TheJacob
TheJacob
Joined: Mar 13, 2010
  • Threads: 3
  • Posts: 47
July 16th, 2013 at 1:09:04 PM permalink
I'm playing on a poker site with an SNG leaderboard where your total is multiplied by your highest streak(games won in a row).

Assuming the probability of winning a game is in 1 in 6 how would I calculate the odds of winning 5 games in a row over a 5000 game sample.

Curious about what kinds of streaks I should expect and also I'd like to be able to look at the leaderboard and see if anything funny is going on.
Mission146
Mission146
Joined: May 15, 2012
  • Threads: 124
  • Posts: 12981
July 16th, 2013 at 1:41:51 PM permalink
Quote: TheJacob

I'm playing on a poker site with an SNG leaderboard where your total is multiplied by your highest streak(games won in a row).

Assuming the probability of winning a game is in 1 in 6 how would I calculate the odds of winning 5 games in a row over a 5000 game sample.

Curious about what kinds of streaks I should expect and also I'd like to be able to look at the leaderboard and see if anything funny is going on.



The first thing that I should mention is that, if there are six players per game, the probability of any individual player winning will very rarely be 1 in 6 because you have skill levels that are going to vary and everything like that. It's difficult to really quantify, but you might have a table where a certain player should win 1,000 out of 2,000 games against all other players (1 in 2) based on skill, where another player should only win one out of one hundred, or so.

For that reason, I think it would be tough to, "See if anything funny is going on," because what you could have is simply a handful of superior players who win a good bit of the time, which is what superior players do.

To the math question, if we assume such a probability, then the probability of winning five in a row is (1/6)^5 = 0.00012860082304526744 = 1/0.00012860082304526744 = 1 in 7776.

I'm going with something a bit easier than a 5,000 game sample. Let's say you have 5,000, "First games," in which you must try to win five in a row, any of the first games can begin the streak, winning more than once in a row (but not five times) is irrelevant to the 5,000 games. If the probability of winning five in a row right off the top is 0.00012860082304526744, but you have 5000 chances, then the probability of winning at least once in exactly 5,000 first games is:

(1-0.00012860082304526744)^5000 = 0.5256890166416966

1-0.5256890166416966 = 0.47431098335830335

So, with 5,000 first games, you're still only 47.431098335830335% to win five in a row once or more. The actual probability of winning for your specific query is even more against you, because those series in which you win 2-4 games in a row still count against your 5,000. To give you an idea, the probability of winning four in a row is:

(1/6)^4 = 0.0007716049382716048 = 1/0.0007716049382716048 = 1 in 1296

So, the probability of you winning four in a row at least one time in 5,000 first games is:

(1-0.0007716049382716048)^5000 = 0.02107824331658854

1-0.02107824331658854 = 97.89217566834114%

So, you'd be a near lock to do four in a row at least once in that many games.
Vultures can't be choosers.
7craps
7craps
Joined: Jan 23, 2010
  • Threads: 18
  • Posts: 1977
July 16th, 2013 at 1:57:02 PM permalink
I get
For a run of 5, #trials = 5000 and P = 1/6
expected # of trials: 9,330
expected # of runs: 0.535429527
distribution of runs
Number of trials5,000
Run Length (X or more)5
trial Probability16.666666667%
EventRun Probability1 in
0 runs of length 5 or more0.5852329961.71
at least 1 run of length 5 or more0.4147670042748592.4110
at least 2 runs of length 5 or more0.1010462977177969.90
at least 3 runs of length 5 or more0.01714463275079758.33
at least 4 runs of length 5 or more0.002218471220281450.76
at least 5 runs of length 5 or more0.0002313388914944,322.66
at least 6 runs of length 5 or more0.00002016819070549,583.03
at least 7 runs of length 5 or more0.000001508866131662,749.32
at least 8 runs of length 5 or more0.00000009877749310,123,763.67
at least 9 runs of length 5 or more0.000000005744204174,088,509.92
at least 10 runs of length 5 or more0.0000000003003113,329,880,344.80
at least 11 runs of length 5 or more0.00000000001425370,158,400,391.04


For a run of 4, #trials = 5000 and P = 1/6
expected # of trials: 1554
expected # of runs: 3.213220165

distribution of runs
Number of trials5,000
Run Length (X or more)4
trial Probability16.666666667%
EventRun Probability1 in
0 runs of length 4 or more0.03985314125.09
at least 1 run of length 4 or more0.9601468594055461.0415
at least 2 runs of length 4 or more0.8313443524428411.20
at least 3 runs of length 4 or more0.6235793325218231.60
at least 4 runs of length 4 or more0.4005589456688002.50
at least 5 runs of length 4 or more0.2213365923695644.52
at least 6 runs of length 4 or more0.1063246558866009.41
at least 7 runs of length 4 or more0.04493086086297222.26
at least 8 runs of length 4 or more0.01689136190379059.20
at least 9 runs of length 4 or more0.005706413431732175.24
at least 10 runs of length 4 or more0.001747700222375572.18
at least 11 runs of length 4 or more0.0004889940689372,045.01
at least 12 runs of length 4 or more0.0001258251672767,947.54
at least 13 runs of length 4 or more0.00002994915379333,389.93
at least 14 runs of length 4 or more0.000006627804986150,879.51
at least 15 runs of length 4 or more0.000001369859587730,001.83
at least 16 runs of length 4 or more0.0000002654820783,766,732.61
at least 17 runs of length 4 or more0.00000004841622520,654,233.21

info found here in links
another BruceZ gem
http://www.pulcinientertainment.com/info/Streak-Calculator-enter.html

added: another by hand method and alternate formula
http://wizardofvegas.com/forum/questions-and-answers/math/13096-what-are-the-odds-of-getting-22-back-to-back-winning-passline-decisions-in-1-000-000-rolls/#post221169

for a run of 5 an alternate formula method is: 0.414582216
for a run of 4 the alternate formula method is: 0.959773133
close enough

N=5,000 and P=1/6
AVG. LENGTH OF LONGEST RUN: 4.478006366
DISTRIBUTION OF LONGEST RUN
"Longest Run Length"Probability1 inor lessor more
0 0.0000000000%#DIV/0!0.0000000%100.0000000%
1 0.0000000000%529,835,250,278,882.000.0000000%~100.0000000%
2 0.0000003236%309,003,340.800.0000003%~100.0000000%
3 3.9853137358%25.093.9853141%99.9999997%
4 54.5379855131%1.8358.5232996%96.0146859%
5 32.9401504625%3.0491.4634500%41.4767004%
6 7.0608053695%14.1698.5242554%8.5365500%
7 1.2283149680%81.4199.7525704%1.4757446%
8 0.2061571842%485.0799.9587276%0.2474296%
9 0.0343939020%2,907.4999.9931215%0.0412724%
10 0.0057323147%17,444.9699.9988538%0.0068785%
11 0.0009552262%104,687.2599.9998090%0.0011462%
12 0.0001591733%628,245.8999.9999682%0.0001910%
13 0.0000265236%3,770,227.5099.9999947%0.0000318%


sim data for the longest run in 5k trials
(I had not used the program for this in a while)
       group         middle    freq  freq/100
---------------------------------------------
2.50 <= x < 3.50 3.00 3920 3.92%
3.50 <= x < 4.50 4.00 54537 54.54%
4.50 <= x < 5.50 5.00 33137 33.14%
5.50 <= x < 6.50 6.00 6941 6.94%
6.50 <= x < 7.50 7.00 1230 1.23%
7.50 <= x < 8.50 8.00 207 0.21%
8.50 <= x < 9.50 9.00 22 0.02%
9.50 <= x < 10.50 10.00 5 0.00%
10.50 <= x < 11.50 11.00 1 0.00%

---------------------------------------------
cumulative
---------------------------------------------
2.50 <= x < 3.50 3.00 3920 3.92%
3.50 <= x < 4.50 4.00 58457 58.46%
4.50 <= x < 5.50 5.00 91594 91.59%
5.50 <= x < 6.50 6.00 98535 98.54%
6.50 <= x < 7.50 7.00 99765 99.76%
7.50 <= x < 8.50 8.00 99972 99.97%
8.50 <= x < 9.50 9.00 99994 99.99%
9.50 <= x < 10.50 10.00 99999 100.00%
10.50 <= x < 11.50 11.00 100000 100.00%


Good Luck to you
winsome johnny (not Win some johnny)
mustangsally
mustangsally
Joined: Mar 29, 2011
  • Threads: 25
  • Posts: 2463
June 27th, 2018 at 8:32:10 PM permalink
Quote: TheJacob

Curious about what kinds of streaks I should expect and also I'd like to be able to look at the leaderboard and see if anything funny is going on.

interesting

using the calculator here
https://sites.google.com/view/krapstuff/home
section 1r.

I get some data
easy to call a few different streaks at once
> runs.r(4,5000,1/6)
[1] "for a run of 4, success probability: 0.9601468594"
[1] "1 in: 1.04151"
[1] "Number of trials: 5000"
> runs.r(5,5000,1/6)
[1] "for a run of 5, success probability: 0.4147670043"
[1] "1 in: 2.41099"
[1] "Number of trials: 5000"
> runs.r(6,5000,1/6)
[1] "for a run of 6, success probability: 0.08536549965"
[1] "1 in: 11.7143"
[1] "Number of trials: 5000"
> runs.r(7,5000,1/6)
[1] "for a run of 7, success probability: 0.01475744595"
[1] "1 in: 67.7624"
[1] "Number of trials: 5000"
> runs.r(8,5000,1/6)
[1] "for a run of 8, success probability: 0.002474296275"
[1] "1 in: 404.155"
[1] "Number of trials: 5000"

hope this helps out for those that come across this thread

Sally
I Heart Vi Hart
Ace2
Ace2 
Joined: Oct 2, 2017
  • Threads: 22
  • Posts: 780
June 28th, 2018 at 12:35:21 AM permalink
Probability of success or probability of failure?
Itís all about making that GTA

  • Jump to: