April 28th, 2014 at 10:42:09 AM
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If a game has two bars with 5 segments that must be populated before you enter the bonus game and knowing the independent probabilities (Pa and Pb) of getting the respective bonus symbols (1 on reel 1, 2 on reel 5 don't have to be on a line like scatters) can one determine on average how many spins are needed. It would seem to be 5/min(Pa, Pb) spins but my simulator says otherwise. For Pa = 1/6 and Pb = 1/7 the number of spins I would expect would be 35 (5/1/7) but my simulator says 40 spins are required on average. That's 5 more plays.

April 28th, 2014 at 12:10:48 PM
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I am assuming the bonus works like this: you need to get 5 of the "reel 1" bonus symbols (probability 1/6 on each spin) and 5 of the "reel 2" bonus symbols (probability 1/7 on each spin) in order to get the bonus.

The reason the answer is more than 35 is, your method assumes that you will always get the 5 bonus symbols with probability 1/6 before you get the 5 with probability 1/7.

Here's how I get my answer of about 40.11022:

Let P(a,b) be the number of spins needed to get from "a" of the 1/6 bonus symbols and "b" of the 1/7 bonus symbols to 5 of each.

P(5,5) = 0

For b = 5, P(a,5) = 1 + (1/6 * P(a+1, 5) + 5/6 * P(a,5))

1/6 * P(a,5) = 1 + 1/6 * P(a+1, 5)

P(a,5) = 6 + P(a+1, 5)

P(5,5) = 0 -> P(4,5) = 6 -> P(3,5) = 12 -> P(2,5) = 18 -> P(1,5) = 24 -> P(0,5) = 30

For a = 5, P(5,b) = 1 + (1/7 * P(5,b+1) + 6/7 * P(5,b))

1/7 * P(5,b) = 1 + 1/7 * P(5,b+1)

P(5,b) = 7 + P(5,b+1)

P(5,5) = 0 -> P(5,4) = 7 -> P(5,3) = 14 -> P(5,2) = 21 -> P(5,1) = 28 -> P(5,0) = 35

For a < 5 and b < 5, P(a,b) = 1 + (5/6 * 6/7 * P(a,b) + 1/6 * 6/7 * P(a+1,b) + 5/6 * 1/7 * P(a,b+1) + 1/6 * 1/7 * P(a+1,b+1))

P(a,b) = 1 + (5/7 * P(a,b) + 1/7 * P(a+1,b) + 5/42 * P(a,b+1) + 1/42 * P(a+1,b+1))

2/7 * P(a,b) = 1 + 1/7 * P(a+1,b) + 5/42 * P(a,b+1) + 1/42 * P(a+1,b+1)

P(a,b) = 7/2 + 1/2 * P(a+1,b) + 5/12 * P(a,b+1) + 1/12 * P(a+1,b+1)

P(a,b) = (42 + 6 * P(a+1,b) + 5 * P(a,b+1) + P(a+1,b+1)) / 12

Work backwards from the already calculated numbers - i.e. calculate P(4,4), then P(3,4), then P(2,4), then P(1,4), then P(0,4), then P(4,3) through P(0,3), then P(4,2) through P(0,2), then P(4,1) through P(0,1), then P(4,0) through P(0,0).

P(0,0) is the number of spins to get from 0 of each to 5 of each; this is the desired solution.

In fact, the exact answer is 2217935 / 55296.

The reason the answer is more than 35 is, your method assumes that you will always get the 5 bonus symbols with probability 1/6 before you get the 5 with probability 1/7.

Here's how I get my answer of about 40.11022:

Let P(a,b) be the number of spins needed to get from "a" of the 1/6 bonus symbols and "b" of the 1/7 bonus symbols to 5 of each.

P(5,5) = 0

For b = 5, P(a,5) = 1 + (1/6 * P(a+1, 5) + 5/6 * P(a,5))

1/6 * P(a,5) = 1 + 1/6 * P(a+1, 5)

P(a,5) = 6 + P(a+1, 5)

P(5,5) = 0 -> P(4,5) = 6 -> P(3,5) = 12 -> P(2,5) = 18 -> P(1,5) = 24 -> P(0,5) = 30

For a = 5, P(5,b) = 1 + (1/7 * P(5,b+1) + 6/7 * P(5,b))

1/7 * P(5,b) = 1 + 1/7 * P(5,b+1)

P(5,b) = 7 + P(5,b+1)

P(5,5) = 0 -> P(5,4) = 7 -> P(5,3) = 14 -> P(5,2) = 21 -> P(5,1) = 28 -> P(5,0) = 35

For a < 5 and b < 5, P(a,b) = 1 + (5/6 * 6/7 * P(a,b) + 1/6 * 6/7 * P(a+1,b) + 5/6 * 1/7 * P(a,b+1) + 1/6 * 1/7 * P(a+1,b+1))

P(a,b) = 1 + (5/7 * P(a,b) + 1/7 * P(a+1,b) + 5/42 * P(a,b+1) + 1/42 * P(a+1,b+1))

2/7 * P(a,b) = 1 + 1/7 * P(a+1,b) + 5/42 * P(a,b+1) + 1/42 * P(a+1,b+1)

P(a,b) = 7/2 + 1/2 * P(a+1,b) + 5/12 * P(a,b+1) + 1/12 * P(a+1,b+1)

P(a,b) = (42 + 6 * P(a+1,b) + 5 * P(a,b+1) + P(a+1,b+1)) / 12

Work backwards from the already calculated numbers - i.e. calculate P(4,4), then P(3,4), then P(2,4), then P(1,4), then P(0,4), then P(4,3) through P(0,3), then P(4,2) through P(0,2), then P(4,1) through P(0,1), then P(4,0) through P(0,0).

P(0,0) is the number of spins to get from 0 of each to 5 of each; this is the desired solution.

In fact, the exact answer is 2217935 / 55296.

April 28th, 2014 at 7:46:17 PM
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Wow and thanks, I'm still trying to digest all of this. In the first line P(a,5) = 1 + (1/6 * P(a+1, 5) + 5/6 * P(a,5), would that make this an infinite progression since replacing the 5/6 * P(a,5) with 5/6 * (1 + (1/6 * P(a+1, 5) + 5/6 * P(a,5))) adds P(a,5) yet again to the equation.

Thanks again,

Thanks again,

April 29th, 2014 at 6:06:30 AM
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Quote:hspumantiWow and thanks, I'm still trying to digest all of this. In the first line P(a,5) = 1 + (1/6 * P(a+1, 5) + 5/6 * P(a,5), would that make this an infinite progression since replacing the 5/6 * P(a,5) with 5/6 * (1 + (1/6 * P(a+1, 5) + 5/6 * P(a,5))) adds P(a,5) yet again to the equation.

You're not adding a P(a,5); you're replacing one with another. You're also removing 5/6 * P(a,5) and adding 5/6 * 5/6 * P(a,5).

If you were to continue the replacements:

P(a,5) = 1 + 1/6 * P(a+1,5) + 5/6 * P(a,5)

= 1 + 1/6 * P(a+1,5) + 5/6 * (1 + 1/6 * P(a+1,5) + 5/6 * P(a,5)

= (1 + 5/6) + 1/6 * P(a+1,5) * (1 + 5/6) + (5/6)

^{2}* P(a,5)

= (1 + 5/6) + 1/6 * P(a+1,5) * (1 + 5/6) + (5/6)

^{2}* (1 + 1/6 * P(a+1,5) + 5/6 * P(a,5))

= (1 + 5/6 + (5/6)

^{2}) + 1/6 * P(a+1,5) * (1 + 5/6 + (5/6)

^{2}) + (5/6)

^{3}* (1 + 1/6 * P(a+1,5) + 5/6 * P(a,5))

...

= 6 + (1/6 * P(a+1,5) * 6) + 0

= P(a+1,5) + 6

April 29th, 2014 at 7:41:11 AM
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I'm not seeing how you get rid of the P(a,5) if you continually replace the P(a,5) with an expression that contains P(a,5) within it.

= (1 + 5/6 + (5/6)2) + 1/6 * P(a+1,5) * (1 + 5/6 + (5/6)2) + (5/6)3 * (1 + 1/6 * P(a+1,5) + 5/6 * P(a,5))

becomes

= (1 + 5/6 + (5/6)2) + 1/6 * P(a+1,5) * (1 + 5/6 + (5/6)2) + (5/6)3 * (1 + 1/6 * P(a+1,5) + 5/6 * ( 1 + 1/6 * P(a+1,5) + 5/6 * P(a,5)))

which then becomes

= (1 + 5/6 + (5/6)2) + 1/6 * P(a+1,5) * (1 + 5/6 + (5/6)2) + (5/6)3 * (1 + 1/6 * P(a+1,5) + 5/6 * ( 1 + 1/6 * P(a+1,5) + 5/6 * 1 + 1/6 * P(a+1,5) + 5/6 * P(a,5))))

etc.

I don't see how you remove the P(a,5) to get to 6 + (1/6 * P(a+1,5) * 6) + 0 unless P(a,5) is progressing P(0,5), P(1,5), P(2,5)...P(4,5).

Thank you,

= (1 + 5/6 + (5/6)2) + 1/6 * P(a+1,5) * (1 + 5/6 + (5/6)2) + (5/6)3 * (1 + 1/6 * P(a+1,5) + 5/6 * P(a,5))

becomes

= (1 + 5/6 + (5/6)2) + 1/6 * P(a+1,5) * (1 + 5/6 + (5/6)2) + (5/6)3 * (1 + 1/6 * P(a+1,5) + 5/6 * ( 1 + 1/6 * P(a+1,5) + 5/6 * P(a,5)))

which then becomes

= (1 + 5/6 + (5/6)2) + 1/6 * P(a+1,5) * (1 + 5/6 + (5/6)2) + (5/6)3 * (1 + 1/6 * P(a+1,5) + 5/6 * ( 1 + 1/6 * P(a+1,5) + 5/6 * 1 + 1/6 * P(a+1,5) + 5/6 * P(a,5))))

etc.

I don't see how you remove the P(a,5) to get to 6 + (1/6 * P(a+1,5) * 6) + 0 unless P(a,5) is progressing P(0,5), P(1,5), P(2,5)...P(4,5).

Thank you,

April 29th, 2014 at 8:07:52 AM
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Quote:hspumantiI'm not seeing how you get rid of the P(a,5) if you continually replace the P(a,5) with an expression that contains P(a,5) within it.

The amount by which you multiply P(a,5) gets smaller and smaller - at first, 5/6, then (5/6)

^{2}, then (5/6)

^{3}, then (5/6)

^{4}, and so on; "eventually" this reaches zero.

Quote:hspumantiI don't see how you remove the P(a,5) to get to 6 + (1/6 * P(a+1,5) * 6) + 0 unless P(a,5) is progressing P(0,5), P(1,5), P(2,5)...P(4,5).

Look at the original equation:

P(a,5) = 1 + 1/6 * P(a+1,5) + 5/6 + P(a,5)

1/6 of the time, one spin from (a,5) will progress to (a+1,5), and 5/6 of the time, it will remain at (a,5).

This is how I get P(a,5) = 1 + 1/6 * P(a+1,5) + 5/6 * P(a,5).

Simply subtract (5/6 * P(a,5)) from both sides of the equation:

P(a,5) - 5/6 * P(a,5) = 1 + 1/6 * P(a+1,5) + 5/6 * P(a,5) - 5/6 * P(a,5)

1/6 * P(a,5) = 1 + 1/6 * P(a+1,5)

Multiply both sides by 6:

P(a,5) = 6 + P(a+1,5)

This is a recursive equation; we are given that P(5,5) = 0, so work backwards:

P(4,5) = 6 + P(5,5) = 6

P(3,5) = 6 + P(4,5) = 12

P(2,5) = 6 + P(3,5) = 18

P(1,5) = 6 + P(2,5) = 24

P(0,5) = 6 + P(1,5) = 30