# Probability that Warriors won't lose consecutive games in season?

I'm trying to determine the probability that the prediction is true, that Warriors won't lose consecutive games during an 82 game season. Assuming that Warriors have an 80% chance of winning every game I was trying to get an approximation using binomial probability (nCr*p^x(1-p)^n-x) but I don't know how to model consecutive lose trials and I'm not sure if this is the best approach. Any help would be appreciated, thanks!

• I think you mean "an 80% chance of winning each game" rather than "every game". You also have to assume that the outcome of each game is independent of the others. – Gordon Smyth Jul 8 '17 at 22:16
• I assume you are trying to answer the 2017 NBA Hackathon application questions. I wrote up a detailed answer to the applications questions (including this one) in this blog post. In short, @kbiolsi's answer is correct, and you can also run a MC simulation to approximate the answer. – Dan Salo Sep 5 '17 at 21:10

A couple of different formulas (one recursive, one not) that will allow you to compute the "chance of getting a run of K or more successes in a row in N Bernoulli trials" can be found at the following link: http://www.askamathematician.com/2010/07/q-whats-the-chance-of-getting-a-run-of-k-successes-in-n-bernoulli-trials-why-use-approximations-when-the-exact-answer-is-known/

The probability that the Warriors will not lose two or more games in a row during an 82-game season given a probability of 0.8 of winning any given game and assuming games are independent of one another is 0.05882.

Here are the probabilities for different numbers of games played:

no. games  probability
2          0.96000
3          0.92800
4          0.89600
5          0.86528
10         0.72666
20         0.51250
30         0.36145
40         0.25492
41         0.24617
50         0.17979
60         0.12680
70         0.08943
80         0.06307
82         0.05882