Is there math on the average times a shooter will make the point?
Basically you have a 33% chance of making the point 25% of the time. You got a 40% chance of making the point 33% of the time. And you have ~45% chance to make the point ~40% of the time.
Multiply. Then add. That's your answer.
luckyQuote: ghettosheckyIs...? I ...
https://wizardofvegas.com/forum/questions-and-answers/gambling/3146-average-number-of-points-hit-per-shooter-in-craps/
"so the average number of points made by the shooter is about 0.68
In other words, if you make even one point, you're having a better-than-average roll.
Not very intuitive, is it?"
a nice read with coffee, in me opinion
not as good if really hungry, for food that is
Sally
Quote: RSThe math isn't too difficult to do yourself.
Basically you have a 33% chance of making the point 25% of the time. You got a 40% chance of making the point 33% of the time. And you have ~45% chance to make the point ~40% of the time.
Multiply. Then add. That's your answer.
That's the probability of making one point.
The expected number of points is 67/98.
This assumes the shooter will keep the dice after a craps on the come-out.
The first question is, what is the probability of making a point?
1/4 of the time, the point will be 4 or 10; there is a 1/3 chance of making this point.
1/3 of the time, the point will be 5 or 9; there is a 2/5 chance of making this point.
5/12 of the time, the point will be 6 or 8; there is a 5/11 chance of making this point.
The chance of making a point is (1/4 x 1/3) + (1/3 x 2/5) + (5/12 x 5/11) = 67/165.
Let P = 67/165, and Q = 1 - P = 98/165.
The expected number of points is:
0 x Q + 1 x P x Q + 2 x P2 x Q + 3 x P3 x Q + ...
= Q x (P + 2 P2 + 3 P3 + 4 P4 + ...)
= Q x P x (1 + 2 P + 3 P2 + 4 P3 + 5 P4 + ...)
= Q x P x (1 + P + P2 + P3 + ...)2
Since -1 < P < 1, this equals
Q x P x (1 / (1 - P))2
= Q x P x (1 / Q)2
= P / Q
In this case, this is 67/98.
Quote: RSThe math isn't too difficult to do yourself.
Basically you have a 33% chance of making the point 25% of the time. You got a 40% chance of making the point 33% of the time. And you have ~45% chance to make the point ~40% of the time.
Multiply. Then add. That's your answer.
I come up with .3945.
Being utterly math challenged, I always "assumed" that the 'average" for making a point once you were out on one-- was 40%, based again as a total math idiot solely on the fact that the "middle" numbers--9&5--will be made 2 out of 5 times. Is my 39.45% a rounding error? due to "fuzzy" percentages on the 6 and 8? ...or what? ... I allus wondered about this, because I know the math can be tricky, and whether my 40% was correct...thanks for any help...tom p
ps o, gawd, please don't tell me it's actually LESS than 40% <sigh>...
pps ok, while you're at it, can you tell me the HE on placing the 5&9 and placing the 4&10?? thanks, tom p
Quote: betwthelinesI come up with .3945.
Being utterly math challenged, I always "assumed" that the 'average" for making a point once you were out on one-- was 40%, based again as a total math idiot solely on the fact that the "middle" numbers--9&5--will be made 2 out of 5 times. Is my 39.45% a rounding error? due to "fuzzy" percentages on the 6 and 8? ...or what? ... I allus wondered about this, because I know the math can be tricky, and whether my 40% was correct...thanks for any help...tom p
ps o, gawd, please don't tell me it's actually LESS than 40% <sigh>...
pps ok, while you're at it, can you tell me the HE on placing the 5&9 and placing the 4&10?? thanks, tom p
The question is not clear to me. What is the OP actually asking?
The average probability of making a point, once one is established, is .406, which can be derived from the "Perfect 1980" : 536 ways to make a point out of 1320 points established. However, that may not be the answer to the question. For any passline decision, the probability of making a point is 536 / 1980 = .2707, compared to 440 / 1980 (.222) for a natural, 220 / 1980 (.1111) for a craps, and 784 /1980 (.3959) for sevening out. 1 / .3959 = 2.526 is the average number of decisions per shooter, and 2.526 * 3.375 = 8.525 is the average number of rolls per shooter. As MathExtremist pointed out in that other thread, .2707 * 2.525 = .68 is the average number of points a shooter is expected to make. You can derive all this stuff and more from the wonderful "Perfect 1980", the craps player's friend.
Cheers,
Alan Shank
Essentially you continue to throw if you roll craps and similarly if you roll a natural. Therefore you may win a few, lose a few between points; but the question only concerned making points rather than how often you win.
Eventually you will establish a point, and then you are odds against making it (since you had the advantage on the come-out roll). The overall chances of making the point are, as has been said, a tad over 40%.
First Roll E (win) E (lose)
Craps 220 0 220
Natural 440 440 0
Point 4/10 330 110 220
Point 5/9 440 176 264
Point 6/8 550 250 300
1980
976 1004
Natural Winner 440
Craps 220
Point Made 536
Point Missed 784
Point Missed 40.606%
Quote: charliepatrick
First Roll E (win) E (lose)
Craps 220 0 220
Natural 440 440 0
Point 4/10 330 110 220
Point 5/9 440 176 264
Point 6/8 550 250 300
1980
976 1004
Natural Winner 440
Craps 220
Point Made 536
Point Missed 784
Point Missed Made 40.606%
Quote: goatcabinThe question is not clear to me. What is the OP actually asking?
The average probability of making a point, once one is established, is .406, which can be derived from the "Perfect 1980" : 536 ways to make a point out of 1320 points established. However, that may not be the answer to the question. For any passline decision, the probability of making a point is 536 / 1980 = .2707, compared to 440 / 1980 (.222) for a natural, 220 / 1980 (.1111) for a craps, and 784 /1980 (.3959) for sevening out. 1 / .3959 = 2.526 is the average number of decisions per shooter, and 2.526 * 3.375 = 8.525 is the average number of rolls per shooter. As MathExtremist pointed out in that other thread, .2707 * 2.525 = .68 is the average number of points a shooter is expected to make. You can derive all this stuff and more from the wonderful "Perfect 1980", the craps player's friend.
Cheers,
Alan Shank
You answered MY question at any rate which was what is the average probability of making a point, once one is established...(after that my eyes glazed over and my jaw went slack but perhaps you answered others' questions there)...anyway prior to that, even, you raised yet another question in my peabrain: what the heck is an "OP"?? ... >/:^) tom p
OP (or "ops") can also mean operator, like an admin or moderator in a non-forum area.
Quote: goatcabinThe question is not clear to me. What is the OP actually asking?
The average probability of making a point, once one is established, is .406, which can be derived from the "Perfect 1980" : 536 ways to make a point out of 1320 points established. However, that may not be the answer to the question. For any passline decision, the probability of making a point is 536 / 1980 = .2707, compared to 440 / 1980 (.222) for a natural, 220 / 1980 (.1111) for a craps, and 784 /1980 (.3959) for sevening out. 1 / .3959 = 2.526 is the average number of decisions per shooter, and 2.526 * 3.375 = 8.525 is the average number of rolls per shooter. As MathExtremist pointed out in that other thread, .2707 * 2.525 = .68 is the average number of points a shooter is expected to make. You can derive all this stuff and more from the wonderful "Perfect 1980", the craps player's friend.
Cheers,
Alan Shank
Thanks so much! This answered my question. Thanks to everyone who contributed as well.
