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.
Best Answer
I had a go of it in Excel. If you manage to get through 60 spins and still have money -- 84.6% of days you won't -- coming out even is the third most common possibility: 2.56% of all days, and 16.7% of all successful days.
Here's my workings. It's quite number-crunchy, which is unfortunate, but it felt the most straightforward. https://docs.google.com/spreadsheets/d/1GZGzHHbSSQFSpk_yRerQY1gnjVSWr0DXfkRJozmoEH4/edit?usp=sharing