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.

[Top]

One of the problems is that you are doing your pay-out odds calculations wrong. Please see my table for the correct calculations:

#of Reds Probabilty Pay Ratio $10 Payout 1000X Payout
-----------------------------------------------------------

0 0.40 2.50 0 0
1 0.24 4.17 41.7 10000
2 0.14 6.94 69.4 10000
3 0.086 11.57 115.7 10000
4 0.052 19.29 192.9 10000
5 0.031 32.15 321.5 10000
6 0.019 53.58 535.8 10000
7 0.011 89.31 893.1 10000
8 0.0067 148.84 1488.44 10000
9 0.0040 248.07 2380.73 10000
10 0.0060 165.38 1653.82 10000

Make sure not to confuse the correct pay ratio odds with win-lose odds. Pay ratio odds are simply calculated by inversing the probability.

Your next problem is that your game as stated is not mathematically balanced. You can do this two ways:

(1) Have them pick a number to aim for before they start the game - and then use the above pay-out ratio to pay that amount. If they get the number wrong then pay nothing. Note: You will have to give them the option of being able to bet on a white ball and modify my first line of the chart accordingly.

(2) Divide the above pay ration by 10. This will give you the odds you need to get a total zero-sum game. That would mean that on 1 red roll instances you would pay $4.17, on 2 red rolls you would pay $6.94, etc.

I hope that helps.

[Top]

Clarification:

I meant to say that you should divide the payout by ten to get the proper payout ratio for game option 2.

So for Option 1 you should pay them $41.7, $69.4, $115.7, etc...

For Option 2 you should pay them one tenth of the above -i.e. $4.17, $6.94, $11.57, etc.... Option 2 is then structured as you originally envisioned the game.

Also the column names are:
1. # of reds in a row picked

2. Probability of # reds in a row picked (all independent, expect last one)

3. Pay ratio odds

4. Payout on $10 bet (if they pick number before game starts -> divide this number by ten if using your original system)

5. Total payout after 1000 trials - using $10 bets

Sorry for the confusion