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6 members have voted

Wizard
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Wizard
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November 23rd, 2021 at 6:55:58 PM permalink
Quote: DJTeddyBear

Put another way, your math is correct if there's only one participant when it hits.
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I agree with that. I was aware of the jackpot sharing rule. What I posted was not the finished product, just the one-player case.
It's not whether you win or lose; it's whether or not you had a good bet.
Wizard
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Wizard
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November 23rd, 2021 at 6:56:57 PM permalink
Quote: DeMango

So you roll aces twice in a row you lose???
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Yep.
It's not whether you win or lose; it's whether or not you had a good bet.
Wizard
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Wizard
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November 23rd, 2021 at 7:00:11 PM permalink
Quote: ThatDonGuy

I get the same numbers you do. To be more specific:

Prob(10) = 4,375 / 306,110,016 (0.00001429224713770883)
Prob(9) = 33,775 / 153,055,008 (0.00022067229580622393)
Prob(8) = 985,565 / 612,220,032 (0.0016098215486029684)
Prob(7) = 9,057,545 / 1,224,440,064 (0.007397295520052497)
Prob(6) = 546,475 / 22,674,816 (0.0241005263284165)
Prob(5) = 897,275 / 15,116,544 (0.05935715200511452)
Prob(Lose) = 42,331 / 46,656 (0.9073002400548698)
link to original post



Thank you for the confirmation! Did you do it with calculus? I did a Markov chain and verified it only for hitting all 10 numbers with your usual method.

I didn't want to fuss with creating and solving 252 equations for all the ways to hit exactly five numbers, but maybe you know something I don't.
It's not whether you win or lose; it's whether or not you had a good bet.
ChumpChange
ChumpChange
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November 23rd, 2021 at 7:08:01 PM permalink
You've got a 96.66% chance of winning $10 or $0. Looks like a crapless one-armed bandit to me.
Ace2
Ace2
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November 23rd, 2021 at 8:48:16 PM permalink
Quote: Wizard

Quote: ThatDonGuy

I get the same numbers you do. To be more specific:

Prob(10) = 4,375 / 306,110,016 (0.00001429224713770883)
Prob(9) = 33,775 / 153,055,008 (0.00022067229580622393)
Prob(8) = 985,565 / 612,220,032 (0.0016098215486029684)
Prob(7) = 9,057,545 / 1,224,440,064 (0.007397295520052497)
Prob(6) = 546,475 / 22,674,816 (0.0241005263284165)
Prob(5) = 897,275 / 15,116,544 (0.05935715200511452)
Prob(Lose) = 42,331 / 46,656 (0.9073002400548698)
link to original post



Thank you for the confirmation! Did you do it with calculus? I did a Markov chain and verified it only for hitting all 10 numbers with your usual method.

I didn't want to fuss with creating and solving 252 equations for all the ways to hit exactly five numbers, but maybe you know something I don't.
link to original post

I don’t think Poisson/calculus would be an efficient method for this problem. Too many combinations to be summed
It’s all about making that GTA
ThatDonGuy
ThatDonGuy
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November 24th, 2021 at 7:22:43 AM permalink
Quote: Wizard

Thank you for the confirmation! Did you do it with calculus? I did a Markov chain and verified it only for hitting all 10 numbers with your usual method.


No - I used brute force computing. Well, I suppose it's identical to a Markov chain.
However, the probability for 10 is simple; there are 10! permutations, each of which has a probability of 1 x 2 x 3 x 4 x 5 x 5 x 4 x 3 x 2 x 1 / 36^10 of occurring, so the result = 3,628,800 x 14,400 / 6^20.

First, I generated all 10! = 3,628,800 permutations of {2, 3, 4, 5, 6, 8, 9, 10, 11, 12}:
1. Start with {2, 3} and {3,2}
2. For each number from 4 to 12 (except 7), replace each previous permutation with multiple permutations where the new number is in each possible position (e.g. replace {2, 3} with {4, 2, 3}, {2, 4, 3}, and {2, 3, 4}; when all of the size-3 permutations are done, replace {4, 2, 3} with {5, 4, 2, 3}, {4, 5, 2, 3}, {4, 2, 5, 3}, and {4, 2, 3, 5}).
Then, for each 10-permutation:
1. Calculate the probability of the first 5 numbers coming up then one of those numbers or 7 (i.e. a 5-roll win).
If the first five numbers are {2, 8, 6, 10, 3}, this is 1/36 x 5/36 x 5/36 x 3/36 x 2/36 x (1 + 5 + 5 + 3 + 2 + 6)/36.
2. Do the same for the 6-roll wins, the 7-roll wins, and so on through the 10-roll wins.
3. Sum the 5-roll win probabilities, then divide by 5! as each permutation of 5 numbers appears in 5! of the 10-permutations; this is the overall probability of a 5-roll win.
4. Sum the 6-roll win probabilities, then divide by 4! as each permutation of 6 numbers appears in 4! of the 10-permutations; this is the overall probability of a 5-roll win.
5. Do the same for the 7-roll through 10-roll wins.

Wizard
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Wizard
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November 24th, 2021 at 11:37:15 AM permalink
My page on the Bonus Craps Progressive is up. I welcome all comments.
It's not whether you win or lose; it's whether or not you had a good bet.
DJTeddyBear
DJTeddyBear
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November 24th, 2021 at 4:29:34 PM permalink
Good article. I see you’ve already incorporated some of my comments.


Intro:
It’s not Stratosphere anymore. It’s Strat. Or actually: STRAT Hotel, Casino & SkyPod

Rules paragraph 1:
The All bet should be the All/Tall/Small bet.


Maybe I am underestimating the intelligence of your readers, but I’d like to see an explanation of what the expected return per $1,000 means.

I’d also like to see a comment, or even a chart, that unless the progressive is at least $1,500 per player, it pays less than a nine hit pay out - or is there a special rule for that?
I invented a few casino games. Info: http://www.DaveMillerGaming.com/ ————————————————————————————————————— Superstitions are silly, childish, irrational rituals, born out of fear of the unknown. But how much does it cost to knock on wood? 😁
redneckrounder
redneckrounder
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December 4th, 2021 at 10:44:09 PM permalink
One thing I didn’t see mentioned is what is the meter rate, i.e. how much of your five dollar bet goes into the progressive? A $1.00 increase for every $5 bet is going to make the breakeven number a lot different then $0.05 for every $5 bet.
DJTeddyBear
DJTeddyBear
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December 5th, 2021 at 3:45:08 AM permalink
Quote: redneckrounder

One thing I didn’t see mentioned is what is the meter rate, i.e. how much of your five dollar bet goes into the progressive?
link to original post

That type of info is rarely, if ever, made public on ANY progressive jackpot.
I invented a few casino games. Info: http://www.DaveMillerGaming.com/ ————————————————————————————————————— Superstitions are silly, childish, irrational rituals, born out of fear of the unknown. But how much does it cost to knock on wood? 😁

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