequence of heads. So my new, hopefully last answers are this:

Restarts: 1023

Flips: 2036 heads + 1023 tails = 3059

[DarkSider]

I don't know the "official" answer as the question i invented myself while looking at Derren Brown - The system :P
But if i was to calculate my way i'd say:
Probability to get 10 heads in a row is 1/1024. So in average, milions ppl trying to get 10 heads in a row would need to start over 1023 times. So that would be my answer for first question.
How many times they fliped the coin:
If they started over 1023 times means the first coin flip in each set was done 1024 times. Half of those are tails so they stoped throwing second flip in that set.
512 second flip. Same half tails and stoped.
256 - 3
128 - 4
64 - 5
32 - 6
16 - 7
8 - 8
4 - 9
2 - 10

So if you add all the flips you get 2046. I'm not sure where Nonny's extra tails are coming from :P

[Meneldil]

The main difficulty in this is is the problem that it's 'in a row' - if you merely tossed 10 coins simultaneously, and a 'success' was whether all 10 were heads, then it would follow a Poisson distribution.

However, being in a row, it becomes considerably more difficult.

I think the best way about doing it is to look at every coin as the start of a group of 10 heads (Obviously the groups overlap). The chance each group is 10 heads is 1/1024, hence the chance each coin is the start of a group of 10 heads is 1/1024. Therefore, pretty straightforwardly, the number of coin tosses needed is just over 1024: 1033 I believe.
I agree that this is somewhat counter-intuitive - to the degree that I'm not completely convinced myself. However, I've done a quick analysis on Excel (well, OpenOffice Calc), using random numbers, and it seems to fit the data.

I'm also completely and utterly certain that Blackwolf is wrong when he says that it's impossible. This is definitely a solveable bit of statistics.

Edit: I'm also going to try and convince you why this is wrong:
Oh ok, I see where you are coming from here.

With regard to number of throws, although it wasn't stated as a definate rule, I'm assuming that once you toss a tail you start again.
So, leaving luck aside. My numbers are based on tossing 10 in a row once, 9 in a row twice, 8 in a row 4 times etc. Only with this assumption is it possible to work out the probabilistic number of throws. At least in my book anyway. :)
Unless it's assumed you throw ten times every attempt then it's 10,240 tosses.
Anything in between restarting on a tail or flipping the 10 come what may, without clear rules, would indeed make it impossible to answer the question.

I'll stick with restarts being the logical 1023 for the reason given, but I think it's just the number of tosses which were causing the problem?

I will amend my total for flips though as it doesn't allow for the times a tail will spoil the sequence of heads. So my new, hopefully last answers are this:

Restarts: 1023

Flips: 2036 heads + 1023 tails = 3059

(Bold mine - these are where I think mistakes are being made).
The first mistake is not realising that 9s are going to appear inside the 10, and 8s inside the 9s, etc. So you're double-counting some. This whole aspect definitely needs some more (mathematically rigorous) work. I'm not at all convinced by this method.
The second part though, is the more glaring error. Surely, chances are that there will be an equal number of heads and tails? Not half the amount of tails, merely because we're looking for runs of heads. I think you've forgotten the chances of two (or more) tails being in a row.

[Martin]

I didn't think they had coins in Romania....

[DarkSider]

I'll woop yo' ass !

[CFalcon]

The main difficulty in this is is the problem that it's 'in a row' - if you merely tossed 10 coins simultaneously, and a 'success' was whether all 10 were heads, then it would follow a Poisson distribution.

That was my initial thought. But a Poisson is used to get the probability of a number of events occuring in a given time, or some other constraining factor to which the mean is related. But that factor cannot limit the number of events, as a Poisson can have infinite events, so the constraint cannot be the number of flips in this case.

It's been a while since I've done statistics, so I'm probably just getting muddled, but I can't see how a Poisson would work here.

Your reasoning for restarting as soon as you get a tails seems good though.

[Meneldil]

That was my initial thought. But a Poisson is used to get the probability of a number of events occuring in a given time, or some other constraining factor to which the mean is related. But that factor cannot limit the number of events, as a Poisson can have infinite events, so the constraint cannot be the number of flips in this case.

It's been a while since I've done statistics, so I'm probably just getting muddled, but I can't see how a Poisson would work here.

Your reasoning for restarting as soon as you get a tails seems good though.

The chances of getting all 10 to be heads would ~ Po(1/1024) - yes?
From that you do need to do a bit more working out on the curve, I agree, but it's relatively straightforward simultaneous equations, I think.

Edit: I meant to do this earlier, but I had a powercut and then forgot. I've been talking complete bollocks in this thread - please ignore my posts.

[Charlie_B]

You're over complicating things Mene. As Cap Falc says, Poisson distributions relate to events occurring in time and therefore it isn't relevant. If the question was simply the probability of tossing 10 coins and getting all heads then it would follow a simple Binomial distribution.

Good question DS, you've got me stumped. For now ;)

Edit: Ok, I agree with DS - 2046. I approached it in a slightly different way but the numbers boil down the same.

[Nonny]

Edit: Ok, I agree with DS - 2046. I approached it in a slightly different way but the numbers boil down the same.


Ditto, I started at the wrong end and got myself into a muddle doing it in bits and pieces. :|

[Nightmare]

You have.
What is propability that you will throw 6 with dice?
1/6 what is propability that you will throw 2 times 6 in row?
1/36
But how many times you must throw dice to get 6 for 2 times = impossible to answer.
It can be anything between 2 and unlimited. You might never throw 2 sixes in row but propability for that to happen gets higher and higher after every throw but it will never reach 100% as such this is impossible question to answer.

I hope that clears it out. Propability that you will throw heads for 10 times in row is 1/2048 or something. But how many times you must throw is impossible to calculate.

That all depends on how the dice are thrown..

[MattM]

Right, DS is correct with 2046; and if you want mathematical proof:

Let X=number of tosses before 10 heads in a row.

Partition the sample space as follows: {T},{HT},{HHT},{HHHT},{HHHHT},{HHHHHT},{HHHHHHT},{HHHHHHHT},{HHHHHHHHT},{HHHHHHHHHT},{HHHHHHHHHH}.

where: T= event that toss is a tails ; H= event toss is head.

Then by the partition theorem for expectation:

E[X] = E[X|{T}]*P({T})+E[X|{HT}]*P({HT})+.....+E[X|{HHHHHHHHHH}]*P({HHHHHHHHHH})

Now, each event in the partition has probability (1/2)^k where k is the length of the chain. Thus, with m=E[X], we have:

m=(1+m)/2+(2+m)/4+(3+m)/8+(4+m)/16+(5+m)/32+(6+m)/64+(7+m)/128+(8+m)/256+(9+m)/512+(10+m)/1024+10/1024

Thus: m = 1023/512+1023*m/1024 => m=1024*1023/512 => m=2*1023 => m=2046.

:)

[Augustus]

The probability of throwing 10 heads in a row is a lot higher than ANY of you getting a girlfriend ;)

[mvdevnull]

took me 17 bushtarion ticks to finally flip 10 heads in a row..

can i answer in ticks cuz that's bout all i know???
:|