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Following Q is from a top 3 test prep company.

Results of dice throw
1. lose $5
2. lose $2
3. no change
4. win $5
5. win $2
6. lose $4

find probability of winning atleast $5 with 2 throws.

--------
given answer = 5/36.

seems to me the answer should be
i.e. 1/6(3/6) + 1/6(3/6) = 1/6

if 1/6 is wrong, can u plz explain why 5/36? thanks!

Tuneman · Answer

rolling the following combinations

3&4
4&3
4&5
5&4

and
 4&4

I bet you forgot about the last one, so did I until I went back to look at it.

oops · Answer

actually, i included the last one twice, since the first throw can be 4 and second throw can be 4 (as you correctly point out). 
but there is also the vice-versa case, albeit the same value

my logic was:

(prob of first throw 4) * (prob of second throw can be any of 3, 4 or 5) = (1/6)*(3/6) = 3/36
+ 
the exact reverse case = (prob of first throw with any of 3, 4 or 5) * (prob of second throw 4) = (1/6)*(3/6)= 3/36 

So, total = 3/36 +3/36 = 1/6

Still confused...

oops · Answer

after thinking quite a bit more:

it seems to me that 4&4 may be considered as one outcome, even though there are two ways of getting it and the answer could be 5/36

i am going with this logic, because if we draw a matrix/table of 1-6 (horizontal) crossing 1-6 (vertical), there is only one slot (out of 36 slots) where 4 crosses 4.

this is so very confusing!!!

anyone has a better insight, please reply.

AK · Answer

Total no. of possibilities = 6 * 6 = 36

favorable conditions for atleast $5 with 2 throws = 5

(no change, win $5 )
( win $5,no change )
(win $2 , win $5)
(win $5,win $2 )
(win $5, win $5)


prob. = 5/36

oops · Answer

My question was (win $5, win $5) can happen in two ways - why is it not counted twice?

KillerSquirrel · Answer

Results of dice throw:

1. lose $5
2. lose $2
3. no change
4. win $5
5. win $2
6. lose $4

find probability of winning atleast $5 with 2 throws.

the secret to solveing probability questions is to look at each event as independent form other events in the question.

the probability to win 5$ in first dice throw is 1/6 (only if you get 4)
that amount will stay 5$ or more, only if you get 5,3,4 in the second throw.

hence :

1/6 * 3/6 = 3/36

the next option is to win 2$ in first and then 5$ in second.

1/6 * 1/6 = 1/36

the final option is to win 0$ in first and 5$ in second.

1/6 * 1/6 = 1/36

total is = 1/36 + 1/36 + 3/36 = 5/36

any other outcome will give less then 5$ needed in both throws

oops · Answer

got confused and kept doublecounting the (win $5, win $5) scenario. 
was repeating the same mistake - was a major mind block.
thanks for the clarification.

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