Quote:JyBrd0403Well if you're suppose to lose 2 in a row in 4 trials and also win 2 in a row in 4 trials you would leave no room for winning once and losing once. It would just be Win-Win-Lose-Lose every 4 trials, so basically all day long that's what would happen. It can't be right.

It's right. First, it's not just win-win-lose-lose. It's an average of all possible outcomes. Second, lose-win-lose-lose would be another repeatable 4 event sequence that also contains 1 and only 1 string of 2 consecutive losses.

What are the potential outcomes in those 4 trials?

WWWW

WWWL

WWLW

WWLL

WLWW

WLWL

WLLW

WLLL

LWWW

LWWL

LWLW

LWLL

LLWW

LLWL

LLLW

LLLL

Of those 16 outcomes, 8 have at least 2 losses. 8 is 50% of 16...

So you're saying you would lose 2 in a row every 8 trials not 4 trials, correct.

Quote:JyBrd0403Quote:RDW4POTUSOf those 16 outcomes, 8 have at least 2 losses. 8 is 50% of 16...

So you're saying you would lose 2 in a row every 8 trials not 4 trials, correct.

No. Of the 16 possible 4-roll sequences, 8 contain at least two consecutive losses. So, a random 4-roll sequence has a 50% chance of containing at least two consecutive losses.

Right, so you would need 2 4-roll sequences to get a loss of 2 in a row, so on average it would be 8 trials not 4 trials.

Quote:JyBrd0403So, a random 4-roll sequence has a 50% chance of containing at least two consecutive losses.

Right, so you would need 2 4-roll sequences to get a loss of 2 in a row, so on average it would be 8 trials not 4 trials.

no, it's 4 trials. 4 trials is the average number of trials before you lose 2 in a row on a 50/50 game.

Unless someone wants to argue with the post from yahoo answers. Personally, I can't do the math below, but it sounds better then 4 trials to lose 2 in a row, 8 trials to lose 3 in a row and 16 trials to lose 4 in a row. Below is the post from yahoo answers.

Let E_n represent the expected number of trials needed to lose n in a row.

We want to get a recursion for E_(n+1) in terms of E_n. To find E_(n+1), let's condition on the outcome of the game just after losing n in a row for the first time.

If the outcome of that game is a win (which occurs with probability 1/2), then we must "start over" and the expected number of trials until losing (n + 1) in a row is E_n + 1 + E_(n + 1).

If the outcome of that game is a loss (which occurs with probability 1/2), then no more trials are needed and the expected number of trials until losing (n + 1) in a row is E_n + 1.

Therefore, overall we have

E_(n + 1) = (1/2)[E_n + 1 + E_(n + 1)] + (1/2)(E_n + 1)

(1/2)E_(n + 1) = E_n + 1

E_(n + 1) = 2(E_n + 1)

Clearly the expected number of trials until the first loss is 2, so E_1 = 2.

Then we have

E_2 = 2(E_1 + 1) = 2(2 + 1) = 6

E_3 = 2(E_2 + 1) = 2(6 + 1) = 14

E_4 = 2(E_3 + 1) = 2(14 + 1) = 30.

So on average, it takes 6 trials to lose 2 in a row, 14 trials to lose 3 in a row, and 30 trials to lose 4 in a row.

In fact, on average, it takes 2^(n + 1) - 2 trials to lose n in a row. Note that this general formula works for n = 1 and satisfies the recursion E_(n + 1) = 2(E_n + 1), so it works for all n by induction.

But there is a simpler way to understand it:

suppose i play a martingale system of exactly 5 steps. Then i have to stake 1, 2, 4, 8, 16, risking every time a total of 31 units to gain 1 unit. If i would always win at the first trial, it would take 31 coin flips to double up.

But given the fact that i expect to win every 2 trials, on average it will not take 31 trials, but 2x31= 62 trials.

The same calculations hold for 2,3,4 etc wins/losses in a row.

Now this is some elementary high school calculation i think hahaha, but i had to think about it myself for a while in order to figure it out. :-))

Losses in a row, # of trials, probability 50% or greater

2 in a row, 4 trials 50.000%

3 in a row, 10 trials 50.781%

4 in a row, 22 trials 51.515%

5 in a row, 44 trials 50.324%

6 in a row, 89 trials 50.302%

7 in a row, 178 trials 50.145%

8 in a row, 356 trials 50.097%

9 in a row, 711 trials 50.039%

You can see that the number of trials is "more or less" doubling as you increase the number of losses by one.

My answer is not in conflict with the one posted above by JyBrd0403. They are answers to different questions.

For example, the probability that you will equal or exceed two losses in a row, out of 6 trials is 67.188%, but 6 is also the expected number of trial to equal or exceed two losses in a row.

Quote:pacomartin

but 6 is also the expected number of trial to equal or exceed two losses in a row.

Does it not refer to the number of exact non overlapping successes/losses in a row which either end or not with a failure?

For example F,F,F,F/,F,F,F,F/,F,F,S is a sequence which contains 2 times 4's in a row, but only 1 streak of 4 or more in a row.

silly

silly

Let x=expected number of future flips starting from the beginning or after a win.

Let y=expected number of future flips starting from one loss.

(1)

x=1+.5x+.5y

y=1+.5x

(2)

.5x-.5y=1

-.5x+y=1

Add the two equations:

.5y=2

y=4

Then plus y=4 into either original equation to get x=6

------------------------

3 loss case

Let x=expected number of future flips starting from the beginning or after a win.

Let y=expected number of future flips starting from one loss.

Let z=expected number of future flips starting from two losses.

(1)

x = 1 + .5y + .5x

y = 1 + .5x + .5z

z = 1 + .5x

We have three equations and three unknowns. After a number of steps, which I won't go through we get x=14. Note that the mdeterm() function in Excel is very helpful in solving simultaneous linear equations.

silly

Quote:mustangsallyHere is a pdf from my stats class

Consecutive Successes: A Recursive Approach.

Thanks. Here is how I would explain that technique, in my own hand-waving manner. Let's assume that we know the four-loss case the answer is 30. What is the answer for the 5-loss case? Let x be answer.

x = 30 + 1 + 0.5x

The 30 is for the 14 flips it will take to three losses in a row, on average.

The 1 is for the next flip, which must be made.

From there there is a 50% chance of a win, sending up back to state x.

So, solving for x:

x = 31 + 0.5x

0.5x = 31

x = 62

Clear as mud?

Yes, Of course.Quote:Wizard

Thanks.

Clear as mud?

That is the beauty of math.

Not always just one way to arrive at an answer.

Boy, I bet the girls never took their eyes off of you when you were a teacher!

Maybe they never learned much because.

Wizard, You are so handsome and so smart!

Back to reality, lunch time is over

Sally

This was a neat trick my stats professor showed us since a run is simple a geometric series. (Yes, One can also use the geometric series formula)

p = 0.5

run length = 2

p^-1 + p^-2

2+4=6 easy!

Statements like this make us mere mortals envious of those with mathematical talents. EASY ???

silly

as good as I assumed.

However I would have another question for you :-))) : suppose i want to play 100 flips with a fair coin. I know that the average number of runs in a given sequence is given by the formula

Aver. Runs= (2HT/N) + 1, where N is the number of coin flips.

Should i have at the start of the 100 sequence 2 heads for example, will the average be influenced?

silly

silly

I think we must specify that the formula 1/p or 1/p^k is for the case when playing for example combos at sportsbets (never do unless you have an edge hahaha). I choose for example 5 bets and place 1 bet.

Maybe you could detail a little bit about the applications, like you wrote in an earlier post......

Quote:mustangsallyNo, this is an incorrect statement. Paco, you should know better> A Typo?

Well, I think this is one of those questions that really depends on the wording. I mean, it depends whether you consider a Win-Win-Win-Win scenario as just one 2 in a row win (one multiple win sequence of 2 or more wins), or if that is consider two 2 in a row win sequences. If you do consider 4 wins in a row as just being 1 winning sequence, your numbers will be different then someone who is considering those 4 wins as two, 2 in a row sequences. The math is going to be affected by the way you word the equation.

So, in this case, if you consider 4 wins in a row, to be, two 2 in a row sequences (say that 10 times fast lol), then it should average out to every 6 trials. If you consider 4 wins in a row to be only one 2 in a row sequence (one multiple win sequence of 2 or more wins) then the answer should be higher, once every 8 trials sounds good to me. It's one of those questions that really depend on the wording. It's like saying the chances of winning 2 in a row on a 50/50 game is 25%, when we all know, it's really 33% (once every 6 trials or once every 3 - 2 trial sequences equals 33%) :)

silly

Quote:mustangsally

And 33%?

NO way. (i sure hope that was a joke)

You are now mixing apples with oranges.

For 25% it is a given that N (the number of trials) is exactly 2... no more or no less. (discrete probability distribution)

The average of 6 assumes no value of N. (continuous probability distribution)

This is the difference between discrete and continuous probability distributions.

You can not mix the 2 as you just did.

Hopefully you now understand the simple math formulas and can make a streak length table where p can be any value. Not just 50%

Sally

Unfortunately, mustangsally, I don't have a clue about the math you're doing lol. However, if the N is the exact number of trials, which is 2 trials, and we expect on average to have 2 wins in a row in 6 trials, then 3 - 2 trial sequences would produce a win of 2 in a row. That equals 33% not 25%. The average of 6 does assume a value of N, it's the 100,000,000,000,000,000,000,000 we agreed to in your previous post. It's just that someone is counting a win-win-win-win sequence as 1 winning sequence, and someone else is counting it as 2 winning sequences.

So, I still come up with a 33% chance of winning 2 in a row if you flip that coin 6 times. I'm not kidding :) Am I wrong?

silly

Maybe the guy who did the programming for your computer was doing some bad math :) Am I wrong? Because, I still come up with 33% chance of winning 2 in a row in 6 trials with a 50/50 chance of winning.

removed

silly

So, I noticed you still haven't answered my question, Am I wrong? I still come up with 33%. And, yes, I have been drinking tonight. In case you were wondering. But, you still haven't answered the question. 25% I hope you're joking :)

-They can't all be wrong. Can they?-

P.S. My email address is JyBrd0403@yahoo.com

6 trials only.

The list are all 64 possible outcomes from

just

6

trials.

The correct answer is:

43/64

67.1585% or whatever it is, would be around 4 wins in a row in 6 trials, I believe.

I'll change the bet, if you lose you have to email me, if I lose we'll negotiate the payoff.

silly

Fraidy Cat :) Flip the coin 6 times. It's just an email.

That 67.1585% is just for 2 heads in a row, correct? It's not for 2 heads in a row, or 2 tails in a row, correct?. Because, if it is for 2 tails, or 2 heads in a row, you would lose the bet and have to email me:)

-They Can't all be wrong. Can They?-

Quote:JyBrd0403That 67.1585% is just for 2 heads in a row, correct?

MustangSally is correct.

Yes the question is what is the probability for two heads in a row only! Not for a streak of 2 of anything.

The probability of getting 2 heads in a row out of n tosses is

16/64 for n=2

24/64 for n=3

32/64 for n=4

38/64 for n=5

43/64 for n=6

...

The first fraction is the easiest calculation. The only possibilities are HH, TT, HT, & TH . So the probability is one in four or (1/2)^2 or p^n with p=0.5, n=2.

43/64 = 67.1585% which is slightly larger than two thirds. 2/3=42.666667/64

p=.5 (assume probability player loses)

q=1-p (probability player wins)

Pkk=p^k

DELTA=q*Pkk

P(j,k)=0 if j<k

For j>k, i== j-k-1 and Q(i,k)=1-P(i,k)

Note j<=2k , Q=1

P(j,k)=P(j-1,k)+Q(i,k)*DELTA

It should be noted that there are other recursive sequences that arrive at the same result. I like this one because it is by far the least computationally intensive and the shortest one to write. There are also approximations for large j .

The original question was what is the probability to loose 2 in a row without any other explanations.

Quote:edwardwhat if my 2 sequence i choose to play is not HH ot TT, but HT or TH?

The original question was what is the probability to loose 2 in a row without any other explanations.

The original question was on average how many trials will it take to lose 2 in a row without a house edge. The answer is 6.

The simpler question is on average how many trials will it take to lose a game of chance without a house edge. The answer is 2, but the reasoning is:

You have a 50% chance of losing in one trial (i.e toss of a coin)

You have a 25% chance of losing in two trials

You have a 1/4 chance of losing in three trials

You have a 1/8 chance of losing in four trials

....

So the expected number of trials is 1+1/2+ 1/4+1/8+... or a standard infinite geometric sequence.

S=1+1/2+ 1/4+1/8+... (equation #1)

1/2*S=1/2+ 1/4+1/8+...(equation #2)

(equation #1)-(equation #2)

1/2*S=1

S=2

Quote:pacomartin

You have a 25% chance of losing in two trials

You have a 1/4 chance of losing in three trials

So the chance of losing in 2 trials is the same as losing in 3 trials?

Quote:edwardSo the chance of losing in 2 trials is the same as losing in 3 trials?

Sorry, that was a mistake. I meant to say

You have a 50% chance of losing in one trial (i.e toss of a coin)

You have a 25% chance of losing in two trials

You have a 1/8 chance of losing in three trials

You have a 1/16 chance of losing in four trials

....

On average it take 3 trials to have a HT sequence or 1.5 trials to have either HT or TH

Quote:edwardI dont think to lose 2 in a row is the same as having a run of length 2, maybe im wrong

On average it take 3 trials to have a HT sequence or 1.5 trials to have either HT or TH

Yes, I agree with you. A run of length 2 could be either a run of hits or losses.

There is a 50% probability of a HT sequence in 3 trials. Agreed!

HHT ,TTT,THT,HHT,HTT,HTH,HTT,HHH

However, that is not the same thing as saying that you can expect to have a HT sequence in just 3 trials. The expected number of trials is 6.

silly

but what is the explanation the probabilities dont add up to 1?

You are mixing up your data.Quote:edwardSally,

but what is the explanation the probabilities dont add up to 1?

We are calculating just the averages where N (number of trials) is not known and can take on any value from 2 and higher (continuous probability distribution)

NOT

each 4 sequences in 2 trials (discrete probability distribution)

added: see my above post for more average wait time for sequences of length 2 and 3

Quote:edwardI dont think to lose 2 in a row is the same as having a run of length 2, maybe im wrong

Why not?

Say you bet for Tails and the next 2 coin flips are HH

You just LOST 2 in a row.

The length of your losing streak (run)

looks like 2 to me

Quote:mustangsallyWhy not?

Say you bet for Tails and the next 2 coin flips are HH

You just LOST 2 in a row.

The length of your losing streak (run)

looks like 2 to me

you are obviously right, i might be a little retarded :-)))

Quote:mustangsallyThe expected number of trials for the sequence HT is 4.

I'm sorry, I am losing it. I once had a brain and I did co-author this paper. But it's gotten rusty.

Quote:Abstract of Journal of Spacecraft and Rockets

"Godunov's Method for Supersonic Tactical Missile Computations," AIAA 85-1812

THIS paper investigates application of a finite volume Godunov method to steady supersonic flow, with the objective of producing a more robust computational method.

The Godunov scheme was originally proposed for unsteady gasdynamics, but can easily be extended to steady supersonic flow. Unlike conventional numerical schemes that presume asmooth flowfield, the Godunov method assumes that the flowfield consists of a number of piecewise constant states with discontinuities occurring midway between the mesh points. The evolution of the flowfield in the vicinity of the discontinuities is determined by solving the associated Riemann problem. The resulting method is first-order accurate,but can be extended to higher order.