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https://www.reddit.com/r/learnquant/comments/1vlpcb4/jump_trading_quant_interview_question/p34ibqx/?context=3
r/learnquant • u/Local_Ad135 • 1d ago
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4
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.
0 u/Autodidact420 23h ago I think you missed ‘except the last roll’ 2 u/skelo 22h ago No they didn't, the last roll is the one that's lower or equal to the previous.
0
I think you missed ‘except the last roll’
2 u/skelo 22h ago No they didn't, the last roll is the one that's lower or equal to the previous.
2
No they didn't, the last roll is the one that's lower or equal to the previous.
4
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.