#1
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A sequence of events
There is an event that happens 1/10 times. It costs $1 for each attempt. If the attempt succeeds, you get $9 back, for a profit of $8. If you play up to 10 times, but stop when you win, is it profitable?
I know this shouldn't be profitable, but let me explain some logic and tell me why it's faulty? In a sequence of 10 events, NOT STOPPING, the event will be successful once. In this sequence, it will evenly be distributed throughout each try, averaging at try #5.5. Visual Display: X = hit 10 trials of 10 sequences X _ _ _ _ _ _ _ _ _ _ X _ _ _ _ _ _ _ _ _ _ X _ _ _ _ _ _ _ _ _ _ X _ _ _ _ _ _ _ _ _ _ X _ _ _ _ _ _ _ _ _ _ X _ _ _ _ _ _ _ _ _ _ X _ _ _ _ _ _ _ _ _ _ X _ _ _ _ _ _ _ _ _ _ X _ _ _ _ _ _ _ _ _ _ X This is average, yes? So if you go through one sequence, stopping when you hit, it should be profitable? |
#2
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Re: A sequence of events
Ok, to clarify more:
In a sequence of 10 tries, you will succeed 1 time, on average. If you have infinite sequences of 10 tries, 1/10 times, the success will be on the first try. 1/10 times, the success will be on the second try, etc. |
#3
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Re: A sequence of events
My calculations:
Note - You STOP at the X. Visual Display: X = hit 10 trials of 10 sequences X _ _ _ _ _ _ _ _ _ +8 _ X _ _ _ _ _ _ _ _ +7 _ _ X _ _ _ _ _ _ _ +6 _ _ _ X _ _ _ _ _ _ +5 _ _ _ _ X _ _ _ _ _ +4 _ _ _ _ _ X _ _ _ _ +3 _ _ _ _ _ _ X _ _ _ +2 _ _ _ _ _ _ _ X _ _ +1 _ _ _ _ _ _ _ _ X _ 0 _ _ _ _ _ _ _ _ _ X -1 8 + 7 + 6 + 5 + 4 + 3 + 2 + 1 + 0 - 1 = 35 35/10 = 3.5 |
#4
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Re: A sequence of events
you're assuming that you will automatically get at least one in ten X, implying that the probability of each trial is not independent. if the probability of each draw is 1/10 than it would be a geometric distribution and the average number of draws before a hit would be 10. meaning a loss of $.10 on each trial.
if a miss implies a greater likelyhood of a hit later on than the probability would be completely differnt, i could do it out but i don't think that's what you were wondering about, if it is than just respond and i'll show you the calculations -little fishy |
#5
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Re: A sequence of events
sorry i responded before i saw your update... i'll do the math out again tomorrow, and if you succeeed exactly once out of every ten trials and yuou can stop after you suceed then yes a 1:8 wager would be EV, more on this tomorrow, but tonight it's bed time
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#6
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Re: A sequence of events
Your EV =
1st trial = (0.1) * $8 = $0.8 2nd trial = (0.9)(0.1) $7 = $0.63 3rd trial = (0.9)^2(0.1) $6 = $0.486 4th trial = (0.9)^3(0.1) $5 = $0.3645 5th trial = (0.9)^4(0.1) $4 = $0.26244 6th trial = (0.9)^5(0.1) $3 = $0.177147 7th trial = (0.9)^6(0.1) $2 = $0.1062882 8th trial = (0.9)^7(0.1) $1 = $0.04782969 9th trial = (0.9)^8(0.1) $0 = doesnt matter 10th trial= (0.9)^9(0.1) -$1 = -$0.03874205 Dont win at all = (0.9)^10 * -$10 = -$3.486784 E(X) = -$0.65 P.S. Theres a 57% chance you will win by the 8th trial and be therefore profitable. |
#8
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Re: A sequence of events
I know the game is -EV. I want to know why my line of thought is wrong.
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#9
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Re: A sequence of events
I wasn't replying to you, I was replying to someone who claimed the game is +EV. I don't understand what you mean yet.
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#10
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Re: A sequence of events
[ QUOTE ]
My calculations: Note - You STOP at the X. Visual Display: X = hit 10 trials of 10 sequences X _ _ _ _ _ _ _ _ _ +8 _ X _ _ _ _ _ _ _ _ +7 _ _ X _ _ _ _ _ _ _ +6 _ _ _ X _ _ _ _ _ _ +5 _ _ _ _ X _ _ _ _ _ +4 _ _ _ _ _ X _ _ _ _ +3 _ _ _ _ _ _ X _ _ _ +2 _ _ _ _ _ _ _ X _ _ +1 _ _ _ _ _ _ _ _ X _ 0 _ _ _ _ _ _ _ _ _ X -1 8 + 7 + 6 + 5 + 4 + 3 + 2 + 1 + 0 - 1 = 35 35/10 = 3.5 [/ QUOTE ] This did not weight the likelihood of the events correctly. |
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