The Gateway to Computer Science Excellence
First time here? Checkout the FAQ!
x
+3 votes
265 views
This question was asked in IITB (RA) admissions 2016.

I have two blue dice, with which I play a game. If I throw a double six (i.e. if I get two six on both the dices) then I win the game. I  separately throw a red dice. If I get a one, then I tell truth about whether I win/loose in the previous game, otherwise I lie. I just rolled the three die. I turn around to you and said, "I won!". What is the probability that I actually won the game?
asked in Probability by Active (2.4k points) | 265 views

2 Answers

+6 votes
Best answer

On blue dice throw chances of winning 1/36

chances of lost 35/36

Next chances of getting 1 in red dice is 1/6 i.e. he is telling truth

other 5/6 times he lies

Now there are 4 possibilities

  • he wins (probability 1/36) and telling truth that he wins (probability 1/6).........i
  • he wins (probability 1/36) and telling false that he lost (probability 5/6)
  • he lost (probability 35/36) and telling truth that he lost  (probability 5/6)
  • he lost (probability 35/36) and telling false that he wins (probability 1/6)...........ii

Now we calculated on he tells he wins i.e. i and ii cases


Now he wins and red dice throws , he tells truth that he wins  1/6 ⨉1/36

"     he lost and red dice thrown , he tells lie that he wins 5/6 ⨉ 35/36

Probability of actually winning chance is

= (1/(36⨉6)) / ((1/(6⨉36))+((5⨉35) / (6⨉36))) =1/176

answered by Veteran (70.1k points)
selected by
isn't question asking probality of actual win not false win ? hence 1/216 is answer?
No , to get actual wining Bayes theorem should be applied
@Srestha I want to know, how you are partitioning the sample space, to apply Bayes Theorem?
@utk now clear?

$P(A)=\frac{1}{6}$

$P(\frac{A}{E1})=\frac{P(A\cap E1)}{P(E1)}=\frac{(\frac{1}{6})(\frac{1}{36})}{(\frac{1}{36})}=\frac{1}{6}$

Similarly, $P(\frac{A}{E2})=\frac{1}{6}$

Then, we have, $P(\frac{E1}{A})=\frac{P(\frac{A}{E1})P(E1)}{P(\frac{A}{E1})P(E1)+P(\frac{A}{E2})P(E2)}=\frac{1}{36}$

 

I am getting this solution. What's wrong? @Srestha

I think u give more priority to blue dice rather than wining probability
Probability for actual win is asked not false win so how is answer 1/176
@srestha,on getting one (1/6),he tells truth always.

so,

"he lost (probability 35/36) and *telling truth that he lost * (probability 5/6)"---shud'nt this be 1/6 cuz he is telling truth that he lost in previous game.

I have few doubts :-

1. What is the meaning of actually win the game?

2. What will be the framing of the question if the solution contain only case 1 and 2 in following ways-

1/6 * 1/36 + 5/6* 1*36 ?

 

0 votes

Note:  1) Brace yourself, spoon feeding is about to begin.

           2) first two dice  ---  implies two blue dice.

              3rd dice          --- implies red dice. 


 

Point 1)  What's asked in question?
ans)        P(person actually won | person claimed that he won) = ?

Point 2)   Why 1/216 isn't the ans.?
ans)         After reading this question, first thought that arrives is....ummm...isn't 1/216 the correct ans. (I bet  most people must have thought this).
                But read point 1 (above), point 3 (below) and now read question 2-3 times.
                Makes Sense now?
                (1/216 is just the probability that person won! and this isn't what's asked in the question!)

Point 3)    Elaborating  srestha's ans:
ans)
        4 cases arise:
        
        person won and  1 came on red dice  ---- person tells the truth i.e. person tells he won.
        person won and  2 or 3 or 4 or 5 or 6 came on red dice  ---- person tells the lie   i.e. person tells he lost.
        person lost  and  1 came on red dice   ---- person tells the truth i.e. person tells he lost.
        person lost  and  2 or 3 or 4 or 5 or 6 came on red dice ---- person tells the lie   i.e. person tells he won.

        

        Therefore, person claims "he won" in 2 cases:
        
        person won and    1 came on red dice
        person lost and    2,3,4,5,6 came on red dice

        Hence, P(person actually won | person claimed that he won) = ??

                                         P(person actually won and person claimed he won)
        =   ---------------------------------------------------------------------------------------------------------------
    P(person actually won and person claimed that he won) + P(person actually lost and person claimed that he won)

                   (6 came on first dice and 6 came on 2nd dice and 1 came on 3rd dice)
        = -----------------------------------------------------------------------------------------------------------------
            (6 came on first dice and 6 came on 2nd dice and 1 came on 3rd dice) + (double 6 didn't come on first 2 dice throws and (2 or 3 or 4 or 5 or 6 came on 3rd dice))

                            1/216
         =  ------------------------------
               1/216 + (35/36)(5/36)

                              1
         =  ------------------------------
                           1+175
        
                               1
          = ------------------------------
                            176

                                  

That's it folks!

-----------------------------------------------------------------------------------

and kudos to srestha coming up with the correct ans.

-----------------------------------------------------------------------------------

answered by Junior (609 points)
edited by


Quick search syntax
tags tag:apple
author user:martin
title title:apple
content content:apple
exclude -tag:apple
force match +apple
views views:100
score score:10
answers answers:2
is accepted isaccepted:true
is closed isclosed:true

29,017 questions
36,844 answers
91,385 comments
34,723 users