FIBS Board backgammon forum

A player  will play n matches and it is known that she win in match i with probability p_win
What is the probability that she will have a winning streak of k matches?

It is 1-P[k,n], where P[k,n] is the probability that she hasn't a streak of k wins in the first n matches.
P[k,n] = 1 for n < k and P[k,k] = 1 - p_win[1]*p_win[2]*...*p_win[k]; k,n >= 1

For n >= k, if she hasn't a streak of k wins in the first n matches she will not have one after first (n+1) matches with the following exception:
she hasn't a streak of k wins in the first (n-k) matches and the last k matches are a loss followed by (k-1) wins and she win in match (n+1).
It results that P[k,n+1] = P[k,n] - P[k,n-k]*(1-p_win[n-k+1])*p_win[n-k+2]*...p_win[n+1]

If p_win = p_win for any i >= 1 then:
P[k,n] = 1 for n < k and P[k,k] = 1 - pow(p_win,k); k,n >= 1
P[k,n+1] = P[k,n] - P[k,n-k]*(1-p_win)*pow(p_win,k)

I attached the file wins_in_a_row.html (with extension changed) which calculate these probabilities using javascript.

keep cool it's only a game

Quote from: sarah on February 25, 2007, 07:33:50 PM
keep cool it's only a game

trying to predict a winning streak on fibs is like to predict how many times sarah will take her false teeth out during an evening

cut it out lews

   Very funny, lewssssss! 

Seriously, Sarah, I was thinking the same thing, sort of.  Like, WOW, is this person obsessing or WHAT!?   

socksey



The average dog is a nicer person than the average person. - Andy Rooney

hey, i liked it and may even use it ... for math nerds, recovered or not, such ramblings are as much fun as the game ... more power to you sobolanul and keep your teeth in tonight sarah ... 

i have no idea what all that means however i feel that if i could understand it it would help me a lot so thank you for posting if i aquire some more intelect i will read the article again. if i learn how to read properly :

  (But down with the av  )

i have made my own mathmatical formulae on winning and losing streaks my computations are as follows :-

if p<v minus the variation of the anomaly of x and x being a factor of three
therefore if x+y is a constant opposable figment of my imagination then x - y to the power of 4.

this will work out the paralax error of and differentials in x> (iejn-jkk) x ??

therefore i conclude that if you do the above then you win win all your games.

please reply if you feel you need greater clarity on any of the above.

yours sincerly

"toon"......

I thought a winning streak was obtained by sending a hundred bucks to Patti 

The odds of winning streaks actually conform to the Binomial Probability Distribution, which is not that hard to compute--particularly in the case of two evenly matched players--all you need to do is count in binary (0, 1, 10, 11, 100, 101, 110, 111 ...).

Substitute 'H' and 'T' for '0' and '1' and you can compute various length trials of flipping a coin and getting Heads or Tails.  Substitute 'W' and 'L' for Win/Lose in the case of two players with equal playing skills and you get:

   1 match in-a-row (trivial)
     possible outcomes:
             W <
             L
     odds:   1/2

   2 matches in-a-row
     possible outcomes:
             WW <
             WL
             LW
             LL
     odds:   1/4

   3 matches in-a-row
     possible outcomes:
             WWW <
             WWL
             WLW
             WLL
             LWW
             LWL
             LLW
             LLL
     odds:   1/8

Yes, there's a pattern developing here.  If the odds are even for each player to win a single match, the odds of a winning streak in n-matches are 1/2^n or 1/16 for n=4, 1/32 for n= 5 ....

For most practical purposes, you can stop here.  This gives a pretty good idea of the trends of winning streaks.  The actual formula for the number of wins (x)  in n matches, (probability of win in single match=p) is:

     P(X=x) = [n,x] * p^x * (1-p)^(n-x) .

[n,x] is the combinatorial or (n! / (x! * (n-x)! ).  Where n=x, this always works out to 1.  Since n-x=0, (1-p)^0 is also 1, and:

     P(X=n) = p^n ,

so here's a chart of streak odds for length=n:

       n =      1        2        3        4        5        6
     p(0.1)  0.10000  0.01000  0.00100  0.00010  0.00001  0.00000 
     p(0.2)  0.20000  0.04000  0.00800  0.00160  0.00032  0.00006 
     p(0.3)  0.30000  0.09000  0.02700  0.00810  0.00243  0.00073 
     p(0.4)  0.40000  0.16000  0.06400  0.02560  0.01024  0.00410 
     p(0.5)  0.50000  0.25000  0.12500  0.06250  0.03125  0.01563 
     p(0.6)  0.60000  0.36000  0.21600  0.12960  0.07776  0.04666 
     p(0.7)  0.70000  0.49000  0.34300  0.24010  0.16807  0.11765 
     p(0. )  0.80000  0.64000  0.51200  0.40960  0.32768  0.26214 
     p(0.9)  0.90000  0.81000  0.72900  0.65610  0.59049  0.53144 
          (for some reason "p(0. produces a smiley"

The only thing missing is a determination of p().  The FIBS ratings formula is as good a method as any, and I'd be happy to explain it here if anyone wants.  It's discussed extensively elsewhere.  (Start with "help formula" on FIBS.)

--
don

Yes don, please explain

What did he say?  What language is that?   

socksey



Thought for the day: Men are like fine wine. They start out as grapes; and it's up to the women to stomp the crap out of them until they turn into something acceptable to have dinner with. - Dave Barry

But here is a harder one for math freaks. 

Let's asume that some one has fibs rating 1600 and that is his/her absolutely correct rating. Then he starts playing. On avarage how many games it takes that he/she hits a big lucky streak (but not necessary winning every game) and gets (let's say) rating 1700 just by luck?

All my opponents say I win with luck, but on the other hand, I always say I win with skill and cunning.  

Formulate that!  

socksey



'Writing. I love the swirl & swing of words as they tangle with human emotions.' - James Michener

Quote from: socksey on October 30, 2007, 05:10:31 AM
What did he say?  What language is that?   

socksey

Don't worry about it, just enjoy the bit that makes most sense:

Quote(for some reason "p(0. produces a smiley"