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 Probability Discussions of probability theory

 09-08-2017, 11:41 AM #1 DMMx69 old hand     Join Date: Jul 2006 Location: This Mortal Coil Posts: 1,678 Odds of a 15 game win streak given 50 games to do it? http://www.espn.com/mlb/story/_/id/2...eak-reaches-15 Guy pays \$75k to insure against a loss of \$1.7MM The insurance company pays out the 1.7MM if the Cleveland Indians won 15 games in a row, any time after August 1st. So, something like, they had 50 games left, and needed a 15-game win streak. If they were 60% to win each game (I know not accurate, but for sake of simplifying problem), what is the chance they have a 15-game win streak? .6^15 = .00047, the chance of them winning 15 in a row at any specific point. They had 36 opportunities to start such a streak, but I'm not sure if multiplying that by 36 is right, since the opportunities overlap? Any help would be great. Thanks in advance.
 09-08-2017, 03:13 PM #2 statmanhal Pooh-Bah   Join Date: Jan 2009 Posts: 4,018 Re: Odds of a 15 game win streak given 50 games to do it? Win Streak Calculator https://www.google.com/search?q=win+...e=utf-8http://
 09-08-2017, 03:32 PM #3 whosnext Pooh-Bah     Join Date: Mar 2009 Location: California Posts: 4,079 Re: Odds of a 15 game win streak given 50 games to do it? We typically want to know the probability that a streak of at least a certain length occurs in a given horizon. For example, if the Indians win 16 games in a row, that is also considered a 15-game winning streak. Due to the overlap issue that OP rightly points out, the derivation of the formula for this probability is a bit complex. The formula is available on the internet, of course. It essentially boils down into a polynomial in terms dealing with the win prob raised to the Kth power (where we seek K wins in a row). If I plugged in the correct terms into the formula, I get that the probability of winning at least 15 games in a row out of a total of 50 games, where the win prob of each game is independently given as 60%: = (15)*(.6)^15 - (38.4)*(.6)^30 + (2.24)*(.6)^45 = .0070443
09-08-2017, 03:47 PM   #4
statmanhal
Pooh-Bah

Join Date: Jan 2009
Posts: 4,018
Re: Odds of a 15 game win streak given 50 games to do it?

Quote:
 Originally Posted by whosnext If I plugged in the correct terms into the formula, I get that the probability of winning at least 15 games in a row out of a total of 50 games, where the win prob of each game is independently given as 60%: = (15)*(.6)^15 - (38.4)*(.6)^30 + (2.24)*(.6)^45 = .0070443
That's what a streak calculator shows.

 09-09-2017, 07:28 AM #5 Monteroy Carpal \'Tunnel   Join Date: Dec 2006 Posts: 9,130 Re: Odds of a 15 game win streak given 50 games to do it? Wonder if the person who wrote the policy gets a raise for the insane markup or fired due to the result.
 09-09-2017, 02:27 PM #6 whosnext Pooh-Bah     Join Date: Mar 2009 Location: California Posts: 4,079 Re: Odds of a 15 game win streak given 50 games to do it? I admit I have gone down the rabbit hole exploring this interesting problem. It is remarkably both simple and complex. There are multiple approaches to solving the problem, including Markov Chains. Anyway, I have fiddled around with one of the solution formulas I found on the internet. I have rearranged terms and obtained a fairly straightforward result: Let P = the prob of winning each single game. Let Q = 1 - P Let N be the total number of games in which the streak can occur. Let K be the number of games in the streak we are interested in. Let W = P^K Let Z = [N/K], the integer value of N/K. Let C(X,Y) be the number of ways Y items can be chosen out of X total items (the choose function). Then the prob of observing a winning streak of K or more games in N total games: = W[1+(N-K)*Q] + Sum [T goes from 2 to Z] (W^T)*((-1)^(T-1))*[C(N-T*K,T-1)*(Q^(T-1)) + C(N-T*K,T)*(Q^T)]
09-14-2017, 11:10 PM   #7
IOWActuary
centurion

Join Date: Jul 2014
Posts: 101
Re: Odds of a 15 game win streak given 50 games to do it?

Quote:
 Originally Posted by whosnext We typically want to know the probability that a streak of at least a certain length occurs in a given horizon. For example, if the Indians win 16 games in a row, that is also considered a 15-game winning streak. Due to the overlap issue that OP rightly points out, the derivation of the formula for this probability is a bit complex. The formula is available on the internet, of course. It essentially boils down into a polynomial in terms dealing with the win prob raised to the Kth power (where we seek K wins in a row). If I plugged in the correct terms into the formula, I get that the probability of winning at least 15 games in a row out of a total of 50 games, where the win prob of each game is independently given as 60%: = (15)*(.6)^15 - (38.4)*(.6)^30 + (2.24)*(.6)^45 = .0070443
What do the powers of 30 and 45 mean in laymens terms? Why does it work out that way?

09-14-2017, 11:51 PM   #8
whosnext
Pooh-Bah

Join Date: Mar 2009
Location: California
Posts: 4,079
Re: Odds of a 15 game win streak given 50 games to do it?

Quote:
 Originally Posted by IOWActuary What do the powers of 30 and 45 mean in laymens terms? Why does it work out that way?
I don't have a good answer, so I will link to the derivation of the exact formula.

If anybody wants to chime in here, go for it.

01-11-2018, 06:38 PM   #9
kraps2312
stranger

Join Date: Jan 2018
Posts: 1
Re: Odds of a 15 game win streak given 50 games to do it?

Quote:
 Originally Posted by whosnext It is remarkably both simple and complex. There are multiple approaches to solving the problem, including Markov Chains.
simple in Excel, been done before
here is my Excel in Google
Streak column Excel
*****
here is R code for the matrix solution

Code:
#####################################################
# runs.m(m,N,p)
#
# Probability of at least 1 run of length m or more
# in N trials with probability p of success.
# Uses Markov chain.
#####################################################

require(expm)

runs.m = function(m,N,p) {
q = 1-p
A = matrix(rep(0,(m+1)^2),nrow=m+1)
A[1,1:m] = q
A[2:(m+1),1:m] = diag(p,m,m)
A[m+1,m+1] = 1
v = c(1,rep(0,m))
(A%^%N %*% v)[m+1]
}
#example
1/runs.m(25,10000000,0.5)
print(runs.m(25,10000000,0.5),9)
> 1/runs.m(25,10000000,0.5)
[1] 7.223312
> print(runs.m(25,10000000,0.5),9)
[1] 0.138440643
and an accurate online javascript streak calculator (thanks BruceZ)
I placed on Google (many others contain error in their code)
Streak Calculator
*****
of course, for simple and some exact answers
Wolfram Alpha (will time-out for large # of trials)
streak of 15 successes in 50 trials p=6/10
Wolfram Alpha

hope this helps out for future streak-type questions

the matrix result (in R) for the OP question
Code:
runs.m = function(m,N,p) {
+   q = 1-p
+   A = matrix(rep(0,(m+1)^2),nrow=m+1)
+   A[1,1:m] = q
+   A[2:(m+1),1:m] = diag(p,m,m)
+   A[m+1,m+1] = 1
+   v = c(1,rep(0,m))
+   (A%^%N %*% v)[m+1]
+ }
> print(runs.m(15,50,0.6),9)
[1] 0.00704428576

Last edited by kraps2312; 01-11-2018 at 06:58 PM. Reason: added result from R

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