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Sorry if this is an easy question but I've only basic Math skills :)

I have 3 people. They could say Yes or No, dependent events.

So I have 8 possible scenarios:

YYY,YYN...NNN

In each scenario I have different probability.

Eg. YYN case:

Player 1: 20% says Yes, 80% says No;
Player 2: 30% says Yes, 70% says No;
Player 3: 25% says Yes, 75% says No;

Eg. NYN case:

Player 1: 20% says Yes, 80% says No;
Player 2: 40% says Yes, 60% says No;
Player 3: 45% says Yes, 55% says No;

Each case is different from the others. As you can see the players' choise depend from the previous players ones.

I'd like to have the probability of each one of the 8 possible scenarios:

P(YYY)=0.15,P(YYN)=0.2 etc.

I tried to multiply each probability but it doesn't sum up to 1:

Case YYN: 0.2*0.3*0.75 = probability of YYN case
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2 Answers 2

up vote 2 down vote accepted

So, if I'm following you correctly:

Player two's response depends on player one's response, and player three's response depends on the responses of players one and two?

If this is the case, you would write for example ''P( P_2=Y | P_1=Y)" to mean the probability that player two says yes given that player 1 said yes. So, with your examples $P( P_2=Y | P_1=N)=.4$?

To find the probability, for example, $P(YNY)$ (that is, the probability that player one says yes and player two says no and player three says yes), you cannot multiply the probabilities that player one says yes, player 2 says yes, and player three says yes. That can be done only when you have independence.

However, you can take the product $$ P(YNY)= P(P_1=Y) \cdot P(P_2 = N | P_1=Y) \cdot P(P_3=Y | P_1=Y\ \text{and}\ P_2=N). $$

This is called the multiplication rule for probabilities.

Your example probabilities do not make perfect sense to me. You might want to start with:

Player one always says yes with probability $a$ and no with probability $1-a$ (for some $a$. )

If player one says yes, then the probability that player two says yes is ...

If player one says no, then the probability that player two says yes is ...

And you would have four cases for the probability that player three says yes.


Example:


Assume

$P_1$ says yes with probability $.2$.

$P_2$ says yes with probability $.1$, if $P_1$ said yes.

$P_2$ says yes with probability $.3$, if $P_1$ said no.

$P_3$ says yes with probability $.4$, if $P_1$ said yes and $P_2$ said yes.

$P_3$ says yes with probability $.5$, if $P_1$ said yes and $P_2$ said no.

$P_3$ says yes with probability $.6$, if $P_1$ said no and $P_2$ said no.

$P_3$ says yes with probability $.7$, if $P_1$ said no and $P_2$ said yes.

Then $$ \begin{alignat*}{2} abc \qquad &P(abc)= P(P_1=a)P(P_2=b|P_1=a) P(P_3=b|P_1=a\ \text{and}\ P_2=b)&\ \\ YYY \qquad & P(P_1=Y)P(P_2=Y|P_1=Y) P(P_3=Y|P_1=Y\ \text{and}\ P_2=Y)& = (.2)(.1)(.4) \\ YYN \qquad &P(P_1=Y)P(P_2=Y|P_1=Y) P(P_3=N|P_1=Y\ \text{and}\ P_2=Y)& =( .2)(.1 )( .6)\\ YNY \qquad &P(P_1=Y)P(P_2=N|P_1=Y) P(P_3=Y|P_1=Y\ \text{and}\ P_2=N)& = ( .2)(.9 )(.5 )\\ YNN \qquad &P(P_1=Y)P(P_2=N|P_1=Y) P(P_3=N|P_1=Y\ \text{and}\ P_2=N)& = ( .2)(.9 )( .5)\\ NNN \qquad & P(P_1=N)P(P_2=N|P_1=N) P(P_3=N|P_1=N\ \text{and}\ P_2=N)& = (.8 )( .7)(.6 )\\ NNY \qquad & P(P_1=N)P(P_2=N|P_1=N) P(P_3=Y|P_1=N\ \text{and}\ P_2=N)& = ( .8)( .7)( .4)\\ NYY \qquad & P(P_1=N)P(P_2=Y|P_1=N) P(P_3=Y|P_1=N\ \text{and}\ P_2=Y)& = ( .8)( .3)(.7 )\\ NYN \qquad & P(P_1=N)P(P_2=Y|P_1=N) P(P_3=N|P_1=N\ \text{and}\ P_2=Y)& = (.8 )(.3 )( .3)\\ \end{alignat*} $$

The sum is (adding in pairs starting from the top) $$ (.2)(.1)+(.2)(.9)+(.8)(.7)+(.8)(.3)=(.2)+(.8)=1. $$

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The answer of the first player will influence the second and so on, so they are dependent. Eg. If we are in the case YYN Player 2 has a probability of answering Y, in the case NYN the player 2 has a different probability. How can I manange this? –  Tom Dwan Jan 5 '12 at 10:44
    
@Giorgio I rewrote my answer based on dependence. Sorry for the initial response. –  David Mitra Jan 5 '12 at 11:03
    
Thank you man! :) So by applying this rule do the probabilities of the 8 events sum up to 1? –  Tom Dwan Jan 5 '12 at 11:09
    
@Giorgio Yes. I included a worked example. –  David Mitra Jan 5 '12 at 11:30
1  
@Giorgio Yes that's correct. I was just using the multiplication rule. Summing in pairs makes calculating the sum easy (you can avoid doing most of the arithmatic). Note the first two entries in the table give: $(.2)(.1)(.4)+(.2)(.1)(.6)= (.2)(.1)\cdot(.4+.6)=(.2)(.1)\cdot 1$. Each of the other two pairs can be simplified in the same manner. –  David Mitra Jan 5 '12 at 13:08

It's right: $0.2(0.3*0.25 + 0.3*0.75 + 0.7*0.25 + 0.7*0.75) + 0.8(0.3*0.25 + 0.3*0.75 + 0.7*0.25 + 0.7*0.75) =$ (0.2 + 0.8)(0.3*0.25 + 0.3*0.75 + 0.7*0.25 + 0.7*0.75) = $ $(0.3*0.25 + 0.3*0.75 + 0.7*0.25 + 0.7*0.75) = $ $0.3(0.25 + 0.75) + 0.7(0.25 + 0.75) = 0.3 + 0.7 = 1$

So they sums to one.

You are proceding in the right way, but if and only if the events are not correlated one to each other.

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They're dependent events. I've edited the answer to better explain it. However thanks for your answer :) –  Tom Dwan Jan 5 '12 at 10:47

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