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?

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.

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

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

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

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.

Re: A sequence of events
 

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.

Re: A sequence of events
 

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

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.

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.