Probability of 20-hand win streak in 7,000 trials: 0.000185852501104.
Probability of 20-hand losing streak in 7,000 trials: 0.001529230036319.
Do you see the enormous difference?
Don
I don’t know what formula you applied. Your figures seem inaccurate.
“Probability of 20-hand win streak in 7,000 trials: 0.000185852501104.”
1 in 5380 hands.
“Probability of 20-hand losing streak in 7,000 trials: 0.001529230036319.”
1 in 654. NO way, Jose!
I got different results for W=0.43 and L=0.48
Number of 20-hand win streaks in 7,000 trials: 0
Number of 20-hand losing streaks in 7,000 trials: 0
How bout p=.5?
Number of 20-hand win streaks in 7,000 trials: 0
Number of 20-hand losing streaks in 7,000 trials: 0
How about coin tossing and shorter streaks?
Number of 5-hand win streaks in 7,000 trials: 55
Number of 5-hand losing streaks in 7,000 trials: 55
Number of 5-hand win streaks in 1,000 trials: 8
Number of 5-hand losing streaks in 1,000 trials: 8
Number of 5-hand win streaks in 100 trials: 1
Number of 5-hand losing streaks in 100 trials: 1
What results you get for p=0.5 and 5-win/loss streaks?
Hopefully a neutral can arbiter our argument.
Sorry, there's nothing to discuss. Have been using this extremely accurate streak calculator since forever.
https://sites.google.com/view/krapstuff/home
Don
Sure I am, a great probability book is on one of the shelves of your impressive library. You guessed it, I am referring to Warren Weaver’s “Lady Luck”. It has a section dedicated to streaks. The formulas are easy to understand and to program into software.
I think you overlooked one important aspect. Lets say coin toss, 5-heads in a row. The pattern must be exactly THHHHHT when you calculate how many 5-head streaks in a number of trials. HTTTTTH for 5-tail streaks. Thats why 20-hand losing streaks are so frequent in your calculations.
“Probability of 20-hand losing streak in 7,000 trials: 0.001529230036319.”
1 in 654. NO way, Jose! You wont get one such a streak in 10,000 uninterrupted hands!
I do use a different program for calculations. It is based on Warren Weaver’s book. I am not at liberty to reveal the software. It might even be forbidden to name the program here. I will publish only a report for coin tossing and 1000 trials. You can see all the streaks and how many times they occur. You can try for yourself. Toss a good coin (no counterfeit allowed!) 1000 times. Just jot down the outcomes: T, H. You’ll see the results come close to the theory. Repeating 1000-toss runs will get closer to theory.
Great book, Warren Weaver’s “Lady Luck”! No offence, just my honest opinion.Code:Probability p = .5 Number of Trials N = 1000 ============================================================================= Streak Lengths 1 2 3 4 5 6 7 8 9 10 TOTAL ============================================================================= Winning Streaks W 125 62 31 16 8 4 2 1 0 0 492 ----------------------------------------------------------------------------- Losing Streaks L 125 62 31 16 8 4 2 1 0 0 492 ============================================================================= Uncounted Elements: 16 ; Distributed as: Win = 8 / Loss = 8 (possible one streak was longer than 8 on either side of the coin.
This is a stupid argument, so I'll end it here and won't write again. To make it as simple as possible, you don't have a clue as to what you're talking about, and your numbers above are absurd and ridiculous. I gave you one site. Here's a second one with identical results to the first: http://www.beatingbonuses.com/calc_streak.htm
If you flip a fair coin 1,000 times, the probability that you'll get a streak of even 9 is 62.4% (that's either heads or tails, so half for one or the other). In other words, it would be unusual NOT to have such a streak.
Write whatever you want. I won't reply. I've been doing this stuff probably since before you were born, and I have no desire to joust with you, especially since you're spouting nonsense.
Don
Problem is, you and your “calculators” do not grasp the concept of streak. You calculate a streak ignoring the existence of other streaks of different lengths. You simply calculate “9 heads in a row”. 0.5^9=0.001953125; 1 in 512. You are just at the beginning of a run, its not a streak.
Warren Weaver’s “Lady Luck” book to the rescue! Read carefully how he establishes a streak (pg 346). For example, 5 heads in a row; THHHHHT. Observe the limits: T. You must have the opposite event surrounding the event we calculate. You have also all kinds of streaks in 1000 trials. THT, HTH, HTTTTTH, etc.
Accurate calculations of number of streaks consider the correct 9-head streak as THHHHHHHHHT. Notice, 11 elements. Naturally the streak begins with T followed by 9 consecutive H and ending with T. In this case, the probability for such configuration is 21.5%. Being under 50% it was rounded down. That’s why it appeared as 0 under streak length 9. So in around 1 in 5 runs of 1000 tosses there will be a 9-H streak. Of course, there could be a 9-T streak instead.
“…I'll end it here and won't write again…Write whatever you want. I won't reply.” Obviously. We all know pride is very hard to swallow. No offence, just my honest opinion (meant to have been my signature).
Lets consider blackjack as per Don’s calculator: p=0.48
Now there is one 9-hand losing streak.Code:Probability p = .48 Number of Trials N = 1000 ============================================================================= Streak Lengths 1 2 3 4 5 6 7 8 9 10 TOTAL ============================================================================= Winning Streaks W 130 62 30 14 7 3 2 1 0 0 475 ----------------------------------------------------------------------------- Losing Streaks L 120 62 32 17 9 5 2 1 1 0 514 =============================================================================
No doubt, the player has more single wins (LWL) than single losses (WLW). A curiosity - the number of 2-win and 2-L streaks are equal (62 in 1000). An exploit?
I’m learning about this streak calculator. As I recall, James989 showed me the calculation of the streak probability numbers here. It was just a set of linear equations to solve for the probability numbers. I summarized all possible results in one equation and posted the conclusion formula here on this site. I will read this part soon.
“ Please don't listen to toolyp. He has no idea what he's talking about.” Please don't listen to DSchles (Don). He has no idea what he's talking about. Please do listen to Warren Weaver. He was a well-respected scholar. His acclaimed book “Lady Luck” is studied at universities worldwide.
I cite from “Lady Luck” published by Dover Publications in 1982. At page 347 you’ll find a table similar to the one you saw in one of my prev posts. He calculated for 1024 trials (power of 2).
The longest streak with an occurrence is 8. Warren Weaver adds: “The remaining 20 will, on the average, be used up in runs longer than 8, which are less likely…”Code:Probability p = .5 Number of Trials N = 1000 ============================================================================= Streak Lengths 1 2 3 4 5 6 7 8 9 10 TOTAL ============================================================================= Winning Streaks W 125 62 31 16 8 4 2 1 0 0 492 ----------------------------------------------------------------------------- Losing Streaks L 125 62 31 16 8 4 2 1 0 0 492 =============================================================================
Who you gonna listen to? Old Don or older Warren Weaver? Yea know, Don said earlier that I should listen to him because he was older. “I've been doing this stuff probably since before you were born.” No offence, just my honest opinion.
I do not read very much, but this blackjackthefroum is my only textbook for knowledge, and it’s also my scientific journal to publish my discoveries. We consider the problem of rolling two dice.
The probability of a result 7 to appear is t=1/6. If we roll them n times, what is the probability of getting a 777 success? Let the probability be p(n), then
p(1) = 0,
p(2) = 0,
p(3) = t^3,
…
p(n) = t^3 + (1-t) p(n-1) + t (1-t) p(n-2) + t^2 (1-t) p(n-3) for n > 3.
Using this recursive formula, I find
p(100)=31.89%,
which agrees with one online streak calculator, but slightly off from another one. So, a streak in this calculator is 777, X777, XYZ777, XYZ7777, and etc.. It seems like you are saying this is not a streak. I guess what matters here is how you define a streak.
Last edited by aceside; 05-03-2024 at 06:55 AM.
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