## Interview Question for Analysts

• 0

Team: Risk analysis
Country: United States
Interview Type: Phone Interview

Comment hidden because of low score. Click to expand.
8
of 8 vote

Is there just one die? If so, the expected reward of a roll is $3.5. So you should reject the first roll if it's less than that (if it's less than or equal to 3). The expected reward is: rewards from first roll + rewards from second roll. First roll = 1/6 * 6 + 1/6 * 5 + 1/6 * 4 = 2.5. Second roll = chance of getting that far * expected value of a roll = 1/2 * (3.5) = 1.75. So the total expected value of this game is 4.25. Comment hidden because of low score. Click to expand. 0 of 0 votes Seems reasonable, intuitive and is very interesting, but is there is a proof? Comment hidden because of low score. Click to expand. 1 of 1 vote Both rolls are independent of each other so each one has an expected reward of$3.5. So if the first roll gives you 1, 2 or 3 then give it another shot. If not then stick with the outcome and make more then expected (4, 5, or 6)

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0
of 0 vote

can i know for which company this question was asked??

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0
of 0 vote

and also did they say how many dice? 2 or 3?

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0
of 0 vote

@Anon -- The company is startup on credit card payment.

S1: Stop after 1st roll and get rewarded: expected reward = (1+2+3+4+5+6)/6 = $3.5 S2: Try 2nd time if a) the first roll is less then 6: (1/6)*6 + (5/6)*(3.5) b) the first roll is less then 5: (1/6)*6 + (1/6)*5 + (4/6)*(3.5) c) the first roll is less than 4: (1/6)*6 + (1/6)*5 + (1/6)*4 + (3/6)*(3.5) d) the first roll is less than 3: (1/6)*6 + (1/6)*5 + (1/6)*4 + (1/6)*3 + (2/6)*(3.5) ... Hence c) is the stragety to maximize the expected reward. Comments? Comment hidden because of low score. Click to expand. 0 of 0 vote I was thinking the following way. Possible outcomes: 1 2 3 4 5 6 First try Probability of doing better on second try 1 1/6 vs 5/6 (2,3,4,5,6 would be better) 2 2/6 vs 4/6 3 3/6 vs 3/6 4 4/6 vs 2/6 So, it seems that anything <=3 should lead to a second try. Comment hidden because of low score. Click to expand. 0 of 0 vote You can a tree to represent the possible outcomes with each branch having a probability. Then you can simply calculate the final probability for a particular leaf (a final outcome). Comment hidden because of low score. Click to expand. 0 of 0 vote Let's say our strategy is reject the reward if it is smaller than a, then our expectation is Ex= \sum_{t=a}^6 t/6 + (a-1)/6 \sum_{t=1}^6 t/6 = (7-a)(6+a)/12 + (7a-7)/12 = (-a^2 + 8a +35)/12 max Ex= 17/4, when a = 4. This is bigger than the first choice whose expectation is 14/4. Comment hidden because of low score. Click to expand. 0 of 0 vote If we have an option to refuse a result and try again then why not wait till you get 6??? :P Comment hidden because of low score. Click to expand. 0 of 0 votes only two attempts! Comment hidden because of low score. Click to expand. 0 of 0 vote The expectation is 3.5 either way. On the first roll if you roll less than 3.5 (i.e 1-3) you should roll again. If you roll higher you would stay. The probability outcomes of the second roll are the same as the first. So expectation here is 3.5. Once you decide to roll again you completely forget what the first dice gave you. Comment hidden because of low score. Click to expand. 0 of 0 vote Define X = face value on dice upon rolling the dice the first attempt Define y = threshold (to be determined). If X > y, then we stop after first die roll. If X <= y, we roll the dice once more. It is clear that 1<=y<=6 Define the average pay off to equal Z. Then Z = { 3.5, X <= y, [(y +1) + ... 6]/(6 -y), X > y } We now compute the expected payoff Z for different values of y y = 1: Z = 3.5 x (1/6) + (2+3+4+5+6)/5 x (5/6) = 47/12 y = 2: Z = 3.5 x(2/6) + (3+4+5+6)/4 x (4/6) = 50/12 y = 3: Z = 3.5 x (3/6)+ (4+5+6)/3 x (3/6)= 51/12 y = 4: Z = 3.5 x (4/6) + (5+6)/2 x (2/6) = 50/12 y = 5: Z = 3.5 x (5/6) + 6 x 1/6 = 47/12 y = 6: Z = 3.5 x (6/6) = 42/12 Based on above, optimal strategy is to choose threshold y = 3. Expected payoff =$51/12 .

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