EV of this particular slot machine

[Griffin]

My ability to calculate probabilities isn't what it used to be. In addition to answering the question below, if anyone can point me to their favorite web site that gives a tutorial on probability calculations ( P(A|B), P(A and B), etc.), I would appreciate it.

On to the question....

I ran across a Javascript slot machine game that has 3 wheels with 10 items on each wheel. If any two wheels pair, you win 5 times your bet. If all three wheels pair, you win 10 times your bet. You can bet $5, $50, or $500 each spin.

Now, I seem to remember that for independent events, (P(A and B) = P(A) * P(B)). If that's correct, then with only two wheels spinning, there's only a 1% chance of them matching. How do you write the equation for three wheels?

Would it be P(two wheels matching) = (P(AandB) OR P(AandC) OR P(BandC))?

Even if that is correct, it doesn't take into account all three wheels matching.

What got me curious about this is that after playing the game for a few minutes, I saw multiple two wheel matches and a couple of three wheel matches. It made me wonder if the game is actually playing according to expected probability, or if the code for the game is buggy.

Thanks for any help,
Griff

[BruceZ]

[ QUOTE ]
Now, I seem to remember that for independent events, (P(A and B) = P(A) * P(B)). If that's correct, then with only two wheels spinning, there's only a 1% chance of them matching.

[/ QUOTE ]

The chance of 2 particular wheels matching is 10% since there are 100 combinations and 10 of them match.

[ QUOTE ]
How do you write the equation for three wheels?

Would it be P(two wheels matching) = (P(AandB) OR P(AandC) OR P(BandC))?


[/ QUOTE ]

That would be 30%, but you are adding the chance of all 3 matching 3 times. The chance of all 3 matching is 1% since there are 1000 combinations and 10 of them match. Since we only want exactly 2 matching, subtract 3*1% from 30% to get 27%. Another way to do it is to observe that there are 10 ways to choose the pair, 9 ways to pick the unpaired item, times 3 ways to choose which wheel is unpaired. 10*9*3/1000 = 27%.

If all of the combinations have the same probability, and if you have described it correctly, then this game has postive EV! The EV is 0.27(5) + 0.01(10) - 0.72(1) = 0.73. So you win 73 cents for every dollar you bet. A $500 bet will average $365. PM me for the location of this game.

[BruceZ]

[ QUOTE ]
In addition to answering the question below, if anyone can point me to their favorite web site that gives a tutorial on probability calculations ( P(A|B), P(A and B), etc.), I would appreciate it.


[/ QUOTE ]

P(A and B) = P(A | B)*P(B)

where P(A | B) means probability that A occurs given that B has occured. If P(A | B) = P(A), then A and B are said to be independent events. If A and B are independent, then P(A and B) = P(A)*P(B).

P(A or B) = P(A) + P(B) - P(A and B)

If P(A and B) = 0, then A and B are said to be mutually exclusive events. This means that they cannot both occur. If and B are mutually exclusive, then P(A or B) = P(A) + P(B).

P(A or B or C or...) =
P(A) + P(B) + P(C) + ...
- P(A and B) - P(A and C) - P(B and C) - ...
+ P(A and B and C) + ...

This is an extension of the above, and it is called the inclusion-exclusion principle (http://forumserver.twoplustwo.com/showthreaded.php?Cat=&Number=417383&page=&view=&sb =5&o=&vc=1).

[Griffin]

Thanks Bruce,

I understood your reply and that explains why I was seeing so many matches. I also see where my error was.

Now, I seem to remember that for independent events, (P(A and B) = P(A) * P(B)). If that's correct, then with only two wheels spinning, there's only a 1% chance of them matching.

The 1% is the probability of one specific number matching (e.g., the probability of 9,9 or 3,3). I was thinking 1% was the probability for any two numbers to match. Now I see why the game is +EV.

Unfortunately, it is just a game, not for real money. If it were for real money I would have bled it dry before telling anyone /images/graemlins/grin.gif .

Thanks for the information on probabilities and the link. I will work through some examples to refresh my memory. /images/graemlins/smile.gif

Griff