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# why cant i figure this out?

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Players A, B, C, D Player B has a 40% chance of beating Player A Player C has a 40% chance of beating Player A Player D has a 40% chance of beating Player A What is the chance that Player A will win? W = % A will win L = % A will lose -> imples B, C xor D won (only one can win) W = 1 - L I would think it would model like this... L = Pr{B wins xor C wins xor D wins} But I can find no rule in prob. theory that utilizes exclusivity with OR, so: L ?= Pr{B wins or C wins or D wins} L = Pr{B wins} plus Pr{C wins} plus Pr{D wins} - Pr{B wins and C wins and D wins} L = Pr{B wins} plus Pr{C wins} plus Pr{D wins} - 0 (mutex) L = .4 + .4 + .4 L ?= 1.2 W ?= 1 - L W ?= 1 - 1.2 W ?= -.2 Not possible...

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Seeing as they all have a 40% chance of beating player A, that's three
So 3 X 40 = 120/3 = 40% average chance

Player A against Player B ( 100% - 40% = 60% chance )

Since 60% chance is a constant for the same three
So 3 X 60 = 180/3 = 60% average chance.

The players against A have a total 40% chance of beating player A and Player A has a 60% chance of beating them all down.

But I'm not entirely sure if this is what you're asking for. :)

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Quote:
 Original post by plywoodPlayers A, B, C, DPlayer B has a 40% chance of beating Player APlayer C has a 40% chance of beating Player APlayer D has a 40% chance of beating Player AWhat is the chance that Player A will win?

What is the chance that Player A will win... what? I mean, what 'game' is this?

Are you asking what the chance is that Player A will win if he plays against Player B, Player C, and Player D - meaning he plays all three other players and beats each one?

Or are you asking what the chance is that Player A will win if he plays against only one of the other players, randomly, and you do not know which one?

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Looking at your calculation, it seems you want A to fight B, C and D in succession. The probability you want is the probability that he wins all three fights.

That's simple. He has 0.6 chance to win each fight. If he wins, he has .6 chance to win the next fight, etc. The chance of winning all three fights is therefore .6^3 = 0.216.

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Hnefi,

Yes you are correct, A fights B, C and D in succession; and it is his probability of winning all three that I am looking for.

Although I don't doubt your method, I am having a difficult time relating your model to the two main theorems from probability theory, namely:

Pr{X and Y} = Pr{X}*Pr{Y|X}
Pr{X or Y} = Pr{X} plus Pr{Y} - Pr{X and Y}

In school I was taught that these two formulas should suffice to solve most any conditional events in probability.

In other words, on my own device I never would have thought to use your method of setting W = (.6)^3, because I am trying to model the problem using conventional theorems. Can you (or someone else) elaborate how I might use these theorems to model the problem and arrive at Hnefi's same 21.6% answer?

thx!

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Consider this theorem:
Pr{X and Y} = Pr{X}*Pr{Y|X}

What it says is that the probability of event X and event Y occurring is the probability of X multiplied by the conditional probability of Y, given X.

The probabilities here are that A beats B, C and D, in succession. So we have three events. Looking at just the first one, we find that Pr{A beats B} is 0.6.

The probability of Pr{A beats C} is not dependent on B, so Pr{A beats C} = Pr{A beats C | A beats B}. By extension, Pr{A wins} = Pr{A beats B} * Pr{A beats C} * Pr{A beats D} = 0.6 * 0.6 * 0.6 = 0.6^3

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Is it me, or are you guys doing it far too complicated? If only one can win, and player A has 60% chance of beating each respective other player, then wouldn't the calculation be this:

0.6 · 0.6 · 0.6 = 0.6³
= 0.216

Player A needs to beat each player in order to win.

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Quote:
 Original post by c4c0d3m0nIs it me, or are you guys doing it far too complicated? If only one can win, and player A has 60% chance of beating each respective other player, then wouldn't the calculation be this:0.6 · 0.6 · 0.6 = 0.6³ = 0.216Player A needs to beat each player in order to win.

Well, yes. That's what the three posts prior to yours was all about. Plywood just wanted a more thorough explanation of how that calculation is connected to what he's been taught about probability theory.

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Quote:
Original post by Hnefi
Quote:
 Original post by c4c0d3m0nIs it me, or are you guys doing it far too complicated? If only one can win, and player A has 60% chance of beating each respective other player, then wouldn't the calculation be this:0.6 · 0.6 · 0.6 = 0.6³ = 0.216Player A needs to beat each player in order to win.

Well, yes. That's what the three posts prior to yours was all about. Plywood just wanted a more thorough explanation of how that calculation is connected to what he's been taught about probability theory.

Oh noez, my bad. It was getting very late, and I must have totally misread half those posts :( Sorry!

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One way to explain it is with dice.

You have a 1 in 6 chance of trowing a six, since there are six sides therefor six combinations.
If you throw two dices the chance that both are sixes are 1 in 36 (6 * 6) since there are 36 diffrent combinations.

and thats how it's connected.

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