Bonus Craps for Crapless Craps
Poll
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6 members have voted
Quote: DJTeddyBearPut 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.
Quote: DeMangoSo you roll aces twice in a row you lose???
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Yep.
Quote: ThatDonGuyI 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)
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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.
I dont think Poisson/calculus would be an efficient method for this problem. Too many combinations to be summedQuote: WizardQuote: ThatDonGuyI 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)
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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.
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Quote: WizardThank 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.
Intro:
Its not Stratosphere anymore. Its 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 Id like to see an explanation of what the expected return per $1,000 means.
Id 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?
That type of info is rarely, if ever, made public on ANY progressive jackpot.Quote: redneckrounderOne thing I didnt see mentioned is what is the meter rate, i.e. how much of your five dollar bet goes into the progressive?
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