Numbered tic tac toe game with cards, 2 options for playing, how much is the profit?

Question

I decided to organize a special tic tac toe game with my friends. I got a board with 9 numbered squares and a bag with 9 cards, with the numbers from 1 to 9.

Here is the board:

Whoever wants to play, pays 1 euro and can choose between 2 options:

• A) Take 3 cards out of the bag and wins if they correspond to three numbers in the same row, column or diagonal. If the player wins, they get the 1 euro back plus 8 euro.

• B) Take 4 cards out of the bag and wins if 3 of them correspond to three numbers in the same row, column or diagonal. If the player wins they get the 1 euro back plus 1 euro.

Am I going to profit? In average, what will be my profit or my loss?

I did:

$$\\$$ For A) the probability of the player winning is $\frac{1}{2}%$, because: $$\frac{^9C_3 + ^9C_3 + ^9C_3}{9\cdot8\cdot7} = .5 = \frac{1}{2}$$

For B) the probability of the player winning is $\frac{1}{8}$. because: $$\frac{^9C_4 + ^9C_4 + ^9C_4}{9*8*7*6} = .125 = \frac{1}{8}$$

I want to calculate how much I win, so what I want to know is the probability of the player losing, so I have:

• A) = $\frac{1}{2}$
• B) = $\frac{7}{8}$

So, a formula to determine my profit would be:

• A) $p = (\frac{x}{2}) - (\frac{x}{2} \cdot 8)$
• B) $p = (\frac{7x}{8}) - \frac{x}{8}$

However, my book says the solution is as follows:

A - The profit is of 12 euro in 84 plays ($\mu \approx 0.14$ euro)

B - The profit is of 30 euro in 126 plays ($\mu \approx 0.24$ euro)

The organizer profits in either choices.

What did I do wrong? How do I solve this?

Answer 1

There are $\binom{9}{n}$ ways of drawing $n$ cards. There is $8$ ways to win in case $A$ and $8\cdot 6$ ways in case $B$ (for each of the eight winning configurations, you have one number left, which can have a value of any of the six remaining numbers). Thus, the probability for the player winning $p_{w}$ is in the two cases $$p_{w,A}=\frac{8}{\binom{9}{3}}=0.0952\dots$$ $$p_{w,A}=\frac{8\cdot 6}{\binom{9}{4}}=0.3809\dots$$

So your expected gain is $$A:\quad 1\times (1-p_{w,A})-8\times p_{w,A}=1-9p_{w,A}=0.1428\dots$$ $$B:\quad 1\times (1-p_{w,B})-1\times p_{w,B}=1-2p_{w,B}=0.2380\dots$$

So you gain something in each case.

Answer 2

I would enumerate the events of interest by ignoring the order in which the cards are drawn. Then the denominator in the first case is $\binom{9}{3}$, and there are only $8$ winning outcomes, for a win probability of $2/21$, and an expected winnings per game of $(2/21)(8) + (19/21)(-1) = -1/7$ (i.e., a loss for the player).

The second game is treated similarly: the denominator is $\binom{9}{4}$, but the numerator is harder to compute. The key observation is that with any four winning cards, there is still only a set of three cards that comprises the winning configuration; the fourth card can be uniquely chosen from any of the $6$ numbers not comprising the winning three cards. So the number of such outcomes is $8(6) = 48$ and the probability of winning is $48/126 = 8/21$. The expected winnings per game is calculated similarly as above, giving $-5/21$, which is actually worse for the player than the first choice.