Math Question...

Question

Let&#x27;s say that I have two variables for the outcome of one game.&#xD;&#xA;When variable one occurs for team A, then team A wins 56% of the games.&#xD;&#xA;When variable two occurs for team A, then team A wins 58% of the games.&#xD;&#xA;What is the probability of a win for team A, when both variables occur for team A? What is the formula? What if there are more than two variables?

DuckofDeath · Answer

Let&#x27;s say that I have two variables for the outcome of one game.&#xD;&#xA;When variable one occurs for team A, then team A wins 56% of the games.&#xD;&#xA;When variable two occurs for team A, then team A wins 58% of the games.&#xD;&#xA;What is the probability of a win for team A, when both variables occur for team A? What is the formula? What if there are more than two variables?

michael550 · Answer

I believe your answer is somewhere around 63% winner when both variables occur. double check my math please

DuckofDeath · Answer

I do not know if you are correct, thats why I asked the question.

Is the formula that you used, (.56)^2 + (.58)^2?

If you did this for 3 variables would the formula be...

(X)^3+(Y)^3+(Z)^3

Firebird27 · Answer

Using your formula z=x^2 &#x2B; y^2 where x=.56 and y=.58, then z=.65

michael550 · Answer

.64823 to be exact.

metslider · Answer

Doesn&#x27;t make sense because if you had a situation where the probs were 80% if the first happened and 90% if the second happened then using that formula would result in a number greater than 100%.I think we may need to know the variables to better understand.  Are they in game variables?  Does the probability of one occurring effect the probability of the other.  Is this something like Team A wins 56% of its home games and 58% of its night games?

Jokeheads · Answer

The way to do this is to think of it like a venn diagram. If the universe of games is n=100, then variable one is v1=56. variable two is v2=58. You need to figure out the probability of v1 & v2 occuring (the middle area of the venn diagram). If you call that area v3, then the probability of win for team A is v1 + v2 - v3.

(This is the case because there is an overlap of v1 and v2 of probability v3. v1 + v2 therefore includes the v3 area counted twice. Subtracting out v3 mitigates that problem)

In your question, you ask what v3 is. That, my friend, you cannot calculate with the available information.

However...
IF (and only if) both events are independent, you can figure this out:
P(win) = 1 - P(lose)
P(lose) = (1-.56)*(1-.58) = .44 * .42 = .1848
P(win) = .8152

81.52% (but only if they are completely independent variables)

The_MathMan · Answer

Sorry but i have some dubt:&#xD;&#xA;i belive (sorry for my errors, but I&#x27;m a mathman, not StatMan  ) that your formula are for two indipendent events (eg: @ the roulette the red/black or odd/even ) so your 81 % is about least one event happens.&#xD;&#xA;But DuckofDeath asked about the same event.&#xD;&#xA;So, if an reliable statistical analysis say that the next roulette number will be red @ 56% and another (different) reliable statistical analisys say that the next roulette number will be red @ 58%, What is the right probability  about this event?&#xD;&#xA;The final formula must to be that if the two variables are 50% and 50% the final probability must be 50%!!!

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