miplet
Joined: Dec 1, 2009
• Posts: 1963
September 13th, 2014 at 9:07:59 PM permalink
Quote: ThatDonGuy

If it's Coverall, I did a Monte Carlo with 50 million runs, and won 1/763 of the time.

Spread sheet transitional matrix (or whatever they are called): 1 in 766.867413 google doc here
“Man Babes” #AxelFabulous
Dieter
Joined: Jul 23, 2014
• Posts: 1608
September 14th, 2014 at 7:22:33 AM permalink
Are there payouts for patterns other than coverall?

Those might make it a playable game.
May the cards fall in your favor.
GWAE
Joined: Sep 20, 2013
• Posts: 9854
September 14th, 2014 at 7:35:50 AM permalink
Quote: miplet

Spread sheet transitional matrix (or whatever they are called): 1 in 766.867413 google doc here

I can not imagine how long it took people to do that before excel existed.
Expect the worst and you will never be disappointed. I AM NOT PART OF GWAE RADIO SHOW
ThatDonGuy
Joined: Jun 22, 2011
• Posts: 4829
September 14th, 2014 at 8:33:31 AM permalink
Quote: miplet

Spread sheet transitional matrix (or whatever they are called): 1 in 766.867413

I started out trying one of those, but didn't have the "House light bulb moment" to figure it out. It never dawned on me that it's really five identical independent problems ("what is the probability that, if you draw a number from {1, 2, ..., 15} 22 times with replacement, each of the numbers from 1 to 5 will be drawn at least once?").

For those of you playing at home:
Assume the card is B1-5, I16-20, N31-35, G46-50, and O61-65
Concentrate on just the Bs:
Let S(n,k) be the probability of k different numbers from 1-5 being drawn after n Bs have been drawn.
S(0,0) = 1
S(n,0) = S(n-1,0) x 10/15
S(n,k) = (S(n-1,k-1) x (6-k)/15) + (S(n-1,k) x (1 - (6-k)/15)), for 0 < k < 5
S(n,5) = (S(n-1,4) x 1/15) + S(n-1,5)
S(22,5) is the probability that you have drawn all of the numbers from 1 to 5 in 22 draws
The probabilities of drawing 16-20, 31-35, 46-50, and 61-65 are the same, so the overall probability of winning is S(22,5)5.
98Clubs
Joined: Jun 3, 2010
• Posts: 1728
September 14th, 2014 at 5:16:53 PM permalink
busted.
Some people need to reimagine their thinking.
DaveG44131
Joined: Sep 11, 2014
• Posts: 6
September 16th, 2014 at 9:11:51 AM permalink
Quote: miplet

Spread sheet transitional matrix (or whatever they are called): 1 in 766.867413 google doc here

I tried to copy your Google doc to analyze the 4x4 variant described in my OP and got 1 in approximately 828. Is this right?
DaveG44131
Joined: Sep 11, 2014
• Posts: 6
September 16th, 2014 at 9:13:27 AM permalink
Quote: Dieter

Are there payouts for patterns other than coverall?

Those might make it a playable game.

Some machines had payouts for patterns other than coverall, but then the payout for coverall was reduced.
miplet
Joined: Dec 1, 2009
• Posts: 1963
September 16th, 2014 at 10:02:53 AM permalink
Quote: DaveG44131

Quote: miplet

Spread sheet transitional matrix (or whatever they are called): 1 in 766.867413 google doc here

I tried to copy your Google doc to analyze the 4x4 variant described in my OP and got 1 in approximately 828. Is this right?

I'm getting 361.0339019 . I added another sheet.
“Man Babes” #AxelFabulous
DaveG44131
Joined: Sep 11, 2014
• Posts: 6
September 16th, 2014 at 11:06:30 AM permalink
Quote: miplet

Quote: DaveG44131

Quote: miplet

Spread sheet transitional matrix (or whatever they are called): 1 in 766.867413 google doc here

I tried to copy your Google doc to analyze the 4x4 variant described in my OP and got 1 in approximately 828. Is this right?

I'm getting 361.0339019 . I added another sheet.

Thanks for checking. I had a feeling I miscopied something.
Dieter
Joined: Jul 23, 2014
• Posts: 1608
September 16th, 2014 at 11:56:24 AM permalink
Quote: DaveG44131

Some machines had payouts for patterns other than coverall, but then the payout for coverall was reduced.

Yeah, but an all or nothing payout eats into your playing money quickly.

Some little wins along the way mean you have a better chance of not going broke before the big one hits.
May the cards fall in your favor.