What am I screwing up?

[asdfasdf32]

You have a bag with 6 red balls and 4 white balls. In the game you are playing, you are trying to consecutively pick out a red balls, with the bag being reset after each turn. Also, in this game, 10 times is the most times you can get red in a row before the game ends. Pretend this is an even money gambling game, $10 to play. If, for instance, you get three red balls in a row, then get a white one on the fourth try, this counts (and pays) as three in a row.

# red balls...odds..............pay-out.........percentage
in a row.......against...........assuming......unique
..................one................$10 buy-in....occurrence

0................1.5000 to 1....$0.00............40.00%
1................0.6667 to 1....$6.67............24.00%
2................1.7778 to 1....$17.78..........14.40%
3................3.6296 to 1....$36.30............8.64%
4................6.7160 to 1....$67.16............5.18%
5................11.860 to 1....$118.60..........3.11%
6................20.433 to 1....$204.33..........1.87%
7................34.722 to 1....$347.22..........1.12%
8................58.537 to 1....$585.37..........0.67%
9................98.229 to 1....$982.29..........0.40%
10..............164.38 to 1....$1643.82........0.60%

Judging by my numbers, lets pretend you did this 1000 times. You would get 1 red ball the first time and miss the second time 24% of the time. So, 240 times out of 1000 you would get paid $6.67. Additionally, 6 times out of 1000 you would get a red ball all ten times, and get paid $1643.82 each time, for a total of $9939.53. If you played 1000 times, it would cost $10,000 to play....so how in the world can the game pay out so much? What am I screwing up?

[sixhigh]

It obviously pays way to much to have a neutral EV. I don't know how you got your numbers - the following ones would give you an EV of zero assuming a wager of 10:

0
4.17
6.94
11.57
19.29
32.15
53.58
89.31
148.84
248.07
165.38

[asdfasdf32]

I got my numbers by saying "If hitting 10 in a row is a 164-to-1 shot, and it costs $10 to play, then even money should pay out $1640." Is this wrong?

[sixhigh]

First I don't get why hitting 10 reds is a 164-1 one shot while hitting exactly 9 is only a 98-1. Hitting exactly 9 reds is less likely.
And when calculating the fair payoff for every shot you have to consider the cases of missing where you will also pay something. I.e. not hitting 10 red balls doesn't necessarily mean you get nothing.

[asdfasdf32]

Hitting ten reds is a 164-to-1 shot because (3/5*3/5*3/5*3/5*3/5*3/5*3/5*3/5*3/5*3/5) = .006, or 164-to-1, right?

[sixhigh]

And the probability of hitting nine reds is (3/5)^9*(2/5) = .004 = 247-1.