JyBrd0403
JyBrd0403
Joined: Jan 25, 2010
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January 19th, 2012 at 12:48:31 PM permalink
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.
dwheatley
dwheatley
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January 19th, 2012 at 12:58:27 PM permalink
Oops. Wrong again.
Wisdom is the quality that keeps you out of situations where you would otherwise need it
JyBrd0403
JyBrd0403
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January 19th, 2012 at 1:04:39 PM permalink
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.
rdw4potus
rdw4potus
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January 19th, 2012 at 1:26:38 PM permalink
Quote: JyBrd0403

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.



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
JyBrd0403
JyBrd0403
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January 19th, 2012 at 1:37:49 PM permalink
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.
rdw4potus
rdw4potus
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January 19th, 2012 at 1:44:49 PM permalink
Quote: JyBrd0403

Quote: RDW4POTUS

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.



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
JyBrd0403
JyBrd0403
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January 19th, 2012 at 1:56:16 PM permalink
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.
rdw4potus
rdw4potus
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January 19th, 2012 at 2:13:03 PM permalink
Quote: JyBrd0403

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.



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
JyBrd0403
JyBrd0403
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January 19th, 2012 at 2:18:34 PM permalink
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.
edward
edward
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January 26th, 2012 at 3:39:58 AM permalink
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. :-))

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