Easy math puzzles

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":

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":

Revised Solution

392/165, or about 2.37575758

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I agree with that answer. Please show your method

Regarding your questions - I don’t think either makes a difference. If bets are ever turned off then it’s like those rolls never happened

Quote: Ace2

Regarding your questions - I don’t think either makes a difference. If bets are ever turned off then it’s like those rolls never happened
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That can’t 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 “off” for three rolls then “always on.”

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":

Revised Solution

392/165, or about 2.37575758

link to original post

I agree with that answer. Please show your method

Regarding your questions - I don’t think either makes a difference. If bets are ever turned off then it’s 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?

(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)

+ …

Quote: unJon

Quote: Ace2

Regarding your questions - I don’t think either makes a difference. If bets are ever turned off then it’s like those rolls never happened
link to original post

That can’t 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 “off” for three rolls then “always on.”
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

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

Quote: Wizard

Show Spoiler

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

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

Wiz’s 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 there’s 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.

Quote: unJon

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

Wiz’s 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 there’s 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.