Suppose a football team plays 8 games. The chance of winning any particular game is 50%. Determine the probability of completing the season without ever having more losses than wins during the season
There’s a 50% chance the team won’t dip under .500 after either ofthe first two games – all depends on whether game #1 is won. At this point, 50% survival according to your rule. If 1-1 after two games, 50% chance it will happen in games 3 or 4. If 2-0 after two games, 0% chance it will happen in games 3 or 4. 1/4 of remaining teams drop out, so 3/8s survival. If 2-2 after four games, 50% chance it will happen in 5 or 6. If not 2-2 after four games, 0% chance it will happen in 5 or 6. 1/8 of remaining teams drop out, so 21/64 survival If 3-3 after 6 games, 50% chance it will happen in 7 or 8. If not, 0% chance it will happen. 1/16 of remaining teams drop out, so 315/1024 survival 30.7617% chance.
This is the solution: You can model it as the number of paths on a grid of lattice points from the origin to (m,n) such that you never cross but may touch the line y = x. In the case when m = n we have what are known as Catalan numbers. If you are interested in seeing the proof of the formulae used below you should write again. In the work that follows we are referring to ACCEPTABLE paths, that is, paths in which at all stages y > x, though we can accept y = x. The required number of acceptable paths is given by: (n-m+1) -------.C(m+n,m) n+1 Using this model with 8 football matches, we have m+n = 8 and there is equal probability at each stage (or game) that either m or n will increase by 1. Number of routes from the origin to (0,8) = (9/9).C(9,0) = 1 " " " (1,7) = (7/8).C(8,1) = 7 " " " (2,6) = (5/7).C(8,2) = 20 " " " (3,5) = (3/6).C(8,3) = 28 " " " (4,4) = (1/5).C(8,4) = 14 ------------ Total = 70 The total number of possible sequences of Win/Lose is 2^8 = 256 and so the required probability = 70/256 = 35/128 (=0.27344) What I cannot understand is why you would have to take a path dependent approach to this porblem.
I’m doing this quick and dirty while I eat lunch, but I don’t check those numbers. 256 unique possible sets of win/loss combos. 1 way to have 8 wins 8 ways to have 7 wins 28 ways to have 6 wins (8C2) 56 ways to have 5 wins (8C3) Prob should be 93/256 or 36.3%
Super I Wrote: ------------------------------------------------------- > I’m doing this quick and dirty while I eat lunch, > but I don’t check those numbers. > > 256 unique possible sets of win/loss combos. > > 1 way to have 8 wins > 8 ways to have 7 wins > 28 ways to have 6 wins (8C2) > 56 ways to have 5 wins (8C3) > > Prob should be 93/256 or 36.3% This was exactly my approach but apparently its wrong. The solution follows a path dependent appraoch which I am trying to understand.
Blastt Wrote: ------------------------------------------------------- > Super I Wrote: > -------------------------------------------------- > ----- > > I’m doing this quick and dirty while I eat > lunch, > > but I don’t check those numbers. > > > > 256 unique possible sets of win/loss combos. > > > > 1 way to have 8 wins > > 8 ways to have 7 wins > > 28 ways to have 6 wins (8C2) > > 56 ways to have 5 wins (8C3) > > > > Prob should be 93/256 or 36.3% > > This was exactly my approach but apparently its > wrong. The solution follows a path dependent > appraoch which I am trying to understand. I think the path dependant approach is wrong. Start with your numbers. 256 possible solutions You came out with 70 possible loss is not greater than win. Subtract out ties (14 per your approach). That leaves 56 where loss is not greater than or equal to win. and 56 where win is not greater than or equal to loss. Add back the ties and you are woefully short of the 256 possible outcomes. I think the path dependant approach (and I remember nothing about it… just guessing) for a case where you may based on performance drop out and not play all 8 games if something happens. Or like modeling outcomes where you have a company that has a certain prob of going bankrupt each year.
Super I Wrote: ------------------------------------------------------- > > > I’m doing this quick and dirty while I eat > > lunch, > > > but I don’t check those numbers. > > > > > > 256 unique possible sets of win/loss combos. > > > > > > 1 way to have 8 wins > > > 8 ways to have 7 wins > > > 28 ways to have 6 wins (8C2) > > > 56 ways to have 5 wins (8C3) > > > > > > Prob should be 93/256 or 36.3% > > > > This was exactly my approach but apparently its > > wrong. The solution follows a path dependent > > appraoch which I am trying to understand. > > > > I think the path dependant approach is wrong. > > Start with your numbers. > > 256 possible solutions > You came out with 70 possible loss is not greater > than win. > > Subtract out ties (14 per your approach). > > That leaves 56 where loss is not greater than or > equal to win. > > and 56 where win is not greater than or equal to > loss. Add back the ties and you are woefully > short of the 256 possible outcomes. > > I think the path dependant approach (and I > remember nothing about it… just guessing) for a > case where you may based on performance drop out > and not play all 8 games if something happens. Or > like modeling outcomes where you have a company > that has a certain prob of going bankrupt each > year. Path-counting approach is correct and your solution is wrong. In fact, you are doing path-counting too by using the combination formula (8C2 etc). But you neglected the condition that at any one stage, the total wins >= total losses. If you think of the win-loss process as a binomial tree, then this condition effectively pruned the tree. Because you did not take this pruning into account, you have over-estimated the number of possible paths. For example, there are only 7 ways to have 7 wins AND not violate the condition. 1) WWWWWWWL - ok 2) WWWWWWLW - ok 3) WWWWWLWW - ok 4) WWWWLWWW - ok 5) WWWLWWWW - ok 6) WWLWWWWW - ok 7) WLWWWWWW - ok 8) LWWWWWWW - not ok. After first game L=1, W=0. Therefore L>W and condition is violated.
Propanol thanks for your explanation. It makes perfect sence now.
sooo, what is the answer? I could be wrong but here are my thoughs: I am assuming that not having more losses than wins means that we can have 4 wins and 4 losses. So as long as we have 4 wins we should be ok! Here are possibilities: 1) WWWW 2) WWWLW 3) WWLWW 4) WWLLWW 5) WLWLWLW 6) WLWWW 7) WLWWLW 8) WLWLWW Now here where I am not sure: Each scneario has a 6.25% probability ( 0.5^4). 6.25% * 8 = 50% probablity! Not sure, Kide CFA
It would help if i really read the question… I looked at season end record, not the interim “performance drop out” concept which I mentioned but didn’t realize they were asking. Nice explanation propanol.
kide, blastt gave the correct answer in his earlier post, which is 70/256. There is no easy way to do this, not that i know of, except to ue the catalan number technique as blastt mentioned. a course in combinatorics would probably help. plus there are several flaws in your thought process. First, not all the probabilities of the 8 scenarios that you gave are 0.5^4. In fact, only scenario 1 is. For scenario 2, for example, the probability of a WWWLW is 0.5^5 if we assume that the outcome of each match is independent, which may not be true in reality because we all know streaks exist. Also, we are not really interested in the probability on the first 4, or 5, matches but the entire 8-match process. Plus, having 4 wins does not guarantee that the condition is satisfied, unless the first 4 matches are all wins, or they come in the form that satisfies the condition. Consider this 4-win situation that does not work: WWWLLLLW. Finally, I’m not sure if CFA is of any use when it comes to problems like this. This involves more math than the rudimentary stuff that CFA demands. And that probably explains why CFA is neither look upon very highly in the quantitative investment field, or at quant strategy driven hedge funds. Actually, this problem has a close analogy to the pricing of a financial instrument, an exotic option in fact. Look at the binomial tree and take a guess; it should be educational.