## Poll

 Yes -- I'm a sucker No votes (0%) Call me maybe. No votes (0%) No -- I only look stupid. 4 votes (66.66%) Total eclipse reminder 4-8-2024 3 votes (50%) I like to ruin magic tricks. No votes (0%) Mamie Van Doren 1 vote (16.66%) I was a Covid hoarder No votes (0%) What is the meaning to Lawyers in Love? 1 vote (16.66%) I prefer Ken Jennings 2 votes (33.33%) I prefer Mayim Bialik. 1 vote (16.66%)

6 members have voted

Wizard
Joined: Oct 14, 2009
• Posts: 24364
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.

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
Joined: Oct 14, 2009
• Posts: 24364
November 23rd, 2021 at 6:56:57 PM permalink
Quote: DeMango

So you roll aces twice in a row you lose???

Yep.
It's not whether you win or lose; it's whether or not you had a good bet.
Wizard
Joined: Oct 14, 2009
• Posts: 24364
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)

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
Joined: Jun 15, 2018
• Posts: 3094
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
Joined: Oct 2, 2017
• Posts: 1686
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)

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 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
Joined: Jun 22, 2011
• Posts: 5417
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}:
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
Joined: Oct 14, 2009
• Posts: 24364
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
Joined: Nov 2, 2009
• Posts: 10735
November 24th, 2021 at 4:29:34 PM permalink

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
Joined: Jul 11, 2021
• Posts: 3
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
Joined: Nov 2, 2009