Wheel Probability

I have a game that I'm designing which has 16 spaces on a wheel, and each position has a value. Once you land on a position, the value is replaced by a "strike". If you land on that position again, you earn a strike - 3rd strike ends your game. Object of the game is to clear the wheel of all 16 values.

From what I recall, in order to figure out the odds of a "winning game" (clearing the wheel completely), you need to figure out the odds with ZERO strikes + ONE strike + TWO strikes.

With ZERO strikes, the first spin is a guarantee, thus it's 16/16. Then the next spin is 15/16.. so the first part of the equation is: (16/16) * (15/16) * (14*16) ... * (1/16) = 1.13423E-06 (We'll call this "Perfect Game" [PG])

So for ONE strike, the second part of the equation is: PG * ( y / 16) | y = 1 to 15 (all added together)

For TWO strikes, the last part would be: PG * (y / 16) * (z / 16) | y = 1 to 15; z = 1 to 15 (matrix all added together)

Thus, for a final answer, I came up with: 7.34E-05, which inverted comes out to a 1 in 13616.34 chance to win.

Can I have someone verify that this is indeed the right way to figure out this problem, and if not, guide me as to where I went wrong? Any help would be greatly appreciated!

Welcome, fellow game designer!
Interesting game you have thought up.

Firstly, you are in luck - there are a number of very fine gaming mathematicians who are members of this board, in fact, the founder of the forum (Mike Shackleford) is perhaps the most primary mathematician in gaming mathematics. Also, "MathExtremist" is a member here at this board, as well as Eliot J. and Charles M., all of whom are top notch.

In addition "DJTeddybear" (Dave M.) is a very fine Roulette game designer, and he too can offer some accurate math calculations and some input on your game design. I am certain that if you PM him, he will give you some very fine advice on this design.

But in designing a game, you also have to have a "payout table" - to pay out people who are playing your game, on occurances that are not too common, but are common enough to see some payout before waiting to see 13,616 spins occur on average. If the goal to winning your game is to clear out every single number offered on the the wheel without getting a triple-hit on a single number, then 13,616 spin is too a long time without any player winning anything at all, clearly.

Currently, the game of Roulette has bets that hit about once in every two spins (even/odd, red/black) for a player to win, or about one in three chances for column bets, ranging up to once in every 36 or 37 spins to win 35:1 (35 to one). Beyond that, players might lose patience.

Do you plan on using this game design as a home-style amusement, or as a potential commerical casino product?
There are certainly a number of wide-ranging outcomes with many payout tables, but just about all of them can be produced using the existing and standard Roulette wheel configuration, which is already very successful.
Best of luck!
My advice: First, give DJTeddybear a message here, asking him for an opinion on this. On Roulette and Roulette style games, - he is very knowledgeable, and a great resource. Gaming is a tough business, and both he and I had been through it all on game design.

Quote: mrgameshow

So for ONE strike, the second part of the equation is: PG * ( y / 16) | y = 1 to 15 (all added together)

This does not seem correct, and I can't really see what makes you think that it is.
Consider probability of "clearing" n numbers with k "strikes" as a function F(n,k), so that F(16,0) = PG.

Then:
F(n,k) = n/N * F(n-1,k) + (N-n)/n*F(n,k-1)
F(1,k) = (N-1)/N * F(1,k-1);
F(1,0) = 1/N

where N = 16
If you create an excel table, with 1..16 rows for n and 0..2 columns for k, put 1/N into the top left corner,
and work your way down, and to the right, using the recursive formula above, then you'll find your answer in the the 16th row from the top.

I'm not as good at math as Dan describes.

For one thing, I'm not sure what you mean by "All added together":

Quote: mrgameshow

So for ONE strike, the second part of the equation is: PG * ( y / 16) | y = 1 to 15 (all added together)

Does that mean PG * ( 1/16 + 2/16 + ... + 14/16 + 15/16 ) ?
Or does it mean PG * ( ( 1 + 2 + 3 + ... + 14 + 15 ) / 16 ) ?

For the record, both are equivalent to PG * 120/16

If that's what you mean, then for two strikes, I get 6.38E-05 or 1 in 15,673.92

For the record, I'm not sure that's the correct way to determine the odds of your game. But I'm at a loss to suggest an alternative. I'm having trouble following weaselman's formula.


FYI: I DID get 1.13423E-06 for the perfect game.

Quote: weaselman

That would be a long game ... :-/

Even before I cracked open Excel, I was going to say the same thing.

Quote: mrgameshow

...

For TWO strikes, the last part would be: PG * (y / 16) * (z / 16) | y = 1 to 15; z =y to 15 (matrix all added together)
...

I highlighted a correction. I get a final answer of 4.4288E-05 or 1 in about 22579

Sorry about the vagueness of the post. I'm trying to remember things back from high school and my Finite math days, so I may be completely off-track :)

When I said "all added together", I meant wouldn't you sub the values 1 to 15 into the equation and add them all together?
eg. (PG * (1/16)) + (PG * (2/16))... + (PG * (15/16)) ? Then in the third part, you'd substitute for (x,y) into the equation -> (1,1) (1,2) .. (15,14)(15,15) ?

My rationale for thinking this is that the odds of the final outcome change as the game continues, so I was trying to take into account each possible scenario along the timeline where a "strike" would be hit, thus adding another turn yet not changing the possibility of the next turn. So the "hiccup" in a perfect game would result in the repeat of a spin with the same odds as the last, adding an addition step to the "perfect game" equation, wouldn't it?

Also - thanks for the welcome! :)

Yes, it's meant as a "bank building risk taker" style of game, where each position on the wheel would be a value to add to the player's bank, with some type of penalty if you risked yourself to your third repeat.

When I ran it in a simulator, I wasn't getting any winners (which was due to a coding oversight on my part..) - and I knew the game wasn't IMPOSSIBLE. Thus, I wanted to try figuring out the odds of the game and see how hard it should / shouldn't be to complete all positions on the wheel.

And it had been quite a while since I tackled this type of problem, so I thought I'd seek some guidance to see if I was heading in the right direction with my train of thought on how to calculate it.

Quote: miplet

I get a final answer of 4.4288E-05 or 1 in about 22579

I do too.

I am a bit amazed I even came up with the correct answer. I am so out of practice with sequences and summations.

for n=16	Formula	Probability	1 in ...
Win with 0 strikes	n!/(nn)	1.13423E-06	881657.9516
Win with 1 strike	(n-1)*(n!/(nn))/2	8.5067E-06	117554.3935
Win with 2 strikes	(n!/(nn))/n2 * ((3(n-1)4+10(n-1)3+9(n-1)2+2(n-1))/24)	3.46471E-05	28862.4598
Any Win	Sum of above	4.4288E-05	22579.47535
Loss	1-(Any Win)	0.999955712	1.00004429

The monster formula above, ((3(n-1)⁴+10(n-1)³+9(n-1)²+2(n-1))/24), is an arithmetic sequence of numbers: 1, 7, 25, 65, 140...

It is the sum of numbers as such:
1 = 1
7 = (1+2) + (2)*2
25 = (1+2+3) + (2+3)*2 + (3)*3
65 = (1+2+3+4) + (2+3+4)*2 + (3+4)*3 + (4)*4
140 = (1+2+3+4+5) + (2+3+4+5)*2 + (3+4+5)*3 + (4+5)*4 + (5)*5

It has to do with the arrangement of when the two strikes might hit in a winning game.

Thanks a lot to those who posted!

The error in my original spreadsheet was that I did a full 'square' matrix of the 2 strike, instead of the 'triangle' -- such as miplet pointed out the error that I should have used (z = y to 15) instead of starting at 1.