Two Dice Challenge
Frankly, I was hoping it would be on the strip, since I’m not planning on renting a car this time.
With that in mind, and considering that that thread is generating several pages of replies every day, I figured we talk about the get-together venue and time, here.
There was an implication that it would be Wednesday, lunchtime, at Red Rock.
Is that true?
Any other ideas?
Quote: DJTeddyBearIn the two dice thread, there was mention of a get-together to prove/disapprove the 1/11 two dice result. And it was implied that the get-together would happen at some point during the table games show week, but maybe not near the show or even not on the strip.
Frankly, I was hoping it would be on the strip, since I’m not planning on renting a car this time.
With that in mind, and considering that that thread is generating several pages of replies every day, I figured we talk about the get-together venue and time, here.
There was an implication that it would be Wednesday, lunchtime, at Red Rock.
Is that true?
Any other ideas?
link to original post
I'm game for a get-together, but I agree to keeping it on the strip due to not renting a car. What's this about proving a dice result?
How could you miss a thread that created over 30 pages in less than a week?Quote: SphinxOfCupsWhat's this about proving a dice result?
link to original post
Anyway, the best post about the challenge is on page 14, here: https://wizardofvegas.com/forum/questions-and-answers/all-other/37156-two-dice-puzzle-part-trois/3/#post850487
Note, PLEASE don’t discuss the problem or the challenge here except to discuss the location and date/time of the challenge.
On that note, I would like it to be on strip since I won’t have a car.
And if we can’t come up with a better location, depending on the number of people, we could do it in my hotel room at Excalibur.
So I guess I should also ask for a head count.
Who’s in?
Quote: DJTeddyBearHow could you miss a thread that created over 30 pages in less than a week?link to original postQuote: SphinxOfCupsWhat's this about proving a dice result?
link to original post
I only hang around the Game Inventors forum.
Quote: DJTeddyBear
On that note, I would like it to be on strip since I won’t have a car.
And if we can’t come up with a better location, depending on the number of people, we could do it in my hotel room at Excalibur.
So I guess I should also ask for a head count.
Who’s in?
link to original post
I would say count me in as a maybe. How much time will there be for exhibitors to go elsewhere for lunch? I was expecting to need to man the booth pretty much all day.
The challenge is scheduled after the days Cutting Edge events. The table game showcase closes at 5:30 that day. I'd like to suggest those who want to meet at a restaurant, that I won't name here, but it is one one of the casinos south of Tropicana. I'll name it by PM on a "need to know" basis. Then, let's go to the Loquacious suite for the challenge.
At this time, can I get a list of names interested in dinner and/or the challenge. So far I've got LoquaciousMoFW, Wizard, and DJTeddyBear.
But if I may, challenge first, then dinner with waitress trivia, then maybe group gambling. Or is there a chance the challenge will take a long time? Hmmm… maybe dinner first is better.
I also suggest, depending on how many are joining us, get a reservation for dinner.
Quote: DJTeddyBearObviously, I’m in for both with no real preference of which is first.
But if I may, challenge first, then dinner with waitress trivia, then maybe group gambling. Or is there a chance the challenge will take a long time? Hmmm… maybe dinner first is better.
I also suggest, depending on how many are joining us, get a reservation for dinner.
link to original post
This challenge should take only an hour or so. I'm open to some tiles after the challenge at the MGM, based on group interest. I can also bring tiles and we can play against each other.
Agreed on the reservation.
Reminder -- trying to get a list of names going.
Quote: WizardQuote: DJTeddyBearObviously, I’m in for both with no real preference of which is first.
But if I may, challenge first, then dinner with waitress trivia, then maybe group gambling. Or is there a chance the challenge will take a long time? Hmmm… maybe dinner first is better.
I also suggest, depending on how many are joining us, get a reservation for dinner.
link to original post
This challenge should take only an hour or so. I'm open to some tiles after the challenge at the MGM, based on group interest. I can also bring tiles and we can play against each other.
Agreed on the reservation.
Reminder -- trying to get a list of names going.
link to original post
I'm on board, definitely looking to integrate myself more with the scene and community. To confirm we're looking at that Wednesday evening?
Quote: SphinxOfCupsTo confirm we're looking at that Wednesday evening?
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Yes.
Quote: Wizard(snip) Here are some tentative details:
- Challenge to take place June 8 in Las Vegas
- LoquaciousMoFW will wager $5 per shake
- I have kindly upped the win for two 2's to 8 to 1.
- LoquaciousMoFW has agreed to keep playing until losing $200. At that point, he/she has the option to back out.
- If for some reason LoquaciousMoFW can't be present, he/she has agreed to front money and have a proxy play for him/her
(snip)
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Under the above rules, I want to bet^^^ on " that either side has lost less than $200 within 125 resolved*** bets" at even money (1/1).
^^^: I would like to wager between $100 and $500.
***: A bet is considered resolved, when it does not end in a push.
Anyone interested, just pm me.
Quote: ksdjdjUnder the above rules, I want to bet^^^ on " that either side has lost less than $200 within 125 resolved*** bets" at even money (1/1).
^^^: I would like to wager between $100 and $500.
***: A bet is considered resolved, when it does not end in a push.
Anyone interested, just pm me.
link to original post
Loq can expect to lose 2/11 units per bet resolved, or $0.91. In 125 resolved bets, he should be down $113.64, well under $200.
Quote: WizardLoq can expect to lose 2/11 units per bet resolved, or $0.91. In 125 resolved bets, he should be down $113.64, well under $200.
With resolved bets paying either $5 or $40, under what circumstances could the bettor be down $113.64...or any fraction of a whole dollar for that matter?
Does this female friend know you refer to her as eye candy? 🤣Quote: WizardI'm happy to say I arranged for a female friend to act as Loq's proxy for the challenge. I do plan to make a video of the challenge and she will definitely add some eye candy to it.
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Quote: coachbellyWith resolved bets paying either $5 or $40, under what circumstances could the bettor be down $113.64...or any fraction of a whole dollar for that matter?
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As I asked in a previous post, when did schools quit teaching what an average is?
Quote: ksdjdjUnder the above rules, I want to bet^^^ on " that either side has lost less than $200 within 125 resolved*** bets" at even money (1/1).
^^^: I would like to wager between $100 and $500.
***: A bet is considered resolved, when it does not end in a push.
I get, within 125 rolls where at least one die is a 2:
The probability that the betting side is behind $200 at some point is 41.58745 %
The probability that the betting side is ahead $200 or more at some point is 1 / 17.5
The probability that neither side ever loses at least $200 is 52.71884 %
Behind $200 at some point
1,000,026,172,663,193,672,068,917,808,364,091,352,612,745,827,399,766,227,575,509,060,806,339,601,868,139,449,193,572,265,814,010,000,000,000,000,000,000,000,000,000,000,000,000,000 / 2,404,634,482,291,387,436,999,035,112,467,943,821,264,376,750,252,588,937,017,049,629,550,269,007,496,589,852,703,863,870,743,390,179,184,382,737,155,521,565,532,657,390,324,999,801
Ahead $200 or more at some point:
6,291,114,915,513,567,923,906,831,411,310,509,736,877,586,729,609,625,721,037,556,887,514,655,502,527,767,082,058,865,876,030,297,363,928,015,959,852,785,969,427,362,876,746,383,105,985,908,612,854,510 / 110,492,304,881,919,948,051,599,917,336,894,646,763,142,837,983,882,725,738,117,023,041,891,695,206,542,537,091,047,761,343,697,223,182,477,961,309,009,064,566,370,099,164,421,021,416,584,499,474,139,961
Either of these happens at some point:
52,242,047,405,889,288,413,301,758,578,995,133,421,038,943,621,711,568,852,833,657,310,810,632,005,947,760,991,861,601,221,542,930,165,889,055,735,462,785,969,427,362,876,746,383,105,985,908,612,854,510 / 110,492,304,881,919,948,051,599,917,336,894,646,763,142,837,983,882,725,738,117,023,041,891,695,206,542,537,091,047,761,343,697,223,182,477,961,309,009,064,566,370,099,164,421,021,416,584,499,474,139,961
Quote: Wizardwhen did schools quit teaching what an average is?
Oh I see...answering a question with a question, eh?
Is the Wizard Jewish. Does it seem like it?
Quote: WizardQuote: coachbellyWith resolved bets paying either $5 or $40, under what circumstances could the bettor be down $113.64...or any fraction of a whole dollar for that matter?
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As I asked in a previous post, when did schools quit teaching what an average is?
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I think that may be in 5th or 6th grade curriculum.
I'll accept whatever punishment is to be meted out.
Wager: I want to bet^^^ even money (1/1) on "that there will be between 0 and 4 " two 2's ", after the first 50 resolved bets" (50 bets that don't involve a push).
^^^: I would like to wager between $100 and $500.
Note: In case my wording isn't clear, I will win the bet if 0 to 4 "two 2's" are rolled, and you will win the bet if 5 (or more) "two 2's" are rolled, in the first 50 resolved bets.
Quote: DieterI'll accept whatever punishment is to be meted out.
Meted out by whom?
Take a month off and get in the game without the stripes on.
Maybe then you can answer why the bettor should expect a condition that cannot occur.
I'm sure I learned "average" (arithmetic mean) in elementary school. (Actually, I was most likely introduced to it via baseball stats outside of school, but I digress.) But I'm pretty sure I never heard the term "Arithmetic Mean" until high school.Quote: ksdjdjEven though "average" can mean more than one thing in Math, I take it to be the "Arithmetic Mean " in relation to this thread and "in general" (but I could be wrong).
I'm guessing it is outlined somewhere in the Two Dice Puzzle thread, but what are the exact terms of this challenge? I'd rather not have to wade through that whole thread to find it, for my mental health's sake (plus I'm lazy!)
Quote: ksdjdjMy previous offer is still open (see link here ) but I also want to put forward the wager below:
Wager: I want to bet^^^ even money (1/1) on "that there will be between 0 and 4 " two 2's ", after the first 50 resolved bets" (50 bets that don't involve a push).
^^^: I would like to wager between $100 and $500.
Note: In case my wording isn't clear, I will win the bet if 0 to 4 "two 2's" are rolled, and you will win the bet if 5 (or more) "two 2's" are rolled, in the first 50 resolved bets.
link to original post
The probability of there being between 0 and 4 "two 2s" in the first 50 resolved bets is about 0.5186.
Quote: ThatDonGuyThe probability of there being between 0 and 4 "two 2s" in the first 50 resolved bets is about 0.5186.
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I'd love to have you at the challenge, giving a running commentary on the odds.
Quote: WizardQuote: ThatDonGuyThe probability of there being between 0 and 4 "two 2s" in the first 50 resolved bets is about 0.5186.
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I'd love to have you at the challenge, giving a running commentary on the odds.
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Actually, where there does appear to be a formula for calculating, say, the probability of reaching +200 before -200 from a particular point, it's somewhat complicated, and the formula itself varies based on the current point.
Since the bets are in steps of $5, let's simplify it to bets of 1, so each bet either wins 8 or loses 1, and the target is +/- 40.
If the current profit is P, starting at 0, the probability of reaching +40 before reaching -40 appears to be, if I am reading Ethier correctly:
1 - (10/11)^(40 + P) * f(40 - P) / f(80)
where f(k) = the sum over all n from 0 to floor((k-1) / 9) of {(-1)^n * combin(n - 8k + 1, n) * (10^8 / 11^9)^n}
It shouldn't be that hard for somebody to whip up an app that can run a simulation for a result from any particular profit point.
What else can "average" mean in math?Quote: ksdjdjEven though "average" can mean more than one thing in Math, I take it to be the "Arithmetic Mean " in relation to this thread and "in general" (but I could be wrong).
Note: See link here , for a fairly simple explanation.
link to original post
Quote: Ace2What else can "average" mean?Quote: ksdjdjEven though "average" can mean more than one thing in Math, I take it to be the "Arithmetic Mean " in relation to this thread and "in general" (but I could be wrong).
Note: See link here , for a fairly simple explanation.
link to original post
link to original post
Average could be mean, median ,or mode.
you can add all the numbers together and divide them to get an average figure.
you can look for the number that half the subjects are on either side of it
you can make the most popular figure the average.
I make a million dollars, my ten employees make $100,000 each.
I can say the average salary is almost $200,000, which is true
My employees can say the average is $100,000 which is true as it is the most popular, and half the employees make it or more, and half make it or less.
Quote: Joeman(snip) I'm guessing it is outlined somewhere in the Two Dice Puzzle thread, but what are the exact terms of this challenge? I'd rather not have to wade through that whole thread to find it, for my mental health's sake (plus I'm lazy!)
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I am pretty sure these two links below sum up the rules and terms etc for the challenge (unless they have changed?):
Link_1
Link_2
Nope. The average is the mean. The median is the median. The mode is the mode.Quote: billryanQuote: Ace2What else can "average" mean?Quote: ksdjdjEven though "average" can mean more than one thing in Math, I take it to be the "Arithmetic Mean " in relation to this thread and "in general" (but I could be wrong).
Note: See link here , for a fairly simple explanation.
link to original post
link to original post
Average could be mean, median ,or mode.
you can add all the numbers together and divide them to get an average figure.
you can look for the number that half the subjects are on either side of it
you can make the most popular figure the average.
I make a million dollars, my ten employees make $100,000 each.
I can say the average salary is almost $200,000, which is true
My employees can say the average is $100,000 which is true as it is the most popular, and half the employees make it or more, and half make it or less.
link to original post
The three are equivalent only in a standard normal distribution...such as the probability distribution of 100 coin flips. Equivalent in this special case but never the same
Quote: Ace2Nope. The average is the mean. The median is the median. The mode is the mode.Quote: billryanQuote: Ace2What else can "average" mean?Quote: ksdjdjEven though "average" can mean more than one thing in Math, I take it to be the "Arithmetic Mean " in relation to this thread and "in general" (but I could be wrong).
Note: See link here , for a fairly simple explanation.
link to original post
link to original post
Average could be mean, median ,or mode.
you can add all the numbers together and divide them to get an average figure.
you can look for the number that half the subjects are on either side of it
you can make the most popular figure the average.
I make a million dollars, my ten employees make $100,000 each.
I can say the average salary is almost $200,000, which is true
My employees can say the average is $100,000 which is true as it is the most popular, and half the employees make it or more, and half make it or less.
link to original post
The three are equivalent only in a standard normal distribution...such as the probability distribution of 100 coin flips. Equivalent in this special case but never the same
link to original post
Billryan is correct that the word average, at least as used in common English as taught to me when I was in school, is broad enough to cover mean, median and mode. As used colloquially it usually is meant as mean.
As a bit of trivia, between the three parts of the puzzle thread, I count about 1,680 posts over almost nine years.
| Current | Prob of Win |
|---|---|
| -195 | 1 / 1053.032 |
| -190 | 1 / 513.821 |
| -185 | 1 / 334.223 |
| -180 | 1 / 244.529 |
| -175 | 1 / 190.795 |
| -170 | 1 / 155.042 |
| -165 | 1 / 129.564 |
| -160 | 1 / 110.507 |
| -155 | 1 / 95.73 |
| -150 | 1 / 83.95 |
| -145 | 1 / 74.349 |
| -140 | 1 / 66.382 |
| -135 | 1 / 59.672 |
| -130 | 1 / 53.949 |
| -125 | 1 / 49.015 |
| -120 | 1 / 44.724 |
| -115 | 1 / 40.96 |
| -110 | 1 / 37.636 |
| -105 | 1 / 34.683 |
| -100 | 1 / 32.044 |
| -95 | 1 / 29.675 |
| -90 | 1 / 27.538 |
| -85 | 1 / 25.603 |
| -80 | 1 / 23.845 |
| -75 | 1 / 22.242 |
| -70 | 1 / 20.777 |
| -65 | 1 / 19.433 |
| -60 | 1 / 18.198 |
| -55 | 1 / 17.06 |
| -50 | 1 / 16.01 |
| -45 | 1 / 15.038 |
| -40 | 1 / 14.138 |
| -35 | 1 / 13.302 |
| -30 | 1 / 12.525 |
| -25 | 1 / 11.801 |
| -20 | 1 / 11.127 |
| -15 | 1 / 10.497 |
| -10 | 1 / 9.908 |
| -5 | 1 / 9.358 |
| 0 | 1 / 8.843 |
| 5 | 1 / 8.359 |
| 10 | 1 / 7.906 |
| 15 | 1 / 7.48 |
| 20 | 1 / 7.08 |
| 25 | 1 / 6.703 |
| 30 | 1 / 6.349 |
| 35 | 1 / 6.016 |
| 40 | 1 / 5.702 |
| 45 | 1 / 5.406 |
| 50 | 1 / 5.126 |
| 55 | 20.569 % |
| 60 | 21.683 % |
| 65 | 22.85 % |
| 70 | 24.073 % |
| 75 | 25.349 % |
| 80 | 26.693 % |
| 85 | 28.11 % |
| 90 | 29.605 % |
| 95 | 31.175 % |
| 100 | 32.817 % |
| 105 | 34.526 % |
| 110 | 36.294 % |
| 115 | 38.113 % |
| 120 | 40.128 % |
| 125 | 42.288 % |
| 130 | 44.55 % |
| 135 | 46.878 % |
| 140 | 49.241 % |
| 145 | 51.613 % |
| 150 | 53.973 % |
| 155 | 56.304 % |
| 160 | 60.277 % |
| 165 | 63.888 % |
| 170 | 67.171 % |
| 175 | 70.155 % |
| 180 | 72.868 % |
| 185 | 75.335 % |
| 190 | 77.577 % |
| 195 | 79.616 % |
Yes, there is a 20% chance of reaching -200 before +200 if you start at +195.
| Profit | # Bets Remain |
|---|---|
| -195 | 5 |
| -190 | 10 |
| -185 | 15 |
| -180 | 20 |
| -175 | 25 |
| -170 | 30 |
| -165 | 35 |
| -160 | 40 |
| -155 | 45 |
| -150 | 50 |
| -145 | 54 |
| -140 | 59 |
| -135 | 64 |
| -130 | 69 |
| -125 | 73 |
| -120 | 78 |
| -115 | 82 |
| -110 | 87 |
| -105 | 91 |
| -100 | 96 |
| -95 | 100 |
| -90 | 105 |
| -85 | 109 |
| -80 | 113 |
| -75 | 117 |
| -70 | 121 |
| -65 | 125 |
| -60 | 129 |
| -55 | 133 |
| -50 | 137 |
| -45 | 140 |
| -40 | 144 |
| -35 | 147 |
| -30 | 151 |
| -25 | 154 |
| -20 | 157 |
| -15 | 160 |
| -10 | 163 |
| -5 | 166 |
| 0 | 169 |
| 5 | 171 |
| 10 | 174 |
| 15 | 176 |
| 20 | 178 |
| 25 | 180 |
| 30 | 182 |
| 35 | 183 |
| 40 | 185 |
| 45 | 186 |
| 50 | 187 |
| 55 | 187 |
| 60 | 188 |
| 65 | 188 |
| 70 | 188 |
| 75 | 188 |
| 80 | 187 |
| 85 | 186 |
| 90 | 185 |
| 95 | 184 |
| 100 | 182 |
| 105 | 179 |
| 110 | 177 |
| 115 | 174 |
| 120 | 170 |
| 125 | 166 |
| 130 | 162 |
| 135 | 157 |
| 140 | 151 |
| 145 | 146 |
| 150 | 140 |
| 155 | 135 |
| 160 | 124 |
| 165 | 113 |
| 170 | 104 |
| 175 | 96 |
| 180 | 88 |
| 185 | 81 |
| 190 | 75 |
| 195 | 69 |
It’s in just a couple hours.Quote: ThatDonGuyEr, when (and where) is the challenge, anyway?
link to original post
For the location, send a PM to the Wiz for the details.
Note: I don’t know if everyone who asks will receive the info. I mean, it might be semi-private.
Right between the Star Trek Convention and the Dungeons & Dragons Annual MeetingQuote: ThatDonGuyEr, when (and where) is the challenge, anyway?
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Quote: Ace2Right between the Star Trek Convention and the Dungeons & Dragons Annual MeetingQuote: ThatDonGuyEr, when (and where) is the challenge, anyway?
link to original post
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Got it - it's "in Vegas."
And you don't want to know how many people whine year after year that the GenCon gaming convention, which gets its name from the fact that originally it was held in the city where the company that originally made D&D was headquartered, is far too big for Indianapolis (seriously, the annual "housing lottery" to see who lets shut out of a hotel that isn't within 10 miles, much less "walking distance," from the convention center, is a sight to behold) and needs to be moved to somewhere with far more concentrated hotel space - say, the Las Vegas strip. Don't worry; it will almost certainly never happen, for two reasons; it needs to be much closer to the eastern half of the country, and yes, they do realize that quite a few of the attendees are under 21.
Recall that LoquaciousMoFW agreed to keep playing until he lost $200, betting $5 at a time. That said, here is a count of every turn.
0 2's: 225
1 2: 104
2 2's: 8
Total: 337
The net win for LoquaciousMoFW was 104*-5 + 8*40 = -200.
The ratio of two 2's to bets resolved was 8/112 = 7.14% or 1 in 14.
The expected number of two 2's in 112 bets resolved is 112/11 = 10.18. So, there was a small shortfall in two 2's, but nothing to be suspicious about. It took quite a while to see the first pair of 2's. As I recall, the proxy was down about $120 by the first time it happened.
The total rolling time I would estimate was a little under an hour.
A video will be forthcoming documenting the event.
I would like to thank LoquaciousMoFW for both the nice suite to perform the challenge and taking the losing side. I also thank the Proxy for her help.
Didn't Coachbelly show?
Quote: WizardRecall that LoquaciousMoFW agreed to keep playing until he lost $200, betting $5 at a time. That said, here is a count of every turn.
0 2's: 225
1 2: 104
2 2's: 8
Total: 337
link to original post
112 resolutions does seem a little low, but it turns out that 3/8 of the time, the bettor loses the entire $200 in 112 or fewer resolutions. The median length is around 148.
Quote: Wizardhere is a count of every turn.
0 2's: 225
1 2: 104
2 2's: 8
Total: 337
link to original post
So, there were 112 trials in which at least one of the two dice was a 2. And only 8 out of 112 times were both dice a 2. If the frequency of two 2s in this challenge was 1 in 6 then we would have expected an average of about 18.7 pairs of 2's out of 112 trials in which at least one of the dice was a two.
I imagine that no one has changed their mind because of this trial, and that the earth continues to spin around the sun.
In other words, if you take the total number of times there was a two and divide that by 6, you should get the number of times both were two. Since there was a 2 on 112 tosses, the NULL hypothesis predicts that 112*(1/6) = 18.7 of those should have been 2-2 and the other 84.3 should show only one two.
Chi-square analysis of this experiment gives a p-Value of 0.004. This is widely recognized as sufficient to negate the NULL hypothesis. Often, p < 0.05 is used as evidence to negate the NULL hypothesis, so this is pretty strong.
Quote: teliotThe NULL hypothesis here was "1/6-th of the time that one of the dice shows a two they will both show a two."
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Meanwhile, here is the same table with expected totals based on correct probabilities.
| Twos | Ways | Actual | Expected |
|---|---|---|---|
| 0 | 25 | 225 | 234.028 |
| 1 | 10 | 104 | 93.611 |
| 2 | 1 | 8 | 9.361 |
| Total | 36 | 337 | 337.000 |
p value = 0.427605218
Well, I maintain that the world is not round.Quote: gordonm888I imagine that no one has changed their mind because of this trial, and that the earth continues to spin around the sun.
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(It’s an oblate spheroid.)
