Help with SIG Interview Question about Options

newwurldmn said:
Nope. You are wrong on the easy question. And didn't even answer the real question.
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$3.00 for each roll ....... My final answer.


EDIT: newwurldmn your question asked "How much are you willing to pay". I'm willing to pay $3.00 - so how can my answer be wrong?


:)
 
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OTM-Options said:
$3.00 for each roll ....... My final answer.


EDIT: newwurldmn your question asked "How much are you willing to pay". I'm willing to pay $3.00 - so how can my answer be wrong?


:)
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Since you are too thick to understand the spirit of the question: make a 10 cent wide market for the contract and then for the contract with the option.
 
newwurldmn said:
Its similar to the dice problem that they have been known to ask. You roll a die and get the payout (if you roll a six, you get $6, if you roll at 2, you get $2). How much are you willing to pay for this contract? The follow up is, how much are you willing to pay up front) for the option of rolling again?
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My math is very rusty so my answer is probably incorrect. For the first part of the question, I used expectancy to solve. Prob of any single outcome of the roll of a die is 1/6. So..
1/6*6 = 1
1/6*5 = .83
1/6*4 = .66
1/6*3 = 0.5
1/6*2 = = 0.33
1/6*1 = .166

Total expectancy for 6 rolls is 3.486
The average I can expect to make is 3.486/6 which gives me 0.581.
Assuming a bid ask spread of 0.05 to 0.10, I'd try and bid at 0.53 or a few ticks up.

From what I remember, dice rolls are independent of one another so I would just pay 0.53 again for a second roll.
 
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Gambit said:
My math is very rusty so my answer is probably incorrect. For the first part of the question, I used expectancy to solve. Prob of any single outcome of the roll of a die is 1/6. So..
1/6*6 = 1
1/6*5 = .83
1/6*4 = .66
1/6*3 = 0.5
1/6*2 = = 0.33
1/6*1 = .166

Total expectancy for 6 rolls is 3.486
The average I can expect to make is 3.486/6 which gives me 0.581.
Assuming a bid ask spread of 0.05 to 0.10, I'd try and bid at 0.53 or a few ticks up.

From what I remember, dice rolls are independent of one another so I would just pay 0.53 again for a second roll.
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Payout will range from $1.00 to $6.00 and you only want to pay $0.53?



:)
 
I think I see my mistake. Instead of dividing all the expectancies by number of outcomes, simply add them all together. The total expectancy for the dice roll game is 3.486 which rounds to 3.5. Then bid under that by a few cents and offer above.

I'll take a crack at the second part later.
 
I can make a binary decision based on probable value, easy. I also understand NPV concepts.

I know I frequent the options board, but I seem to not know the way to value optionality. Can someone illuminate me on the fundamental equation?
 
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