A friend of mine was playing the bar game Pig Wheel recently and posed some interesting questions to me. He was playing with others as a group of four and, acting collectively, they came out about even after two hours, which surprised him. That got me thinking about the game.

So the game:

There are 45 numbers on a wheel, you place a bet on a number, the wheel is spun, and if you win, the payout is 40-to-1.

Let x = number of spots bet on.

Let y = amount place on each spot (assuming evenly distributed - which doesn't change any of the math below)

$$\frac{x}{45}\left(40-x\right)y - \frac{45-x}{45}xy = -\frac{xy}{9}$$

Here $xy$ is the total amount bet on the spin. So you're losing on average roughly 11 cents on dollar you put in (per spin).

On to what has stumped me:

They were betting \$10 on 8 numbers every spin and had a bank of \$400. Let's say the group saw a spin every two minutes, for a total of 60 spins. What is probability they come out even at the end of those 60 spins i.e. what is the probability that they 'succeed' on 12 spins?

Thoughts:

$$\binom{60}{12}\left(\frac{8}{45}\right)^{12}\left(\frac{37}{45}\right)^{48}\approx .116$$

But this is an overestimation, since it includes junk sequences such as 48 failures followed by 12 successes, which clearly would not be possible in this real-life example. It seems like quite a significant overestimation since the reverse of most 'good' sequences are 'bad' sequences but not vice versa.

I've thought about this more and have more I could say but I'll stop here for now and toss it out for others to think over.