Well /sci/?

>>7834757
5 flips so possible combinations are 2^5

Only 1 out come t t t h h

So probability is 1/32

>>7834764
>5 flips
It could be HTHTHTHTHTHTHTHH

>>7834767
The preceding flips are irrelevant.

>>7834771
in general, yes. not in the way you wrote it.

>>7834771
Ok listen dumbo.

Obviously the probability that the first instance of two consecutive heads precedes the first instance of three consecutive tails should be greater than [math]\frac{1}{2} [/math].

Furthermore, the sum of the probabilities that the first instance of two consecutive heads precedes the first instance of three consecutive tails and that the first instance of three consecutive tails precedes the first instance of two consecutive heads should be 1.

>>7834773
to add to this, consider 6 flips.

there's 4 outcomes
t t t x h h
t t t h h x

so the probability is 4/64 = 1/16. see?

I now realize that you might think it's asking for immediate precedence, as in t t t h h somewhere in the sequence. I don't think this is the case

>>7834773
The problem is that this question has no meaning unless you want to know the chance of it happening on a certain interval.

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Just gotta straighten this out.

>>7834779
Ah, yes, that may be what he was thinking.

t h t h h t h... satisfies the conditions of the problem.

>>7834784
It certainly still has meaning if you're asking about it happening in a random infinite sequence of tails and heads. Why wouldn't it?

>>7834777
>>7834779
Those are both incorrect.

Probability doesn't add to 1
Ttttxhh and ttthhx only look at the flips that matter

>>7834789
it doesn't, he's right. we might want to ask for the limit but it doesn't make sense to ask about it in infinite sequence

Assuming it's not
>immediately precedes
it should be 2/3, no?

>>7834795
Listen, why the hell not?

>>7834797
That's what I got.

>>7834793
The probabilities certainly add up to 1.

E.g. the probability that in an infinite random sequence of heads and tails, there is no instance of two consecutive heads, is zero.

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>>7834786
HT is just T
HTH is just H
HTT is just TT
THT is just T
TTH is just H
TH is just H

So: ugly picture, but it's really simple

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A = C/2
B = (A+D)/2
C = (B+1)/2
D = (B+1)/2

A = 2/5
B = 3/5
C = 4/5
D = 4/5

>>7834806
Ah, you're right. Much simpler.

>>7834805
It's asking for the first instances. In an infinite string there's a 1/4 chance that it never happens

>>7834757
The 1st sentence means the coin never stops flipping, thus there is never a result of heads or tails.

>>7834807
please simplify your states, see >>7834806

TTT = 0
TT = H/2
T = H/2 + TT/2
HH = 1
H = 1/2 + T/2

Answer = T/2 + H/2

>>7834816
T = 3H/4
H = 1/2 + 3H/8
=> 8H = 4 + 3H
=> H = 4/5
=> T = 3/5

Answer: 7/10

Let E stand for be the event of two consecutive heads happening before three consecutive tails.
Let s0 be the probability of E after having flipped no coins.
Let h1 be the probability of E after a sequence of 1 consecutive heads, without E having happened before.
Let h2 be the probability of E after a sequence of 2 consecutive heads, without E having happened before.
Let t1 be the probability of E after a sequence of 1 consecutive tails, without E having happened before.
Let t2 be the probability of E after a sequence of 2 consecutive tails, without E having happened before.
Let t3 be the probability of E after a sequence of 3 consecutive tails, without E having happened before.

Then we have:
s0 = 1/2 h1 + 1/2 t1
h1 = 1/2 h2 + 1/2 t1
h2 = 1
t1 = 1/2 h1 + 1/2 t2
t2 = 1/2 h1 + 1/2 t3
t3 = 0

Solving this system yields:

s0 = 7/10
h1 = 4/5
h2 = 1
t1 = 3/5
t2 = 2/5
t3 = 0

>>7834816
>>7834817
Looks good to me.

>>7834814
In an infinite string the probability that some instance of two consecutive heads occurs is obviously 1, as is the probability that some instance of three consecutive tails occur.

If you don't really know probability, note that probability 1 [math] \neq [/math] "always happens".

>>7834757
9:45ths?

>>7834851
You're right.
Good thing it has nothing to do with the discussion

>>7834818
Wrong

If s0 is probability of E after flipping 0 coins then s0 obviously = 0

>>7834757
50/50 happens or it doesn't.

>>7834912
>>7827381
Literally you.

>>7834757
The answer is 1 - sum from k=3 to inf of a(k)/2^k

where a(n) = a(n-2)+a(n-3), a(0) = a(2) = 0, a(1) = 1

This is approximately 70.1%

>>7834757
Like all events its 50/50. Either it happens or it doesn't.

Elementary t.b.h.

>>7835158
Wait no, it's more like 70.0%

100%

If you're Hillary Clinton in Iowa.

>>7835158
>>7835177
Can some CSfag confirm this with a shitty algorithm?

>>7835300
*I meant probability that 2 will be heads.

>>7835300
>That's the probably that 3 of the tosses will be tails. In OP's question, order matters.
Huh? It is the probability of 3 consecutive tails NOT occurring before two consecutive heads. It takes into account order.

>>7834764 is gibberish. There are obviously more ways for consecutive heads to occur without 3 consecutive tails than in 5 flips.

>>7835300
Not only does the calculation leading to 1/32 make no sense, but if you think about it this is saying that 3 consecutive tails will occur much much faster than only two consecutive heads. That's silly.

>>7835326
>>7835312
You're right. I was wrong. I'm going to delete my other post. This is what I get for trying to do word math problems at 1AM.

>>7834757
if ur hillary clinton 100% and then 0%

>>7835237
No, but some mathfag can confirm that the answer is precisely 7/10 by elegant means.

>>7834817

>>7835993
I don't understand what T and H refer to here.

>>7836041

heads and tails

>the city of you

>>7836080
Heads and tail are not values. Obviously they refer to probabilities, but in what context?

There are 5 state probabilities after each flip. State W: got TT before
HHH. State L: got HHH before TT. States HT,TH and HH: didn't get TT or
HHH, last two flips were HT,TH or HH respectively. Note W+L+HT+TH+HH=1.
With respect to the variables in the order listed above, the transition
matrix is A =
1 0 0 1/2 0
0 1 0 0 1/2
0 0 0 1/2 0
0 0 1/2 0 1/2
0 0 1/2 0 0

In the limit as n goes to infinity, A^n becomes
A^infinity=
1.0 0.0 0.8 0.6 0.4
0.0 1.0 0.2 0.4 0.6
0.0 0.0 0.0 0.0 0.0
0.0 0.0 0.0 0.0 0.0
0.0 0.0 0.0 0.0 0.0

for n=3, the state probabilites are
p_3 = [ 3/8; 1/8; 2/8; 1/8; 1/8];

Multiplying A^infinity * p_3 gives
p_infinity = [0.7 0.3 0 0 0]

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>>7834757
Too lazy to read the thread to see if it has been solved yet, but here

>>7836435
>[math] \frac{1}{2} [/math]
Yeah and I suppose the probability that the first instance of two consecutive heads occurs before the first instance is a billion consecutive tails is also [math] \frac{1}{2} [/math]

\facetiousness

reasoning: if two heads occurs first, three tails become certain over infinite tosses

3x tails over 3 flips = 1/8, in 3 flips probability of 2 consecutive heads is 2/8 (hht, thh). Relevant results of total 8 possibilities = 3/8, 5/8 would just lead to more flips. So total pool of possible results we consider = 3. therefore, it is 2/3 probable that we get 2 consecutive heads before 3 consecutive tails

>>7837628
actually I forgot hhh also counts, so probability is 3/4.

>>7836560
Did you not read the bit under "for finite n"?

>>7834757
I've always hated this part of math, mostly because I'm so shit at it.

>>7834757
happened in 700025 of 1 million trials, so the 70% people have it right

>>7834757
directly after? only hhttt? or hhthttt is acceptable?

>>7834757
Two negative binomial random variables with parameters (2,1/2) and (3,1/2) respectively. Find the cdf of their difference using their generating functions, and evaluate it at zero. Done.

>>7837793
Nvm, I neglected that the events are consecutive.

A coin is flipped repeatedly.
What is the probability that the first two consecutive heads precede the third consecutive head?

>>7837714
Yes, the answer is exactly [math] \frac{7}{10} [/math]. There are elegant ways of showing this without using a simulation.

>>7834757
Jeez, guise, it's just P(two heads) - P(three tails) KQED

>>7836435
>implying thth...hh is the only way you can get consecutive heads before *three* consecutive tails

Try again.

>>7837628
>>7837630
This is wrong because you are assuming you can split up the flips into independent groups of 3. In reality there is continuous memory through this system.

>>7838590
Nope.

>>7838598
Shit, misread the problem.

>>7834915
you got memed son

Ok guys, are you ready to have the problem solved as clearly and unambiguously as possible? The answer is [math] \frac{7}{10} [/math], and here's why.

Let [math] P_{TH}, P_{HT}, P_{TT} [/math] be the probabilities that the first instance of two consecutive heads precedes the first instance of three consecutive tails GIVEN the previous two flips were TH, HT, or TT respectively. Let [math]P[/math] be the solution to the given problem.

Then we have the system of equations:

[math] P_{TH} = \frac{1}{2} \cdot 1 + \frac{1}{2} \cdot P_{HT} [/math]

(either the next flip is H or T, with [math]\frac{1}{2}[/math] chance each, and in these two cases either there are two consecutive heads or, upon the next flip, the preceding two flips will have been HT)

Similarly, also

[math] P_{TT} = \frac{1}{2} \cdot P_{TH} + \frac{1}{2} \cdot 0 [/math]

[math] P_{HT} = \frac{1}{2} \cdot P_{TH} + \frac{1}{2} \cdot P_{TT} [/math]

Finally, the first two flips are either HH, HT, TH, or TT with probability [math]\frac{1}{4} [/math] each, so:

[math] P = \frac{1}{4} \cdot 1 + \frac{1}{4} \cdot P_{TH} + \frac{1}{4} \cdot P_{HT}+ \frac{1}{4} \cdot P_{TT} [/math]

Solving this system of four unknowns in four variables, we find:

[math] P = \frac{7}{10} [/math]

>thanks sci for doing my homework, hehe

Time to draw out a tree diagram I guess.

>>7838953
As OP, I knew the answer, and posted it in >>7838834. It is not the kind of problem that would appear as homework; it requires too much creativity. I just thought it was a fun problem, so shared it with /sci/.

>>7838834
Surely you mean 3/10 right? I haven't read the proof but 7/10? You have to get 3 tails in a row, the addum that you need it directly after the first 2 heads probably changes the chances, but barely. It's got to be near at least close to 1/8, at the problems core you need 3 consistant coin flips, 7/10 is so high it doesn't even make sense, you can yell at me and try to make me read your post all you want

>>7839033
Not OP, but the problem is asking for the chance that two consecutive heads will be reached before three consecutive tails (not directly before). So you can flip coins until you reach two consecutive heads or until you reach three consecutive tails. The former will occur 0.7 of the times you perform the experiment, which is not surprising since two consecutive heads is "easier" to get than three consecutive tails. Hopefully this helps you understand the question.

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>>7839077
Well this is awkward, i thought it meant that those 3 tails had to immediately preceed those first 2 heads.
Incidentally, the answer to that is probably 7/80

>>7839093
>is probably 7/80
What is the chance if that being true?

>>7839093
I don't understand how that interpretation makes sense though. Either it means, "what is the chance of getting HHTTT in unlimited flips?" which is 1. Or it means "What is the chance that the first consecutive heads is immediately followed by three consecutive tails?" Which is simply 1/8.

>>7839126
It's only a small difference, hence why the answer is so close to 1/8, but technically you could get 3 tails before 2 heads, so that would be a fail too

>>7839126
It's that the first [math] instance [/math] of two consecutive heads precedes the first [math] instance [/math] of three consecutive tails. So e.g. hthtthhthtthttt... suffices.

>>7838834
Do you mean PTH is the prob of getting HH before TTT given the last two flips were TH AND we haven't gotten HH or TTT yet?


Having trouble following this I'm afraid. In PTH = 1/2 * 1 + 1/2 * PHT
Can you expain what the "1" means?

This is me: >>7836343
It works, but is clunkier than yours as you have to find a limit.

>>7834757
english isnt my first language:
does procede mean that it has to be HHTTT or can it also be HH something TTT
do they have to be right next each other?

>>7839238
Your first sentence is the correct interpretation, yes.

In [math] P_{TH} = \frac{1}{2} \cdot 1 + \frac{1}{2} \cdot P_{HT} [/math], we are taking the expected value of [math]P_{TH}[/math] in terms of the two possibilities for the next flip, heads or tails, each of which has probability [math]\frac{1}{2}[/math].

There is a [math]\frac{1}{2}[/math] chance of the next flip being H, in which case since the previous two flips were TH, we have THH, i.e. two consecutive heads. So in that case, the probability of the first instance of two consecutive heads preceding the first instance of three consecutive tails is 1 — we just got two consecutive heads! On the other hand, if the next flip is T ([math]\frac{1}{2}[/math] chance), we now have THT, so the probability of [the problem] is [math]P_{HT}[/math].

Each of the equations expresses the expected value of the variable in terms of the other variables.

>>7836086
The probability of reaching two heads first given that the last few flips were the ones shown.

>>7839326
Ok, I can see it now. Very nice. I don't think I would have thought of "looking ahead" like that, if you know what I mean. It's more natural for me to look backward and express the probabilities in terms of probabilities of previous flips.

Thanks for posting it.

>>7834757
It could land in a infinite way.
There is just probability