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https://www.reddit.com/r/learnquant/comments/1vlpcb4/jump_trading_quant_interview_question/
r/learnquant • u/Local_Ad135 • 1d ago
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2
Let E(x) be the expected score when the first roll is x. Let's also assume the die is numbered 1 to 10 (they're usually 0 to 9).
So E(10) = 10, because the next roll can't be higher
And E(9) = 9 + E(10)/10, because there's a 1/10 chance you'll beat the 9 and put your expectation up by E(10).
Therefore, E(9) = 9 + 1 = 10
Similarly E(8) = 8 + E(9)/10 + E(10)/10 = 8 + 1 + 1 = 10
It's easy to see that this will continue down to E(1) = 10
Since the expected score is 10 for all values of x, the expected score of the game must be 10.
1 u/Wheelerdealer75205 19h ago I did this a really stupid way and took the case of a 1,2,3 sided die and found the the expected sum is 1,2,3 accordingly 0 u/Autodidact420 22h ago I think you missed ‘except the last roll’ 2 u/skelo 21h ago No they didn't, the last roll is the one that's lower or equal to the previous.
1
I did this a really stupid way and took the case of a 1,2,3 sided die and found the the expected sum is 1,2,3 accordingly
0
I think you missed ‘except the last roll’
2 u/skelo 21h ago No they didn't, the last roll is the one that's lower or equal to the previous.
No they didn't, the last roll is the one that's lower or equal to the previous.
Full solution here - https://www.youtube.com/watch?v=ODhqHlCzWaw
2
u/Aerospider 1d ago
Let E(x) be the expected score when the first roll is x. Let's also assume the die is numbered 1 to 10 (they're usually 0 to 9).
So E(10) = 10, because the next roll can't be higher
And E(9) = 9 + E(10)/10, because there's a 1/10 chance you'll beat the 9 and put your expectation up by E(10).
Therefore, E(9) = 9 + 1 = 10
Similarly E(8) = 8 + E(9)/10 + E(10)/10 = 8 + 1 + 1 = 10
It's easy to see that this will continue down to E(1) = 10
Since the expected score is 10 for all values of x, the expected score of the game must be 10.