Quick probability question

[JasonP530]

How do I calculate the odds of a prolonged losing streak? For example, if I win X% of sessions, how would I calculate the odds of losing 10/15 or something like that? Also, is using win rate to determine the probability of a losing streak just the matter of using the Standard deviation and the win rate?

Thanks in advance.

Jason

[Paul2432]

If your sessions are always the same length and you win x% of your sessions, then the probability of losing Y consecutive sessions is ((100-x)/100)^Y.

Win rate and SD together can be used to estimate the likelihood of different losing streaks.

Note that in both of the above, the problem is that you probably don't know X or your win rate very accurately unless you have quite a bit of data (say 400,000 hands).

Paul

[JasonP530]

Thanks for the reply. How would you calculate the odds of losing say 12 of 20 sessions with a win rate of 70% of the sessions? I think it is some type of combination, but I can figure out the exact formula.

[SumZero]

The probability of x wins and n-x loses given that each iid trial has a p% win percentage is choose(n,x)*p^x*(1-p)^(n-x).

Imagine x is 4, n is 7, and p is 2/3. p() = choose(7, 4) * (2/3)^4 * (1/3)^3 = 35 * 16/81 * 1/27 = 0.256. (the choose part is to represent the number of different ways you can mix the 4 wins and 3 loses together. There are 7! different ordering of the tries (think all the possible orderings of the letters ABCDEFG) but since 4 of them are wins and the wins are identicle you divide by 4! (think now all the possible orderings of the letters ABCWWWW) and since 3 of them are loses you need to divide by 3! (which gives all the possible orderings of WWWWLLL which is what we want)).

So in this case there is about a 25.6% chance that you will have exactly 4 wins in the 7 tries with the win percentage of 2/3.

If you want to know the odds that you win at least 4 then you just repeat the step for 5, 6, and 7 and add them all up. (Or, if the numbers were less symetric, like if you want to know the chance you win at least 2 tries, you could calculate 1 - the chance you didn't do it (so in the at least 2 chances one it would be 1 - (p(0)+p(1)) which is quicker to calculate then p(2)+p(3)+...+p(n)).