On average how many trials will it take to lose 2 in a row on a 50/50 game of chance?
January 19th, 2012 at 12:48:31 PM
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How about 3 in a row, 4 in a row. etc. It may seem elementary but someone said it would be 4 trials for 2 in a row, 8 trials for 3 in a row, and 16 trials for 4 in a row. That can't be right.
January 19th, 2012 at 12:58:27 PM
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Oops. Wrong again.
Wisdom is the quality that keeps you out of situations where you would otherwise need it
January 19th, 2012 at 1:04:39 PM
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Well 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.
January 19th, 2012 at 1:26:38 PM
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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 as the clock ticked and the day passed, opportunity met preparation, and luck happened." - Maurice Clarett
January 19th, 2012 at 1:37:49 PM
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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.
So you're saying you would lose 2 in a row every 8 trials not 4 trials, correct.
January 19th, 2012 at 1:44:49 PM
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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.
"So as the clock ticked and the day passed, opportunity met preparation, and luck happened." - Maurice Clarett
January 19th, 2012 at 1:56:16 PM
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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.
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.
January 19th, 2012 at 2:13:03 PM
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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.
"So as the clock ticked and the day passed, opportunity met preparation, and luck happened." - Maurice Clarett
January 19th, 2012 at 2:18:34 PM
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Nevermind, I found the answer on yahoo answers. 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.
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.
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.
January 26th, 2012 at 3:39:58 AM
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The yahoo answer is correct.
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. :-))
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. :-))
January 26th, 2012 at 7:10:50 AM
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The exact answer to how many trials must you do until you equal or exceed a 50% probability is below (given no house edge):
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.
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.
January 26th, 2012 at 8:02:14 AM
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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.
January 26th, 2012 at 9:28:22 AM
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January 26th, 2012 at 9:32:08 AM
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January 26th, 2012 at 10:21:09 AM
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Here is the 2-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.
(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
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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.
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.
"For with much wisdom comes much sorrow." -- Ecclesiastes 1:18 (NIV)
January 26th, 2012 at 12:11:26 PM
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January 26th, 2012 at 12:19:27 PM
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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?
"For with much wisdom comes much sorrow." -- Ecclesiastes 1:18 (NIV)
January 26th, 2012 at 12:31:19 PM
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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
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January 26th, 2012 at 12:43:01 PM
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" Since a run of any length is not an independent event one can easily just add all the steps along the way.
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 ???
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 ???
January 26th, 2012 at 1:14:46 PM
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January 26th, 2012 at 1:35:13 PM
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I appreciate you attempt to educate me. Unfortunately I am still lost. Evidently my 11 grade education by the Xaverian Brothers was not
as good as I assumed.
as good as I assumed.
January 26th, 2012 at 2:05:01 PM
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Thank You MustangSally and all who provided different methods for the question.
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?
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?
January 27th, 2012 at 10:40:30 AM
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January 27th, 2012 at 10:48:42 AM
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January 28th, 2012 at 3:34:58 PM
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Good formulas Sally.
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......
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......
January 29th, 2012 at 2:26:42 PM
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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%) :)
January 29th, 2012 at 5:34:10 PM
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January 29th, 2012 at 5:57:53 PM
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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?
January 29th, 2012 at 6:18:22 PM
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January 29th, 2012 at 6:28:31 PM
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67.1875% I thought we agreed it was 33% or lower (25%). I think that's a 67.1875% chance that you won't win 2 in a row in 6 trials, which would make sense.
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.
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.
January 29th, 2012 at 7:08:49 PM
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The exact probability as a fraction is 43/64
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January 29th, 2012 at 7:26:36 PM
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Ah! Now I see the problem... You're saying it's 43/64, correct? How many trials were in those 64 sequences? It's 64 x 6 = 384 trials. So, it would be 43/384 not 43/64. Providing that 43 is the correct number of 2 in a rows:)
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
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
January 29th, 2012 at 7:47:27 PM
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Yes, you are wrong.
6 trials only.
The list are all 64 possible outcomes from
just
6
trials.
The correct answer is:
43/64
6 trials only.
The list are all 64 possible outcomes from
just
6
trials.
The correct answer is:
43/64
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January 29th, 2012 at 8:02:58 PM
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I think you're wrong too :) Now we have a bet. Flip that coin 6 times and then tell me you come up with 2 wins in a row is 67.1585% or whatever it is :) If you lose, you have to email me :) If it does come up to 67.1585% or whatever it is, I'll email you :) Providing you give me your email address.
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.
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.
January 29th, 2012 at 9:03:49 PM
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January 29th, 2012 at 9:28:09 PM
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That's very interesting, but I thought we had a bet. Flip that coin 6 times and tell me the results are 67.1585%. And, we haven't even negotiated the payoff yet ;)
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?-
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?-
January 30th, 2012 at 3:55:49 AM
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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
January 30th, 2012 at 4:00:22 AM
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The recursive sequence of probability P is the special case where p=1/2, and k=2
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 .
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 .
January 30th, 2012 at 4:04:55 AM
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what 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 what is the probability to loose 2 in a row without any other explanations.
January 30th, 2012 at 4:13:21 AM
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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
January 30th, 2012 at 4:58:12 AM
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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?
January 30th, 2012 at 5:03:17 AM
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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
....
January 30th, 2012 at 5:21:22 AM
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I 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
On average it take 3 trials to have a HT sequence or 1.5 trials to have either HT or TH
January 30th, 2012 at 6:45:29 AM
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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.
January 30th, 2012 at 8:07:44 AM
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removed
silly
silly
I Heart Vi Hart
January 30th, 2012 at 8:16:46 AM
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Sally,
but what is the explanation the probabilities dont add up to 1?
but what is the explanation the probabilities dont add up to 1?
January 30th, 2012 at 8:29:13 AM
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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
I Heart Vi Hart
January 30th, 2012 at 9:03:04 AM
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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
I Heart Vi Hart
January 30th, 2012 at 9:37:18 AM
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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 :-)))
January 30th, 2012 at 10:20:54 AM
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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.
