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ThatDonGuy
ThatDonGuy
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December 22nd, 2021 at 5:50:56 PM permalink
Quote: Ace2

I hope I phrased the question properly

With 24 ways to roll a point and 6 ways to roll a seven, the expectation is to roll a point number 4 times (including repeats) before rolling a seven. Therefore the average number of distinct (not repeated) points rolled must be at least 1.
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You phrased it correctly; I read it wrong.

The number I got is the number of distinct point numbers made before sevening out. You want the number of distinct place bets won before a seven is rolled.

Questions:
1. Are the bets placed before the shooter's first comeout?
2. Should the shooter make a point, are the bets still active during the next comeout (as opposed to being "off" until another point number is established)?

Assuming the answer to both of these is "yes":

392/165, or about 2.37575758

Ace2
Ace2
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December 22nd, 2021 at 6:22:56 PM permalink
Quote: ThatDonGuy

Quote: Ace2

I hope I phrased the question properly

With 24 ways to roll a point and 6 ways to roll a seven, the expectation is to roll a point number 4 times (including repeats) before rolling a seven. Therefore the average number of distinct (not repeated) points rolled must be at least 1.
link to original post


You phrased it correctly; I read it wrong.

The number I got is the number of distinct point numbers made before sevening out. You want the number of distinct place bets won before a seven is rolled.

Questions:
1. Are the bets placed before the shooter's first comeout?
2. Should the shooter make a point, are the bets still active during the next comeout (as opposed to being "off" until another point number is established)?

Assuming the answer to both of these is "yes":

392/165, or about 2.37575758


link to original post

I agree with that answer. Please show your method

Regarding your questions - I donít think either makes a difference. If bets are ever turned off then itís like those rolls never happened
Itís all about making that GTA
unJon
unJon
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December 23rd, 2021 at 1:26:15 AM permalink
Quote: Ace2



Regarding your questions - I donít think either makes a difference. If bets are ever turned off then itís like those rolls never happened
link to original post



That canít be generally right since the problem ceases when the shooter seven outs. For example, assume your same question, but after placing the bets, the bettor turns them ďoffĒ for three rolls then ďalways on.Ē
The race is not always to the swift, nor the battle to the strong; but that is the way to bet.
ThatDonGuy
ThatDonGuy
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December 23rd, 2021 at 6:43:53 AM permalink
Quote: Ace2

Quote: ThatDonGuy

Quote: Ace2

I hope I phrased the question properly

With 24 ways to roll a point and 6 ways to roll a seven, the expectation is to roll a point number 4 times (including repeats) before rolling a seven. Therefore the average number of distinct (not repeated) points rolled must be at least 1.
link to original post


You phrased it correctly; I read it wrong.

The number I got is the number of distinct point numbers made before sevening out. You want the number of distinct place bets won before a seven is rolled.

Questions:
1. Are the bets placed before the shooter's first comeout?
2. Should the shooter make a point, are the bets still active during the next comeout (as opposed to being "off" until another point number is established)?

Assuming the answer to both of these is "yes":

392/165, or about 2.37575758


link to original post

I agree with that answer. Please show your method

Regarding your questions - I donít think either makes a difference. If bets are ever turned off then itís like those rolls never happened
link to original post


My method is a 64-state Markov chain calculated with brute force. I thought about a Poisson-based method, but is there an easy way to calculate, say, all of the results where 4 of the numbers are rolled and then a 7?
Ace2
Ace2
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December 23rd, 2021 at 8:43:06 AM permalink
(1/3 + 2/5 + 5/11)*2 = 392/165 =~ 2.376 distinct point numbers will be rolled before a seven, on average

1/3 chance of rolling at least one four before a seven (times 1)

+ 2/5 chance of rolling at least one five before a seven (times 1)

+ Ö
Last edited by: Ace2 on Dec 23, 2021
Itís all about making that GTA
Ace2
Ace2
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unJon
December 23rd, 2021 at 9:25:32 AM permalink
Quote: unJon

Quote: Ace2



Regarding your questions - I donít think either makes a difference. If bets are ever turned off then itís like those rolls never happened
link to original post



That canít be generally right since the problem ceases when the shooter seven outs. For example, assume your same question, but after placing the bets, the bettor turns them ďoffĒ for three rolls then ďalways on.Ē
link to original post

The expected number of total point numbers rolled before a seven is 4. This is true whether or not come out rolls are counted
Itís all about making that GTA
teliot
teliot
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December 23rd, 2021 at 9:30:47 AM permalink
Quote: Ace2

I hope I phrased the question properly

With 24 ways to roll a point and 6 ways to roll a seven, the expectation is to roll a point number 4 times (including repeats) before rolling a seven. Therefore the average number of distinct (not repeated) points rolled must be at least 1.
link to original post

Computer assisted ... about 2.37575
Climate Casino: https://climatecasino.net/climate-casino/
Wizard
Administrator
Wizard
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December 24th, 2021 at 12:21:22 PM permalink

I seem to agree with the masses. Here is my table showing the probability of 0 to 6 points made.

Total points made Probability Expected
6 0.062168 0.373009
5 0.101016 0.505079
4 0.129245 0.516979
3 0.151531 0.454594
2 0.170057 0.340114
1 0.185983 0.185983
0 0.200000 0.000000
Total 1.000000 2.375758

ďExtraordinary claims require extraordinary evidence.Ē -- Carl Sagan
unJon
unJon
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December 24th, 2021 at 12:30:01 PM permalink
Quote: Wizard


I seem to agree with the masses. Here is my table showing the probability of 0 to 6 points made.

Total points made Probability Expected
6 0.062168 0.373009
5 0.101016 0.505079
4 0.129245 0.516979
3 0.151531 0.454594
2 0.170057 0.340114
1 0.185983 0.185983
0 0.200000 0.000000
Total 1.000000 2.375758


link to original post



Can someone explain to me what Iím missing? I assume we can drop spoilers at this point.

Wizís table above says the probability of 0 points is 0%. But how is that possible? If the place bets are working from the come out, then thereís a chance of a 7 killing them all before a point number is rolled. Likewise if they are off until a point is established, there is still likewise a probability of a 7 out killing them before another point number rolled.

It seems to me that everyone has an implicit assumption that the six point numbers are made and always on, but rebought if a seven winner is rolled.
The race is not always to the swift, nor the battle to the strong; but that is the way to bet.
ThatDonGuy
ThatDonGuy
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unJon
December 24th, 2021 at 1:09:15 PM permalink
Quote: unJon

Can someone explain to me what Iím missing? I assume we can drop spoilers at this point.

Wizís table above says the probability of 0 points is 0%. But how is that possible? If the place bets are working from the come out, then thereís a chance of a 7 killing them all before a point number is rolled. Likewise if they are off until a point is established, there is still likewise a probability of a 7 out killing them before another point number rolled.


No, it says the expected value of rolling 0 point numbers is 0, and the probability of doing so is 1/5.

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