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kurosh
09-27-2005, 11:53 PM
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
09-28-2005, 12:07 AM
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
09-28-2005, 12:11 AM
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
09-28-2005, 12:12 AM
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
09-28-2005, 12:14 AM
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
09-28-2005, 12:17 AM

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
09-28-2005, 12:18 AM
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 (http://www.google.com/search?hl=en&amp;lr=&amp;q=.1+*+8+%2B+.9+*+.1+*+7+%2B+.9%5 E2+*+.1+*+6+%2B+.9%5E3+*+.1+*+5+%2B+.9%5E4+*+.1+*+ 4+%2B.9%5E5+*+.1+*+3+%2B+.9%5E6+*+.1+*+2+%2B+.9%5E 7+*+.1+*+1+%2B+.9%5E8+*+.1+*+0+%2B+.9%5E9+*+.1+*+% 28-1%29+%2B++%289%2F10%29%5E10+*+%28-10%29&amp;btnG=Search).

kurosh
09-28-2005, 12:19 AM
I know the game is -EV. I want to know why my line of thought is wrong.

jason_t
09-28-2005, 12:21 AM
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
09-28-2005, 12:32 AM
[ 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.

Justin A
09-30-2005, 03:50 PM
[ QUOTE ]
I know the game is -EV. I want to know why my line of thought is wrong.

[/ QUOTE ]

Because you'll miss all ten tries about 35% of the time.