On average how many trials will it take to lose 2 in a row on a 50/50 game of chance?
Quote: edwardyou are obviously right, i might be a little retarded :-)))
I thought "retarded" had to do with a distributor in an engine (a classic car that is) where the timing was "backed off" from the normal specs, like 10degrees before top dead center being normal but some grease monkey set it to 2 degrees AFTER top dead center for smog reasons.
(yes, I can work on cars also)
Hey, we all learn about probability at different rates.
No one ever said probability is easy. (maybe me!) LOL
I was absolutely failing in prob/stats class until the last two weeks of the semester where the professor gave us gambling questions to go along with theory.
It all then made it clear.
The lights went on for me!
My professor's last words to our class was, "Even if you can not prove one formula like a/(1-r) (the sum of an infinite geometric series), as long as you can do the math accurately using the proper formula, you will be ahead of most"
Don't be sorry.Quote: pacomartinI'm sorry, I am losing it. I once had a brain and I did co-author this paper. But it's gotten rusty.
You may just need a vacation and a good back rub.
I was surprised you did not link to your losing streak graph pdf. I have it in my computer. :)
Someone was pointing out the problem with it being 33%. All the math says it's 25%. The math you're doing is based on 8 possible outcomes, or what I would consider 8 trials, mistakenly, of course lol
You see, if 25% was the correct answer for a win of 2 or more then you would expect it to take 8 trials to produce 2 wins in a row. You guys blew that theory up with the math proving that it is an expected 6 trials rather than 8. The old win/loss matrix would show that as 8 trials, of course, they then say that, that's just a list of possible outcomes not actual trials. Right.
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So, yes I guess the simplest way would be for me to just come out and say that the math of .5 x .5 = .25 or 25% is just flat out wrong :) That's my statement. It's actually 33%, and all the math showing that 33% is incorrect, I would say is in error.
So, another simple way I can prove my point, as always without using a whole lot of math lol is to flip a coin. So, my hypothesis is this, in just 2 trials, the next 2 coin flips, I think I have a 33% of winning twice in a row, I'll choose heads so that's a 33% of getting 2 heads in a row. The only parameters being that you must bet on each coin flip, so, if a Tails comes up first that would count as a failure and the next coin flip would start a new sequence. Count the number of trials, and count the number of times you get a run of 2 heads in a row. Just for clarification, 4 wins in a row would count as two 2 wins in a row sequences. Divide the number of times you get 2 in a row by the number of trials, and the answer is going to be 33%:) Am I wrong?
So, do we still have a bet mustangsally :)
Quote: JyBrd0403
So, another simple way I can prove my point, as always without using a whole lot of math lol is to flip a coin. So, my hypothesis is this, in just 2 trials, the next 2 coin flips, I think I have a 33% of winning twice in a row, I'll choose heads so that's a 33% of getting 2 heads in a row. The only parameters being that you must bet on each coin flip, so, if a Tails comes up first that would count as a failure and the next coin flip would start a new sequence. Count the number of trials, and count the number of times you get a run of 2 heads in a row. Just for clarification, 4 wins in a row would count as two 2 wins in a row sequences. Divide the number of times you get 2 in a row by the number of trials, and the answer is going to be 33%:) Am I wrong?
So, do we still have a bet mustangsally :)
Sorry, that's wrong, it would be divide the number of times you get 2 wins in a row by 1/2 (half) the total number of trials equals 33%. Or, whatever the math is that makes the thing below equal 25%. Same math, but we're getting 2 wins in a row in 6 trials instead of 8 trials.
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Quote: JyBrd0403Okay, so where are we at? :)
Oh Yeah, we're all agree that on average it will take 6 trials to win 2 in a row.
Remember at the start of this thread I got argued with the 4 trial thing. Thank goodness for Yahoo.
Though, I think,mustangsally, would have straightened that out, and made it a lot easier too.
So, now it's just that I'm saying that if you have established that a 2 win sequence occurs once every 6 trials,
We all agree that it takes on average 6 trials to win 2 in a row
It does not follow from the above statement that a 2 win sequence occurs once every 6 trials
The odds of two wins in a row out of 6 trials are 43/64 * 6 , or 4.0312500
Think about what is the difference between the above two statements.
Forget about counting and all that stuff, if youre saying the probability is 0.33 or 1/3, you are saying the average wait time is 3.
Quote: edwardActually, this should be 6th grade math, or? hahahah, damn
I sense that you may not be totally convinced, so I made a quick and dirty simulation. Change the word in first line to recalculate the sheet.
Since there is only 100 simulations, the average may only come close to 6.00 .
Simulation Spreadsheet
It's not 6th grade math. Just because the question sounds easy, the mathematics may not be.
Quote: pacomartinWe all agree that it takes on average 6 trials to win 2 in a row
It does not follow from the above statement that a 2 win sequence occurs once every 6 trials
The odds of two wins in a row out of 6 trials are 43/64 * 6 , or 4.0312500
Think about what is the difference between the above two statements.
Semantics, of course you won't have a 100% guarantee of winning 2 in a row in the next 6 trials. But, we're talking averages here, on average it will be 6 trials to produce 2 wins in a row.
So, in other words, the math your doing involving probable outcomes as opposed to actual trials, is incorrect. The probable outcome math you're doing is saying in the next 2 trials you have a 25% chance of winning 2 in a row. When you do actual trials, a simulation, or whatever the math is that mustangsally, and the yahoo guy did, it is in fact, a 33% chance of winning 2 in a row in the next 2 trials.
So, I'm asking this, is the math for actual trials correct, or is the math of probable outcomes correct. I'm personally going with reality and dumping the old math. I say it's 33%. So, we need to determine if I have a 33% chance of winning 2 in a row, in the next 2 trials (discrete probability distribution, I guess), or if reality is wrong and you have a 25% chance of winning 2 in a row because of this probable outcome math.
I'm still going with 33% chance of winning 2 in a row in the next 2 trials. I hope you didn't leave the thread mustangsally, I'm still not joking :)
-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."-
I also will agree on the 6 trials for a run of length 2.Quote: JyBrd0403Quote: pacomartinWe all agree that it takes on average 6 trials to win 2 in a row
It does not follow from the above statement that a 2 win sequence occurs once every 6 trials
The odds of two wins in a row out of 6 trials are 43/64 * 6 , or 4.0312500
Think about what is the difference between the above two statements.
Semantics, of course you won't have a 100% guarantee of winning 2 in a row in the next 6 trials. But, we're talking averages here, on average it will be 6 trials to produce 2 wins in a row.
The probability of 2 in a row in the next 6 trials was shown to be 43/64 by sledge hammer counting and Sally earlier linked to pacomartin's thread on how to do the math yourself with a recursive formula-actually 2 of them, his and the Wizards-, a Markov Chain approach or even an online script by another math maven BruceZ.
Snap out of it. The beginning of your thread was good stuff, it is now breaking down into loud snickers, and they are aimed at you.
Here is the distribution for a run of length 2 for 2 to 80 trials. I used the Wizard's formula in my Excel spreadsheet.
43/64 = 67.1875%
That is my answer.
What are you not understanding here?
| trial | 2+ | relative |
|---|---|---|
| 0 | 0 | probability |
| 1 | 0 | |
| 2 | 25.000000% | 25.000000% |
| 3 | 37.500000% | 12.500000% |
| 4 | 50.000000% | 12.500000% |
| 5 | 59.375000% | 9.375000% |
| 6 | 67.187500% | 7.812500% |
| 7 | 73.437500% | 6.250000% |
| 8 | 78.515625% | 5.078125% |
| 9 | 82.617188% | 4.101563% |
| 10 | 85.937500% | 3.320313% |
| 11 | 88.623047% | 2.685547% |
| 12 | 90.795898% | 2.172852% |
| 13 | 92.553711% | 1.757813% |
| 14 | 93.975830% | 1.422119% |
| 15 | 95.126343% | 1.150513% |
| 16 | 96.057129% | 0.930786% |
| 17 | 96.810150% | 0.753021% |
| 18 | 97.419357% | 0.609207% |
| 19 | 97.912216% | 0.492859% |
| 20 | 98.310947% | 0.398731% |
| 21 | 98.633528% | 0.322580% |
| 22 | 98.894501% | 0.260973% |
| 23 | 99.105632% | 0.211132% |
| 24 | 99.276441% | 0.170809% |
| 25 | 99.414629% | 0.138187% |
| 26 | 99.526425% | 0.111796% |
| 27 | 99.616870% | 0.090445% |
| 28 | 99.690041% | 0.073171% |
| 29 | 99.749238% | 0.059197% |
| 30 | 99.797129% | 0.047891% |
| 31 | 99.835874% | 0.038745% |
| 32 | 99.867219% | 0.031345% |
| 33 | 99.892578% | 0.025359% |
| 34 | 99.913094% | 0.020516% |
| 35 | 99.929691% | 0.016598% |
| 36 | 99.943119% | 0.013428% |
| 37 | 99.953982% | 0.010863% |
| 38 | 99.962771% | 0.008789% |
| 39 | 99.969881% | 0.007110% |
| 40 | 99.975633% | 0.005752% |
| 41 | 99.980287% | 0.004654% |
| 42 | 99.984052% | 0.003765% |
| 43 | 99.987098% | 0.003046% |
| 44 | 99.989562% | 0.002464% |
| 45 | 99.991555% | 0.001994% |
| 46 | 99.993168% | 0.001613% |
| 47 | 99.994473% | 0.001305% |
| 48 | 99.995528% | 0.001056% |
| 49 | 99.996382% | 0.000854% |
| 50 | 99.997073% | 0.000691% |
| 51 | 99.997632% | 0.000559% |
| 52 | 99.998084% | 0.000452% |
| 53 | 99.998450% | 0.000366% |
| 54 | 99.998746% | 0.000296% |
| 55 | 99.998986% | 0.000239% |
| 56 | 99.999179% | 0.000194% |
| 57 | 99.999336% | 0.000157% |
| 58 | 99.999463% | 0.000127% |
| 59 | 99.999565% | 0.000103% |
| 60 | 99.999648% | 0.000083% |
| 61 | 99.999716% | 0.000067% |
| 62 | 99.999770% | 0.000054% |
| 63 | 99.999814% | 0.000044% |
| 64 | 99.999849% | 0.000036% |
| 65 | 99.999878% | 0.000029% |
| 66 | 99.999901% | 0.000023% |
| 67 | 99.999920% | 0.000019% |
| 68 | 99.999935% | 0.000015% |
| 69 | 99.999948% | 0.000012% |
| 70 | 99.999958% | 0.000010% |
| 71 | 99.999966% | 0.000008% |
| 72 | 99.999972% | 0.000007% |
| 73 | 99.999978% | 0.000005% |
| 74 | 99.999982% | 0.000004% |
| 75 | 99.999985% | 0.000003% |
| 76 | 99.999988% | 0.000003% |
| 77 | 99.999990% | 0.000002% |
| 78 | 99.999992% | 0.000002% |
| 79 | 99.999994% | 0.000001% |
| 80 | 99.999995% | 0.000001% |
Quote: JyBrd0403So, in other words, the math your doing involving probable outcomes as opposed to actual trials, is incorrect. The probable outcome math you're doing is saying in the next 2 trials you have a 25% chance of winning 2 in a row. When you do actual trials, a simulation, or whatever the math is that mustangsally, and the yahoo guy did, it is in fact, a 33% chance of winning 2 in a row in the next 2 trials.
So, I'm asking this, is the math for actual trials correct, or is the math of probable outcomes correct. I'm personally going with reality and dumping the old math. I say it's 33%. So, we need to determine if I have a 33% chance of winning 2 in a row, in the next 2 trials (discrete probability distribution, I guess), or if reality is wrong and you have a 25% chance of winning 2 in a row because of this probable outcome math.
I'm still going with 33% chance of winning 2 in a row in the next 2 trials. I hope you didn't leave the thread mustangsally, I'm still not joking :)
What are you missing?
Are not there 4 possible outcomes of 2 coin tosses? 2^2
HH
TT
HT
TH
Let me see. If I bet Tails for the very next 2 flips, there is only one out of the four possible outcomes that has me winning 2 in a row.
Maybe I should choose Heads instead.
Ok.
1 way to get HH out of 4.
1/4
This makes perfect sense to me.Quote: JyBrd0403
-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."-
TT
HT
TH
That equals 25%. Unfortunately, we only have 6 trials to produce 2 heads and only 6 trials to produce 2 tails in a row. As a side note. Notice if you put these back to back, it would look like this HHTTHTTH , notice the TT's. The 6 trials in actuality would look like this.
HH
TT
HT
is one scenario.
HH
TT
TH
is the other scenario. Or in other words 2 heads in a row, 2 tails in a row, and one non 2 in a row. So, if I picked Heads for the next 2 flips, there is only 1 out of 3 possible outcomes that has me winning 2 in a row 1/3 or 33%.
If I picked Tails for the next 2 flips, there is only 1 out of 3 possible outcomes that has me winning 2 in a row 1/3 or 33%.
Maybe I should bet there will be a non 2 in a row. There's still only 1 out of 3 possible outcomes that have a non 2 in a row. 1/3 or 33%.
So, I still come up with 33%, based on the fact that we only have 6 trials to work with here not 8 trials. Yep, I still think it's 33%. Why is that a difficult concept, besides the fact it would mean the answer of 25% would be wrong. That's my point, this is really getting silly. It's obviously 1 out of 3 possible outcomes or 33%.
NO man!Quote: JyBrd0403HH
TT
HT
TH
That equals 25%. Unfortunately, we only have 6 trials to produce 2 heads and only 6 trials to produce 2 tails in a row.
That is your misunderstanding.
The 4 possible outcomes are for 2 flips only.
There are no 6 flips here.
You are talking about 2 in a row in 2 flips.
1 in 4.
Done
Next...
2 in a row in 6 flips is a totally different question.
You can not mix up the 2.
In 6 flips do you agree that there are a possible 64 outcomes, 64 possible paths the coin flips could take?
You need to understand this or you be lost for a long time... maybe forever.
Please explain to me how, on average you can get 2 heads in a row, and 2 tails in a row on average to occur every 6 trials (which we've already agreed upon) doing the math that you're doing. You can't, it would have to be on average every 8 trials.
HH
TT
HT
TH
You need 8 trials to average 25% of the time you'll get 2 heads, and 25% of the time you'll get 2 tails. That is on average 2 heads in a row every 8 trials, and on average 2 tails in a row every 8 trials. That's on average, I don't really care how many times you flip the coin. It's just the average that I'm looking for. So, on average it would be 1 in 4, or 2 in 8, or 3 in 12, or 4 in 16 to get 25%. Please explain how you can get those averages to occur every 6 trials instead of every 8 trials. You can't.
Quote: JyBrd0403There are 4 possible outcomes for 2 flips only. So what?
Please explain to me how, on average you can get 2 heads in a row, and 2 tails in a row on average to occur every 6 trials (which we've already agreed upon) doing the math that you're doing. You can't, it would have to be on average every 8 trials.
HH
TT
HT
TH
You need 8 trials to average 25% of the time you'll get 2 heads, and 25% of the time you'll get 2 tails. That is on average 2 heads in a row every 8 trials, and on average 2 tails in a row every 8 trials. That's on average, I don't really care how many times you flip the coin. It's just the average that I'm looking for. So, on average it would be 1 in 4, or 2 in 8, or 3 in 12, or 4 in 16 to get 25%. Please explain how you can get those averages to occur every 6 trials instead of every 8 trials. You can't.
Oh, yes I can. No Problem.
But not to rude people like you.
I first asked you a simple question of possible outcomes in 6 coin flips. You could not even answer a simple question.
That is pure rudeness in my book.
Let us see who steps up next.
Have a great time in your universe!
Quote: 7crapsOh, yes I can. No Problem.
But not to rude people like you.!
LOL!
Quote: JyBrd0403There are 4 possible outcomes for 2 flips only. So what?
Then for more then 2 flips the occurence of 2 signs of same type is conditioned by the fact that the previous sign is the same. So you will have 2 waiting times which you add up (one for single win and one for double win) :-))))))))))))))
Quote: 7craps2 in a row in 6 flips is a totally different question.
What 7craps is saying is critical. For the question of the average number of trials to get 2 losses (or heads), you could have some trials that last some arbitrary length (easily over 10). But many times it will be under 6 trials.The average of them together will be 6.
See simulation spreadsheet if you are having trouble visualizing.
If you lay out exactly 6 trials, then you will have 43 out of a possible 64 (67.18750%) where there are two losses in a row.
If you lay out exactly 2 trials, then you will have 1 out of a possible 4 (25.000%) where there are two losses in a row.
If you lay out exactly 3 trials, then you will have 3 out of a possible 8 (37.500%) where there are two losses in a row.
If you lay out exactly 4 trials, then you will have 8 out of a possible 16 (50.000%) where there are two losses in a row.
If you lay out exactly 5 trials, then you will have 19 out of a possible 32 (59.375%) where there are two losses in a row.
It isn't semantics. It's a different question.
Quote: pacomartinWhat 7craps is saying is critical. For the question of the average number of trials to get 2 losses (or heads), you could have some trials that last some arbitrary length (easily over 10). But many times it will be under 6 trials.The average of them together will be 6.
See simulation spreadsheet if you are having trouble visualizing.
If you lay out exactly 6 trials, then you will have 43 out of a possible 64 (67.18750%) where there are two losses in a row.
If you lay out exactly 2 trials, then you will have 1 out of a possible 4 (25.000%) where there are two losses in a row.
If you lay out exactly 3 trials, then you will have 3 out of a possible 8 (37.500%) where there are two losses in a row.
If you lay out exactly 4 trials, then you will have 8 out of a possible 16 (50.000%) where there are two losses in a row.
If you lay out exactly 5 trials, then you will have 19 out of a possible 32 (59.375%) where there are two losses in a row.
It isn't semantics. It's a different question.
I think part of the confusion is in terminology. If you ask for the probability of a 6 in one roll of a die, that is 1/6. But the average (expected) roll of a die is worth 3 1/2. One is a question about probabilities of certain outcomes in a distribution, the other is a question about the mean of that distribution. They are separate inquiries. If the word "average" isn't used precisely, there will be confusion.
Game A produces 2 wins in a row on average every 6 trials.
Game B produces 2 wins in a row on average every 8 trials.
Which game is the 50/50 game?
What is the percentage chance of winning 2 in a row on game A?
What is the percentage chance of winning 2 in a row on game B?
The last two questions only mean something if you specify the number of trials. If you mean two trials for game A then the answer is 1/4.
Great questions.Quote: JyBrd0403Let me try this a different way. Let's say you have 2 games.
Game A produces 2 wins in a row on average every 6 trials.
Game B produces 2 wins in a row on average every 8 trials.
Which game is the 50/50 game?
This thread was starting to look ugly. and still is a little.
2 formulas for the average number of trials to see a run of a specific length are (n = streak length)
average # of trials = ((p^-n)-1)/(1-p)
or
average # of trials = (1/p^n)*((1-p^n)/(1-p))
All you need to do is solve for p
I have already done that back in school.
Let us see who else can do it.
Streak probability is NOT 6th grade stuff. Not even high school level.
Because those kids just want to know the answer. answers only. Most do not give a damn about the solution.
This forum should care about solutions.
That is how we learn.
added: not all solutions are easy to understand. The Wizard used a matrix solution in his ask the wizard column for 2 in a row. maybe he wanted to be different.
He was different. Like using a shot gun to kill an ant.
Quote: JyBrd0403Let me try this a different way. Let's say you have 2 games.
Game A produces 2 wins in a row on average every 6 trials.
Game B produces 2 wins in a row on average every 8 trials.
Which game is the 50/50 game?
What is the percentage chance of winning 2 in a row on game A?
What is the percentage chance of winning 2 in a row on game B?
Using the previously-cited formula for mean wait time:
((1/p^n) - 1) / (1 - p)
Game A, 2 in a row: ((1/p^2) - 1) / (1 - p) = 6
Game B, 2 in a row: ((1/p^2) - 1) / (1 - p) = 8
After some manipulation,
Game A: 7p^2 - 6p^3 = 1
Game B: 9p^2 - 8p^3 = 1
Solving for p, and restricting to 0 < p < 1 (because 1 is a solution to both equations, and both equations have solutions < 0)
Game A, p(win) = 0.5
Game B, p(win) = (1+sqrt(33))/16, or about 0.421535
FYI, Wolfram Alpha is awesome.
Edit: I don't type fast enough...
:)Quote: MathExtremist
FYI, Wolfram Alpha is awesome.
Edit: I don't type fast enough...
you did just fine.
Quote: MathExtremistYeah, but I've completely forgotten how to factor polynomials using my own head. It's been soooo long.
Now I have the same problem as You have, as i've learned this maybe 20 years ago and used it. But i was able to find the solution and I'm very proud ! hahahah
So, we have
9x^2 - 8x^3=1 <=> 8x^3 - 9x^2 + 1 =0
now write the last one as 8x^3 - 8x^2 -x^2 + 1 = 0 <=> 8x^2(x-1) - (x^2 - 1) <=> 8x^2(x-1) - (x-1)(x+1) <=> (x-1)(8x^2 -x - 1) = 0
from here we have X1 = 1;
x2 and x3 follow from the ecuation 8x^2 -x-1=0,
delta = b^2 - 4ac=33 =>x2,x3 = (1 +-sqrt33)/16
hahaha i've solved it. Damn!
Now this was for sure some 9th grade math :-)))))))))))))))))))))))
Quote: edwardNow this was for sure some 9th grade math :-)))))))))))))))))))))))
I taught high school math about 20 years ago, and it was the fairly smart kids who could do cubic equations in 9th grade. Sometimes if you can easily see one solution, (it's pretty obvious that x=1 makes " 8x^3 - 9x^2 + 1 " equal to zero, then you can divide by x-1).
I think a lot of the stuff we do will be lost to future generations as equation solvers and even symbolic calculus integrators will be built into telephones. It's difficult to understand what an impact Napier had in his day on computation by working out the logarithm tables.
Knowing how to factor sure helped James Martin when playing the British gameshow Countdown. If you don't know how to play the game the rules are the numbers: 25, 50, 75, & 100 plus two random numbers are selected. In this incident the random numbers were 3 and 6. A final target is chosen by computer (in this 952), and the objective of the game is to get as close as the target as possible using basic arithmetic functions in 30 seconds.
The loser came up with (6+3)*100 + 50 + 75/25 = 953 missing the target by 1. He used all six numbers, but that is not a requirement. The winner got exactly 952. If you can factor, then try it yourself before you watch the video. The host was gobsmacked.
The final answer is you can't tell what the percentage chance of winning 2 in a row is for either game? Seriously? I can go back to Yahoo for the answer.
I endorse that strategy and suggest you do it very soon.Quote: JyBrd0403I can go back to Yahoo for the answer.
Quote: JyBrd0403Let me try this a different way. Let's say you have 2 games.
Game A produces 2 wins in a row on average every 6 trials.
Game B produces 2 wins in a row on average every 8 trials.
What is the percentage chance of winning 2 in a row on game A?
What is the percentage chance of winning 2 in a row on game B?
pacomartin mentioned that the percentage chance of winning is based on the number of trials. He is absolutely correct.
Here are the numbers and your answers
Col A x = number of trials
Col B is the relative frequency prob[X=x]
Col E is the cumulative frequency prob[X<=x]
Game A p = 0.5 run length 2
x prob[X=x] prob[X<x] prob[X>=x] prob[X<=x] prob[X>x]
2 0.2500000000 0.0000000000 1.0000000000 0.2500000000 0.7500000000
3 0.1250000000 0.2500000000 0.7500000000 0.3750000000 0.6250000000
4 0.1250000000 0.3750000000 0.6250000000 0.5000000000 0.5000000000
5 0.0937500000 0.5000000000 0.5000000000 0.5937500000 0.4062500000
6 0.0781250000 0.5937500000 0.4062500000 0.6718750000 0.3281250000
7 0.0625000000 0.6718750000 0.3281250000 0.7343750000 0.2656250000
8 0.0507812500 0.7343750000 0.2656250000 0.7851562500 0.2148437500
9 0.0410156250 0.7851562500 0.2148437500 0.8261718750 0.1738281250
10 0.0332031250 0.8261718750 0.1738281250 0.8593750000 0.1406250000
11 0.0268554688 0.8593750000 0.1406250000 0.8862304688 0.1137695312
12 0.0217285156 0.8862304688 0.1137695312 0.9079589844 0.0920410156
13 0.0175781250 0.9079589844 0.0920410156 0.9255371094 0.0744628906
14 0.0142211914 0.9255371094 0.0744628906 0.9397583008 0.0602416992
15 0.0115051270 0.9397583008 0.0602416992 0.9512634277 0.0487365723
16 0.0093078613 0.9512634277 0.0487365723 0.9605712891 0.0394287109
17 0.0075302124 0.9605712891 0.0394287109 0.9681015015 0.0318984985
18 0.0060920715 0.9681015015 0.0318984985 0.9741935730 0.0258064270
19 0.0049285889 0.9741935730 0.0258064270 0.9791221619 0.0208778381
20 0.0039873123 0.9791221619 0.0208778381 0.9831094742 0.0168905258
21 0.0032258034 0.9831094742 0.0168905258 0.9863352776 0.0136647224
22 0.0026097298 0.9863352776 0.0136647224 0.9889450073 0.0110549927
23 0.0021113157 0.9889450073 0.0110549927 0.9910563231 0.0089436769
24 0.0017080903 0.9910563231 0.0089436769 0.9927644134 0.0072355866
25 0.0013818741 0.9927644134 0.0072355866 0.9941462874 0.0058537126
26 0.0011179596 0.9941462874 0.0058537126 0.9952642471 0.0047357529
27 0.0009044483 0.9952642471 0.0047357529 0.9961686954 0.0038313046
28 0.0007317141 0.9961686954 0.0038313046 0.9969004095 0.0030995905
29 0.0005919691 0.9969004095 0.0030995905 0.9974923786 0.0025076214
30 0.0004789131 0.9974923786 0.0025076214 0.9979712917 0.0020287083
31 0.0003874488 0.9979712917 0.0020287083 0.9983587405 0.0016412595
32 0.0003134527 0.9983587405 0.0016412595 0.9986721931 0.0013278069
33 0.0002535885 0.9986721931 0.0013278069 0.9989257817 0.0010742183
34 0.0002051574 0.9989257817 0.0010742183 0.9991309391 0.0008690609
35 0.0001659759 0.9991309391 0.0008690609 0.9992969150 0.0007030850
36 0.0001342773 0.9992969150 0.0007030850 0.9994311923 0.0005688077
37 0.0001086326 0.9994311923 0.0005688077 0.9995398249 0.0004601751
38 0.0000878856 0.9995398249 0.0004601751 0.9996277105 0.0003722895
39 0.0000711010 0.9996277105 0.0003722895 0.9996988115 0.0003011885
40 0.0000575219 0.9996988115 0.0003011885 0.9997563334 0.0002436666
41 0.0000465362 0.9997563334 0.0002436666 0.9998028696 0.0001971304
42 0.0000376486 0.9998028696 0.0001971304 0.9998405181 0.0001594819
43 0.0000304583 0.9998405181 0.0001594819 0.9998709764 0.0001290236
44 0.0000246413 0.9998709764 0.0001290236 0.9998956178 0.0001043822
45 0.0000199352 0.9998956178 0.0001043822 0.9999155530 0.0000844470
46 0.0000161279 0.9999155530 0.0000844470 0.9999316809 0.0000683191
47 0.0000130478 0.9999316809 0.0000683191 0.9999447287 0.0000552713
48 0.0000105559 0.9999447287 0.0000552713 0.9999552846 0.0000447154
49 0.0000085399 0.9999552846 0.0000447154 0.9999638245 0.0000361755
50 0.0000069089 0.9999638245 0.0000361755 0.9999707334 0.0000292666
51 0.0000055894 0.9999707334 0.0000292666 0.9999763228 0.0000236772
52 0.0000045219 0.9999763228 0.0000236772 0.9999808448 0.0000191552
53 0.0000036583 0.9999808448 0.0000191552 0.9999845031 0.0000154969
54 0.0000029596 0.9999845031 0.0000154969 0.9999874627 0.0000125373
55 0.0000023944 0.9999874627 0.0000125373 0.9999898571 0.0000101429
56 0.0000019371 0.9999898571 0.0000101429 0.9999917942 0.0000082058
57 0.0000015672 0.9999917942 0.0000082058 0.9999933614 0.0000066386
58 0.0000012679 0.9999933614 0.0000066386 0.9999946293 0.0000053707
59 0.0000010257 0.9999946293 0.0000053707 0.9999956550 0.0000043450
60 0.0000008298 0.9999956550 0.0000043450 0.9999964848 0.0000035152
61 0.0000006713 0.9999964848 0.0000035152 0.9999971562 0.0000028438
62 0.0000005431 0.9999971562 0.0000028438 0.9999976993 0.0000023007
63 0.0000004394 0.9999976993 0.0000023007 0.9999981387 0.0000018613
64 0.0000003555 0.9999981387 0.0000018613 0.9999984942 0.0000015058
65 0.0000002876 0.9999984942 0.0000015058 0.9999987817 0.0000012183
66 0.0000002327 0.9999987817 0.0000012183 0.9999990144 0.0000009856
67 0.0000001882 0.9999990144 0.0000009856 0.9999992026 0.0000007974
68 0.0000001523 0.9999992026 0.0000007974 0.9999993549 0.0000006451
69 0.0000001232 0.9999993549 0.0000006451 0.9999994781 0.0000005219
70 0.0000000997 0.9999994781 0.0000005219 0.9999995778 0.0000004222
71 0.0000000806 0.9999995778 0.0000004222 0.9999996584 0.0000003416
72 0.0000000652 0.9999996584 0.0000003416 0.9999997237 0.0000002763
73 0.0000000528 0.9999997237 0.0000002763 0.9999997764 0.0000002236
74 0.0000000427 0.9999997764 0.0000002236 0.9999998191 0.0000001809
75 0.0000000345 0.9999998191 0.0000001809 0.9999998537 0.0000001463
76 0.0000000279 0.9999998537 0.0000001463 0.9999998816 0.0000001184
77 0.0000000226 0.9999998816 0.0000001184 0.9999999042 0.0000000958
78 0.0000000183 0.9999999042 0.0000000958 0.9999999225 0.0000000775
79 0.0000000148 0.9999999225 0.0000000775 0.9999999373 0.0000000627
80 0.0000000120 0.9999999373 0.0000000627 0.9999999493 0.0000000507
Game B p = 0.421535 run length 2
x prob[X=x] prob[X<x] prob[X>=x] prob[X<=x] prob[X>x]
2 0.1776917562 0.0000000000 1.0000000000 0.1776917562 0.8223082438
3 0.1027884618 0.1776917562 0.8223082438 0.2804802180 0.7195197820
4 0.1027884618 0.2804802180 0.7195197820 0.3832686798 0.6167313202
5 0.0845237995 0.3832686798 0.6167313202 0.4677924792 0.5322075208
6 0.0739583316 0.4677924792 0.5322075208 0.5417508108 0.4582491892
7 0.0633928637 0.5417508108 0.4582491892 0.6051436746 0.3948563254
8 0.0547047924 0.6051436746 0.3948563254 0.6598484670 0.3401515330
9 0.0471027293 0.6598484670 0.3401515330 0.7069511962 0.2930488038
10 0.0405866743 0.7069511962 0.2930488038 0.7475378705 0.2524621295
11 0.0349636528 0.7475378705 0.2524621295 0.7825015234 0.2174984766
12 0.0301220358 0.7825015234 0.2174984766 0.8126235591 0.1873764409
13 0.0259501939 0.8126235591 0.1873764409 0.8385737531 0.1614262469
14 0.0223563338 0.8385737531 0.1614262469 0.8609300869 0.1390699131
15 0.0192601361 0.8609300869 0.1390699131 0.8801902231 0.1198097769
16 0.0165927556 0.8801902231 0.1198097769 0.8967829787 0.1032170213
17 0.0142947824 0.8967829787 0.1032170213 0.9110777611 0.0889222389
18 0.0123150627 0.9110777611 0.0889222389 0.9233928238 0.0766071762
19 0.0106095189 0.9233928238 0.0766071762 0.9340023426 0.0659976574
20 0.0091401802 0.9340023426 0.0659976574 0.9431425228 0.0568574772
21 0.0078743338 0.9431425228 0.0568574772 0.9510168566 0.0489831434
22 0.0067837977 0.9510168566 0.0489831434 0.9578006543 0.0421993457
23 0.0058442926 0.9578006543 0.0421993457 0.9636449469 0.0363550531
24 0.0050349020 0.9636449469 0.0363550531 0.9686798489 0.0313201511
25 0.0043376058 0.9686798489 0.0313201511 0.9730174547 0.0269825453
26 0.0037368800 0.9730174547 0.0269825453 0.9767543347 0.0232456653
27 0.0032193502 0.9767543347 0.0232456653 0.9799736848 0.0200263152
28 0.0027734943 0.9799736848 0.0200263152 0.9827471792 0.0172528208
29 0.0023893862 0.9827471792 0.0172528208 0.9851365653 0.0148634347
30 0.0020584741 0.9851365653 0.0148634347 0.9871950395 0.0128049605
31 0.0017733909 0.9871950395 0.0128049605 0.9889684304 0.0110315696
32 0.0015277896 0.9889684304 0.0110315696 0.9904962200 0.0095037800
33 0.0013162022 0.9904962200 0.0095037800 0.9918124222 0.0081875778
34 0.0011339181 0.9918124222 0.0081875778 0.9929463402 0.0070536598
35 0.0009768789 0.9929463402 0.0070536598 0.9939232192 0.0060767808
36 0.0008415885 0.9939232192 0.0060767808 0.9947648077 0.0052351923
37 0.0007250348 0.9947648077 0.0052351923 0.9954898425 0.0045101575
38 0.0006246230 0.9954898425 0.0045101575 0.9961144655 0.0038855345
39 0.0005381174 0.9961144655 0.0038855345 0.9966525829 0.0033474171
40 0.0004635921 0.9966525829 0.0033474171 0.9971161750 0.0028838250
41 0.0003993881 0.9971161750 0.0028838250 0.9975155631 0.0024844369
42 0.0003440759 0.9975155631 0.0024844369 0.9978596390 0.0021403610
43 0.0002964239 0.9978596390 0.0021403610 0.9981560629 0.0018439371
44 0.0002553714 0.9981560629 0.0018439371 0.9984114344 0.0015885656
45 0.0002200044 0.9984114344 0.0015885656 0.9986314388 0.0013685612
46 0.0001895355 0.9986314388 0.0013685612 0.9988209742 0.0011790258
47 0.0001632862 0.9988209742 0.0011790258 0.9989842605 0.0010157395
48 0.0001406723 0.9989842605 0.0010157395 0.9991249328 0.0008750672
49 0.0001211902 0.9991249328 0.0008750672 0.9992461230 0.0007538770
50 0.0001044063 0.9992461230 0.0007538770 0.9993505293 0.0006494707
51 0.0000899468 0.9993505293 0.0006494707 0.9994404761 0.0005595239
52 0.0000774899 0.9994404761 0.0005595239 0.9995179660 0.0004820340
53 0.0000667581 0.9995179660 0.0004820340 0.9995847241 0.0004152759
54 0.0000575126 0.9995847241 0.0004152759 0.9996422367 0.0003577633
55 0.0000495475 0.9996422367 0.0003577633 0.9996917842 0.0003082158
56 0.0000426856 0.9996917842 0.0003082158 0.9997344698 0.0002655302
57 0.0000367739 0.9997344698 0.0002655302 0.9997712437 0.0002287563
58 0.0000316810 0.9997712437 0.0002287563 0.9998029247 0.0001970753
59 0.0000272934 0.9998029247 0.0001970753 0.9998302182 0.0001697818
60 0.0000235135 0.9998302182 0.0001697818 0.9998537317 0.0001462683
61 0.0000202571 0.9998537317 0.0001462683 0.9998739888 0.0001260112
62 0.0000174516 0.9998739888 0.0001260112 0.9998914404 0.0001085596
63 0.0000150347 0.9998914404 0.0001085596 0.9999064751 0.0000935249
64 0.0000129525 0.9999064751 0.0000935249 0.9999194276 0.0000805724
65 0.0000111587 0.9999194276 0.0000805724 0.9999305862 0.0000694138
66 0.0000096133 0.9999305862 0.0000694138 0.9999401995 0.0000598005
67 0.0000082819 0.9999401995 0.0000598005 0.9999484814 0.0000515186
68 0.0000071349 0.9999484814 0.0000515186 0.9999556164 0.0000443836
69 0.0000061468 0.9999556164 0.0000443836 0.9999617632 0.0000382368
70 0.0000052955 0.9999617632 0.0000382368 0.9999670587 0.0000329413
71 0.0000045621 0.9999670587 0.0000329413 0.9999716208 0.0000283792
72 0.0000039303 0.9999716208 0.0000283792 0.9999755511 0.0000244489
73 0.0000033860 0.9999755511 0.0000244489 0.9999789371 0.0000210629
74 0.0000029171 0.9999789371 0.0000210629 0.9999818542 0.0000181458
75 0.0000025131 0.9999818542 0.0000181458 0.9999843672 0.0000156328
76 0.0000021650 0.9999843672 0.0000156328 0.9999865322 0.0000134678
77 0.0000018652 0.9999865322 0.0000134678 0.9999883974 0.0000116026
78 0.0000016069 0.9999883974 0.0000116026 0.9999900043 0.0000099957
79 0.0000013843 0.9999900043 0.0000099957 0.9999913886 0.0000086114
80 0.0000011926 0.9999913886 0.0000086114 0.9999925812 0.0000074188
Got to plug my friend Rick Parris at Peanut Software for the excellent free Winstats program.
And for p = 0.5
the histogram.
See how trial 3 and 4 bars are equal.
That feature exists across the board for all streak lengths.
for a run of 5, trials 6,7,8,9 and 10 all have the exact same probability.
Quote: mustangsallypacomartin mentioned that the percentage chance of winning is based on the number of trials. He is absolutely correct.
So, you're saying with the information available that game A wins 2 in a row on average every 6 trials, and game B wins 2 in a row on average every 8 trials, you can't determine the percentage chance of winning 2 in a row for either game A or game B? I can.
Game A has a 33% chance of winning 2 in a row, and Game B has a 25% chance of winning 2 in a row.
I took whoever's advice it was to hurry up and ask yahoo, I'm pretty sure there's enough information to determine the percentage chance of winning 2 in a row for both game A and game B.
So, do we still have a bet mustangsally :)
P.S. A quick second point. The math pacomartin did is the percentage chance of winning in the next n trials. It's not what the average percentage chance of winning 2 in a row. I'm going to go ahead and say pacomartin's answer is the incorrect answer for the "average" percentage chance of winning 2 in a row for game A.
Really?Quote: JyBrd0403So, you're saying with the information available that game A wins 2 in a row on average every 6 trials, and game B wins 2 in a row on average every 8 trials, you can't determine the percentage chance of winning 2 in a row for either game A or game B? I can.
Game A has a 33% chance of winning 2 in a row,
In how many trials are you talking about?
2. are you taking about just 2 trials?
3
4
5
6 maybe this one?
Quote: JyBrd0403and Game B has a 25% chance of winning 2 in a row.
Really?
In how many trials?
2
3
4
5
6 maybe this one?
At least you make many people laugh.
You have all the answers for every trial 2 to 80 inclusive.
Averages and exact probability distributions are 2 different things.
Averages come from the distribution.
Make sure you ask the Yahoo members about that.
Quote: JyBrd0403Infinite trials mustangsally Infinite:) For as long as you play the game the "average" percentage chance of winning 2 in a row will remain the same. I mean, unless someones cheating. That's kinda one way to determine whether they're cheating or not.
Over infinite trials, the percentage chance of winning 2 in a row is 100% assuming it's possible to win at all (e.g. p > 0). And why did you put "average" in quotes? What are you trying to average, anyway?
FYI, the chances of a player in Game A winning the next two games are 0.25, while the chances of a player in Game B winning the next two are 0.177692. Note the distinction between winning the next two in a row, that is, two wins in two trials, as opposed to winning two in a row anywhere over infinite trials.
Is it possible for anyone, besides me, to state in percentage terms the average of winning 2 in a row in 6 trials? Because, I'm still guessing that in percentage terms that would be 33%. Just state that fact in percentage terms.
On average you will win 2 in a row every 6 trials or stated in percenatage terms __%
Sorry for being short with you, but I'm drunk and have to get up early tomorrow :)
Quote: pacomartinI taught high school math about 20 years ago, and it was the fairly smart kids who could do cubic equations in 9th grade. Sometimes if you can easily see one solution, (it's pretty obvious that x=1 makes " 8x^3 - 9x^2 + 1 " equal to zero, then you can divide by x-1).
I think a lot of the stuff we do will be lost to future generations as equation solvers and even symbolic calculus integrators will be built into telephones. It's difficult to understand what an impact Napier had in his day on computation by working out the logarithm tables.
Knowing how to factor sure helped James Martin when playing the British gameshow Countdown. If you don't know how to play the game the rules are the numbers: 25, 50, 75, & 100 plus two random numbers are selected. In this incident the random numbers were 3 and 6. A final target is chosen by computer (in this 952), and the objective of the game is to get as close as the target as possible using basic arithmetic functions in 30 seconds.
The loser came up with (6+3)*100 + 50 + 75/25 = 953 missing the target by 1. He used all six numbers, but that is not a requirement. The winner got exactly 952. If you can factor, then try it yourself before you watch the video. The host was gobsmacked.
I forgot almost everything, 20 or more years back i would make like you sugested f(1) and divide by x-1.(probably, if there is not another even faster method)
But now i was able only to write on paper the longer solution, which i have put in the post. This is very basic math i know, unfortunately by not using it anymore i forgot. I was just contempt that i was able to remember the x^2-1 =(x-1)(x+1) thing and the formula for calculating the 2 solutions x2,x3= (-b -/+sqrtdelta)\2a.
PS: i don't like these types of gameshows, they somehow suggest that you must be able to do these calculations in 30 seconds. If you cannot, then you're garbage. Lets face it, to be able to do these calculations very quickly requires alot of exercise.
Quote: JyBrd0403
Is it possible for anyone, besides me, to state in percentage terms the average of winning 2 in a row in 6 trials? Because, I'm still guessing that in percentage terms that would be 33%. Just state that fact in percentage terms.
Average does not mean 100%. So in 6 trials the percentage chance to win 2 in a row is about 67%.
2/6 = 0.333(3) but has nothing to do with your original question.
Probability of A winning 2 in a row on average every 6 trials = 2/6=1/3
33.33%
Probability of B winning 2 in a row on average every 8 trials = 2/8=1/4
25 %
It couldn't get any easier. Those are the correct answers in case you were wondering.
Quote: JyBrd0403Haven't a clue what you're talking about guy. This is for a 50/50 game of chance, so I don't now what p > 0 means. The percentage chance of winning 2 in a row once in infinity would be extremely close to 100%, I agree. So what? I've been asking what the "average" chance of winning 2 in a row would be (stated in percentage terms) for a 50/50 game of chance. And, before we go on to the how many trials thing again, it's infinite trials. On average it's once every 6 trials, or stated in percentage terms, 33% period.
Is it possible for anyone, besides me, to state in percentage terms the average of winning 2 in a row in 6 trials? Because, I'm still guessing that in percentage terms that would be 33%. Just state that fact in percentage terms.
On average you will win 2 in a row every 6 trials or stated in percenatage terms __%
Sorry for being short with you, but I'm drunk and have to get up early tomorrow :)
Drunk or not, your question has no meaning.
If you're playing a 50/50 game, the probability of winning 2 games in a row over 6 games is 67.1875%.
If you're playing a 50/50 game, the probability of winning 2 games in a row over 2 games is 25%.
If you're playing a 50/50 game, the probability of winning 2 games in a row over a million games is very close to 100%.
If you're playing a 50/50 game, and you consider all possible number of games before you see the 2 wins in a row, the mean (average) number of games you will play before you see those 2 wins in a row is 6 games. Sometimes you win 2 games in a row right away, sometimes you don't win 2 in a row until the 100th and 101st games. The weighted average of all of those possibilities is 6 games.
But dividing 2 by 6 is meaningless here. It doesn't represent any relevant concept. It would be like saying "the average total of two dice is 7", which it is, but then rolling the dice, getting a 4, and saying "that's 4/7 or 57.14%". That's meaningless. It's just a 4.
In other words, stating things in percentage terms is improper for the question you're asking.
I thought I clarified earlier that this is for infinite trials, and here we are back with these 6 trials 2 trials stuff again.
So, with your crazy math what is the chances of winning 2 in a row once in infinity. Answer 100%
Hey, I got a question while we're waiting for infinity, what's the average number of trials it takes to win 2 in a row on a 50/50 game. Answer is guess what 33%. That's some pretty damn helpful information. Don't you think mustangsally?
As far as the dice thing, I think you might need a drink, I don't have a clue what you're talking about.
See as far as useless information is concerned, I would have to put the 6 trials thing and 2 trials thing pretty high up there, no offense intended to anyone. See the math for that is saying that in 6 trials 67.7% of the time a 2 in a row will occur. (I won't get into this with you guys, but that's wrong too) So, 67.7% of the time a 2 in a row occurs, that means 33.3% of the time it goes past 6 trials before you see a win of 2 in a row. That's very unhelpful for anyone who's let's say gambling or trying to use this information in any sort of productive way. Why, because in infinite trials, the answer is 100%. I dare say, after just a 1000 trials I can pretty much give you a 99.99% chance that you would have had at least 1 2 in a row by then. It's useless info. Well, I shouldn't say useless, but it's useless for my purposes. What's more important is the average number of trials.
If, I go 18 trials without winning 2 in a row then I win 2 in a row three times back to back and it averages once every 6 trials. That's a damn good bit of information to have. At least I think so.
Quote: JyBrd0403Hey, I got a question while we're waiting for infinity, what's the average number of trials it takes to win 2 in a row on a 50/50 game. Answer is guess what 33%.
33% is not "a number of trials". The answer is 6 trials. If you want to put that in percentage form, it's 600%. But that's not the same as 33%.
You're not suggesting that the "average number of trials it takes to win 2 in a row" is 1/3 of a trial, are you? That's what 33.33...% equals.
Quote: MathExtremist33% is not "a number of trials". The answer is 6 trials.
No, you're correct it is 6 trials, or 33% of the time. Finally we agree on something.
I should have said while we're waiting for infinity, what was the average number of times a 2 in a row occurred, the answer would be 33% of the time.
If you're playing a 50/50 game, and you consider all possible number of games before you see the 2 wins in a row, the mean (average) number of games you will play before you see those 2 wins in a row is 6 games. Sometimes you win 2 games in a row right away, sometimes you don't win 2 in a row until the 100th and 101st games. The weighted average of all of those possibilities is 6 games. -
I apologize if this was in your original post. I might have stopped reading after seeing the 6 trials and 2 trials thing. Yes, in case you're wondering, I'm drunk again tonight LOL
Quote: JyBrd0403I should have said while we're waiting for infinity, what was the average number of times a 2 in a row occurred, the answer would be 33% of the time.
That's a different question. You would, for example, first need to define whether 3 in a row counts as just one instance of 2 in a row, or whether it counts as two instances of 2 in a row, or whether it doesn't count at all. For example, how many times does 2 wins in a row occur in the following sequence:
W W L W W W L L L W W W W
1, 3, and 6 are all plausible answers, and which you pick will make a big difference.
p is a probability of event happening (normally it is the probability of player losing, but can be 50% for coin toss, or of a player winning)
P is the probability of a streak of length k (player losses by default) out of j trials
E is the expected number of trials, or an average number of trials for a streak of length k.
P(j,k)=0 if j<k
E(1)=1/p
P(j,k)=P(j-1,k)+(1-P(j-1-k,k) )*(1-p)*p^k for j=k....
E(k) = ( 1 + E(k-1) )/(1-p) for k=2....
Exactly.Quote: MathExtremistThat's a different question. You would, for example, first need to define whether 3 in a row counts as just one instance of 2 in a row, or whether it counts as two instances of 2 in a row, or whether it doesn't count at all. For example, how many times does 2 wins in a row occur in the following sequence:
W W L W W W L L L W W W W
1, 3, and 6 are all plausible answers, and which you pick will make a big difference.
It all depends on the counting method used and the type of statistical problem that one is trying to solve.
And the math for each is different. (most are concerned with N, G and E, but that does not mean one is better than the other.)
counting method/ frequency
N(n,k) 4
M(n,k) 6
G(n,k) 3
E(n,k) 1
E under counts the runs of length 2, at first glance if someone that parlayed one time after a win did win 3 parlays of length 2.
But E still says only 1.
It is also how one interprets the data.
Many trip up right there and are forever lost.
from:
K. S. Kotwal · R. L. Shinde
Joint distributions of runs in a sequence
of higher-order two-state Markov trials
Received: 8 June 2004 / Revised: 11 April 2005 / Published online: 21 July 2006
© The Institute of Statistical Mathematics, Tokyo 2006
The pdf can be found online
Quote: mustangsallyExactly.
It all depends on the counting method used and the type of statistical problem that one is trying to solve.
And the math for each is different. (most are concerned with N, G and E, but that does not mean one is better than the other.)
counting method/ frequency
N(n,k) 4
M(n,k) 6
G(n,k) 3
E(n,k) 1
E under counts the runs of length 2, at first glance if someone that parlayed one time after a win did win 3 parlays of length 2.
But E still says only 1.
It is also how one interprets the data.
Many trip up right there and are forever lost.
I understand all this, believe it or not, but, I don't understand why it's difficult to simply state that on average you'll win 2 in a row 33% of the time.
If I had said, on average you win 2 in a row 25% of the time, everyone would have said that's absolutely correct. Right? We wouldn't be having this conversation of what do you mean by 2 in a row, and it depends how you count it etc.
But, for some reason it's real difficult for you to state that if you win 2 in a row once every 6 trials, that this means that on average you would win 2 in a row 33% of the time.
I guess what I'm saying is the normal question people ask is "What is the chance of winning 2 in a row". My answer would be 33%.
Quote: JyBrd0403But, for some reason it's real difficult for you to state that if you win 2 in a row once every 6 trials, that this means that on average you would win 2 in a row 33% of the time.
You keep saying this. It's still wrong. For starters, 1 out of 6 is 16.67%, not 33%. Now consider that the chances of winning 2 in a row over 2 trials is 25%, so how could the chances be lower over more trials?
Hint: learn the difference between expected wait time and probability.
