A sequence of events

[kurosh]

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?

[kurosh]

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.

[kurosh]

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

[Little Fishy]

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

[Little Fishy]

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

[Luzion]

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.

[jason_t]

This game is not +EV. The EV of the game is

EV = .1 * 8 + .9 * .1 * 7 + .9^2 * .1 * 6 + ... + .9^9 * .1 * (-1) + (.9)^10 * (-10) = -.651.

[kurosh]

I know the game is -EV. I want to know why my line of thought is wrong.

[jason_t]

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.

[jason_t]

[ 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.