What percentage of games does a team have to win for there to be at least a 20% chance the team doesn't lose two games in a row over the course of the season. For the sake of the problem, assume that the team has a 90% chance of winning each game and that there are 50 games in a season.

I am new to probability and am trying to learn it this summer before I take a class that requires knowledge of it. I saw this problem in my textbook and I am not really sure how to approach it so any help would be appreciated!

  • $\begingroup$ If I'm not mistaken, the team having a $90\%$ chance of winning each game is irrelevant to the problem. $\endgroup$ – Shuri2060 Jul 23 '17 at 22:32
  • $\begingroup$ I feel like it is relevant but I could be wrong. Why do you think the chance of winning is irrelevant @Shuri2060? $\endgroup$ – Chris Deering Jul 23 '17 at 22:35
  • $\begingroup$ The question asks what percentage of games a team has to win for .... ie. you choose how many games they win. Whether they have a $1\%$ or $99\%$ chance of winning doesn't make a difference. It's a bit like asking 'How many $6$s do I need from a double dice roll in order to have a chance of getting a sum of $11$?'. The answer to that is you need to roll at least one $6$, regardless of how the dice may be biased. $\endgroup$ – Shuri2060 Jul 23 '17 at 22:37
  • $\begingroup$ Oh I see. That makes sense. But even though you know you need to roll a six, doesn't that require knowing the probability that a six is rolled? $\endgroup$ – Chris Deering Jul 23 '17 at 22:42
  • $\begingroup$ Once you decide the number of games the team wins and loses, the chances of games being lost twice in a row is up to how the wins and losses are ordered, and not the probabilities of wins and losses themselves. $\endgroup$ – Shuri2060 Jul 23 '17 at 22:44

if you lose $n_l$ games your season could be represented by a string of 50 characters consisting of $n_l$ "L"s and $(50-n_l)$ "W"s

out of the 50 positions choose $n_l$ of them to be "l"s and fill the rest with "W"s

There are $\binom{50}{n_l}$ equally probable strings

To count the number of these strings in which the "l"s are all isolated imagine that you begin forming your string with $(50 - n_l)$ "W"s and there are now $(51-n_l)$ places to position the $n_l$ "L"s

There are $\binom{51-n_l}{n_l}$ equally probable strings of this type. Note that $\binom nk =0$ whenever $k>n$ so there are zero strings of this type if $n_l>25$

We can now write the probability that a team goes through a season without back to back losses as a function of $n_l$

$$ P(n_l) = \dfrac{\binom{51-n_l}{n_l} }{\binom{50}{n_l} } $$

plugging in some values for $n_l$ ...

$$P(5)=0.65 : P(8) =0.27: P(9) = 0.18$$

So you must lose fewer than 9 games which means winning at least 42 games (84% of the games) will give you better than 20% chance of having no back to back losses.

  • $\begingroup$ Thank you so much! Follow up question, if you begin formatting your string with (50 - $n_l$) Ws, why are there (51 - $n_l$) places to position the Ls? $\endgroup$ – Chris Deering Jul 24 '17 at 14:12
  • $\begingroup$ The Ls must be inserted either between two Ws or at the beginning or end of the string. E.g 4 wins and 2 losses "1W2W3W4W5" - there are $\binom 52$ ways to replace two of the 5 numbers with Ls $\endgroup$ – WW1 Jul 24 '17 at 16:52

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