How much does this happen?
Taking all numbers together (4,5,6,8,9,10), how often can a craps player expect his/her established point to hit - percentage wise?
Also, what percentage can a craps player expect that his/her pass line bet will hit 7 on the come out roll? In other words, in 1000 rolls, how many of those can be expected to be a win on the pass line on the come out?
You may use that link to answer the first question. I can answer it, but I don't know if you want me to assume an established point, or take it from the Come Out roll?
If you mean, in 1,000 rolls (including those that are not the Come Out roll) in the second question, I don't know. If you mean, "In 1,000 Come Out Rolls:"
1,000 * [0.222222 - 2/36 (for Yo-Eleven)]
1,000 * [0.222222 -.0555556]
1,000 * .1666667 (Rounding up to true value of percentage of seven occuring 6/36 or 1/6)
1: 166.67 wins on 7 for 1,000 Come Out rolls. (You may also use 1,000/36 * 6-Ways to Make Seven)
1,000 * .05555555 = 1: 55.55 wins on 11 for 1,000 Come Out Rolls. (You may also use 1,000/36 *2-Ways to make the dealer say, "Yo!"
Statistically, you should win 1: 222 combined in 1,000 Come Out Rules.
______________________
Statistically, you should also play the wrong way, Don't Pass, Baby!
You will only win 1,000 * (1/36 + 2/36)
1,000 * (.02777777778 (Snake Eyes, One Time!) + .0555555555)
1,000 * .0833333333333 1 : 83 Come Out Rolls, but you statistically have a better chance of winning after the Come Out Roll than with the Pass Line.
The House Edge is .05% less per bet made and .01% less per bet resolved than playing the right way.
The other players will love you, too. They love it when you make a little gun out of your fingers and shoot at the dice when your Don't Bet wins.
(I'm kidding about the last part, as you probably know. )
Quote: Mission146https://wizardofodds.com/ask-the-wizard/craps/probability/
You may use that link to answer the first question. I can answer it, but I don't know if you want me to assume an established point, or take it from the Come Out roll?
If you mean, in 1,000 rolls (including those that are not the Come Out roll) in the second question, I don't know. If you mean, "In 1,000 Come Out Rolls:"
1,000 * [0.222222 - 2/36 (for Yo-Eleven)]
1,000 * [0.222222 -.0555556]
1,000 * .1666667 (Rounding up to true value of percentage of seven occuring 6/36 or 1/6)
1: 166.67 wins on 7 for 1,000 Come Out rolls. (You may also use 1,000/36 * 6-Ways to Make Seven)
1,000 * .05555555 = 1: 55.55 wins on 11 for 1,000 Come Out Rolls. (You may also use 1,000/36 *2-Ways to make the dealer say, "Yo!"
Statistically, you should win 1: 222 combined in 1,000 Come Out Rules.
______________________
Statistically, you should also play the wrong way, Don't Pass, Baby!
You will only win 1,000 * (1/36 + 2/36)
1,000 * (.02777777778 (Snake Eyes, One Time!) + .0555555555)
1,000 * .0833333333333 1 : 83 Come Out Rolls, but you statistically have a better chance of winning after the Come Out Roll than with the Pass Line.
The House Edge is .05% less per bet made and .01% less per bet resolved than playing the right way.
The other players will love you, too. They love it when you make a little gun out of your fingers and shoot at the dice when your Don't Bet wins.
(I'm kidding about the last part, as you probably know. )
Well maybe, please explain further....
Here is what the wiz says
"Of those 100 points established, on average 45.45 would be on a 6 or 8, 40 would be on a 5 or 9, and 33.33 would be on a 4 or 10. You could expected on average 18.94 points made on a 6 or 8, 13.33 on a 5 or 9, and 8.33 on a 4 or 10."
So if I want to know in a 100 rolls how many times a point will win, the answer is 13.53% (the average of the 3 averages above)?
Quote: Mission146
If you mean, in 1,000 rolls (including those that are not the Come Out roll) in the second question
Yes, this was a question, not how many come out rolls will win.
If I were betting 5$ pass line bets, and kept the 5$ wins in a pile from every time a 7 was rolled on a come out roll, how much would I expect to have after 1000 rolls? (This excludes 11 wins, and pass line wins after the point was established).
Quote: slackyhackyYes, this was a question, not how many come out rolls will win.
If I were betting 5$ pass line bets, and kept the 5$ wins in a pile from every time a 7 was rolled on a come out roll, how much would I expect to have after 1000 rolls? (This excludes 11 wins, and pass line wins after the point was established).
Quote:Since 1 = 165/165, the total is (165 + 392)/165 or 557/165. This fraction is approximately 3.376. In other words, if you played many, many Crap games (each decision is a game), kept track of the total number of rolls, and divided the number of rolls by the number of games, the result would be approximately 3.376. According to Stewart Ethier from the University of Utah as quoted in Peter Griffin's book (Extra Stuff: Gambling Ramblings, Huntington Press, Las Vegas, 1991, p. 168), this result was first published in the American Mathematical Monthly in 1909.
http://catlin.casinocitytimes.com/article/how-long-is-a-craps-roll-1240
In 1,000 rolls you would have approximately 1000/3.376 or 296.20853 decisions. If there are 296.20853 decisions, then there will be 296.20853 Come-Out rolls.
We know that a seven will occur on the come out roll 6/36 or 1/6 or 16.66667% times.
296.20853 * .16666667 = 49.3681 sevens on Come Out Rolls
49.3681 * 5 = 246.84 or $246.84
That is how much you would win on Sevens after the Come Out roll alone in 1,000 rolls, on average.
Quote: slackyhacky
Well maybe, please explain further....
Here is what the wiz says
"Of those 100 points established, on average 45.45 would be on a 6 or 8, 40 would be on a 5 or 9, and 33.33 would be on a 4 or 10. You could expected on average 18.94 points made on a 6 or 8, 13.33 on a 5 or 9, and 8.33 on a 4 or 10."
So if I want to know in a 100 rolls how many times a point will win, the answer is 13.53% (the average of the 3 averages above)?
No, because the question assumes the point has been made.
In 100 rolls:
(100/36 *1) + (100/36 *1) + (100/36 *2) + (100/36 * 2) + 100/36 *6) or
2.77778 + 2.77778 + 5.55556 + 5.55556 + 16.66667 or
33.333348 rolls fail to establish a point in the first place.
(100/36 * 3) + (100/36 * 3) + (100/36 * 4) + (100/36 * 4) + (100/36 *5) + (100/36 * 5) or
8.33333 + 8.33333 + 11.11111 + 11.11111 + 13.88889 + 13.88889 or
66.66666 Comes Out rolls out of 100 establish a point.
Of the 66.66666 points established:
(8.33333 * 2)/66.6---(11.11111 * 2)/66.6---(13.88889 * 2)/66.6
.25025 are on 4 or 10 (25.025%) ---.33367 are on 5 or 9 (33.367%) and .41708 are on 6 or 8 (41.708%)
66.6 * .25025 = 16.66665 (Points on 4 or 10)
66.6 * .33367 = 22.22242 (Points on 5 or 9)
66.6 * .41708 = 27.777528 (Points on 6 or 8)
___________
Out of 100 rolls, you will establish a point 66.6 times.
___________
In the chart to which I linked you, the Wizard has the probability of a point of 4 or 10 winning at 0.055556 5 or 9 winning at 0.088889 and 6 or 8 winning at 0.126263, so the probability of making a point and winning is .270708 or 27.0708%.
That was the purpose of the chart.
However, your question was in 100 total rolls, not points made.
We'll go back to the 3.376 rolls per decision.
In 100 rolls you would have 100/3.376 = 29.62085 decisions.
Of those decisions, .055556 * 29.62085 are making a point of 4 or 10 = 1.64562 decisions
.088889 * 29.62085 are making a point of 5 or 9 = 2.63297 decisions
.126263 * 29.62085 are making a point of 6 or 8 = 3.74002 decisions
.11111 * 29.62085 lose on 4 or 10 point = 3.29117 decisions
0.133333 * 29.62085 lose on 5 or 9 point = 3.94945 decisions
0.151515 * 29.62085 lose on 6 or 8 point = 4.48800 decisions
0.222222 * 29.62085 win on come out = 6.58240 decisions
0.111111 * 29.62085 lose on come out = 3.29120 decisions
In 100 rolls, there will be 29.62085 decisions. Of those decisions, you will establish and make a point 8.01861 Times
You will establish and fail to make a point 11.72862 times
You will win on the come out 6.58240 times
You will lose on come out 3.29120 times
If you add the results you get 29.62083 with differences due to rounding.
In total, you will win 14.60101 and lose 15.01982 times.
PROOF:
If you divide 14.60101/29.62083 = 49.29% (The probability of winning a Pass Line Bet)
Quote: Mission146No, because the question assumes the point has been made.
In 100 rolls:
(100/36 *1) + (100/36 *1) + (100/36 *2) + (100/36 * 2) + 100/36 *6) or
2.77778 + 2.77778 + 5.55556 + 5.55556 + 16.66667 or
33.333348 rolls fail to establish a point in the first place.
(100/36 * 3) + (100/36 * 3) + (100/36 * 4) + (100/36 * 4) + (100/36 *5) + (100/36 * 5) or
8.33333 + 8.33333 + 11.11111 + 11.11111 + 13.88889 + 13.88889 or
66.66666 Comes Out rolls out of 100 establish a point.
Of the 66.66666 points established:
(8.33333 * 2)/66.6---(11.11111 * 2)/66.6---(13.88889 * 2)/66.6
.25025 are on 4 or 10 (25.025%) ---.33367 are on 5 or 9 (33.367%) and .41708 are on 6 or 8 (41.708%)
66.6 * .25025 = 16.66665 (Points on 4 or 10)
66.6 * .33367 = 22.22242 (Points on 5 or 9)
66.6 * .41708 = 27.777528 (Points on 6 or 8)
___________
Out of 100 rolls, you will establish a point 66.6 times.
___________
In the chart to which I linked you, the Wizard has the probability of a point of 4 or 10 winning at 0.055556 5 or 9 winning at 0.088889 and 6 or 8 winning at 0.126263, so the probability of making a point and winning is .270708 or 27.0708%.
That was the purpose of the chart.
However, your question was in 100 total rolls, not points made.
We'll go back to the 3.376 rolls per decision.
In 100 rolls you would have 100/3.376 = 29.62085 decisions.
Of those decisions, .055556 * 29.62085 are making a point of 4 or 10 = 1.64562 decisions
.088889 * 29.62085 are making a point of 5 or 9 = 2.63297 decisions
.126263 * 29.62085 are making a point of 6 or 8 = 3.74002 decisions
.11111 * 29.62085 lose on 4 or 10 point = 3.29117 decisions
0.133333 * 29.62085 lose on 5 or 9 point = 3.94945 decisions
0.151515 * 29.62085 lose on 6 or 8 point = 4.48800 decisions
0.222222 * 29.62085 win on come out = 6.58240 decisions
0.111111 * 29.62085 lose on come out = 3.29120 decisions
In 100 rolls, there will be 29.62085 decisions. Of those decisions, you will establish and make a point 8.01861 Times
You will establish and fail to make a point 11.72862 times
You will win on the come out 6.58240 times
You will lose on come out 3.29120 times
If you add the results you get 29.62083 with differences due to rounding.
In total, you will win 14.60101 and lose 15.01982 times.
PROOF:
If you divide 14.60101/29.62083 = 49.29% (The probability of winning a Pass Line Bet)
That all sounds like a bunch of crap(s).
Actually thanks...
