A Variation On The 5 Count Method Of Playing Craps
I have a friend who likes to qualify a shooter by counting the rolls the player has made before deciding to bet on that player. My friend knows the 5-count but prefers to count rolls instead. We have had discussions about this and decided heuristically that the two methods are the same. If you do not want to employ the 5-count for whatever reason you can do what my friend does instead. This will help you to preserve your bankroll and keep you at the table longer hoping to find a rhythmic roller or controlled shooter who can have a "hot" hand.
If you want to know why you should employ this technique at craps, read my first article entitled, "The 5-Count Changes the Odds of the Game!"
What I have found is you can expect the same results as using the 5-count if you employ the following method of counting the rolls from the first time a player starts shooting the dice. What you need to do is count every roll made by the player when he first gets the dice. If he successfully makes it through 7 rolls without sevening out then the player has "qualified" for you to bet on him. Make your first wager when the player has completed the 7th roll. Your first bet on this player will be when he is making the 8th roll. It does not matter what numbers are thrown, how many points are made, not made or even if the player has established a point yet. Wait 7 rolls, and then bet on that shooter from that point on.
Following is what I wrote to my friend and the technical explanation for my above conclusions.
I was doing some research on the length of rolls, 5-count and length of the average craps roll. I was reading Dr. Catlin's articles in which he explained how he calculated some of these math problems. I was going to figure out how many shooters are eliminated if one were to use the 6-count or 7-count. These are the type of crazy things old mathematicians like to play with in their spare time. OK, so maybe I am just a little bit wacky. But when one has spent a lifetime solving problems, using math and logic to figure things out, these sorts of things are intriguing to me.
While doing that I came across his calculations of how many rolls it takes for a shooter to seven out. Below is his chart showing the various probabilities.
|
Roll |
Probability of Seven Out |
|
1 |
0.00000000 |
|
2 |
0.11111111 |
|
3 |
0.22788066 |
|
4 |
0.33264746 |
|
5 |
0.42387109 |
|
6 |
0.50278913 |
|
7 |
0.57095589 |
|
8 |
0.62980865 |
|
9 |
0.68060930 |
|
10 |
0.72445344 |
Obviously no one can seven out on their first roll so the probability of sevening out by the first roll is zero. 24 out of 36 rolls will yield a point number on the first roll and the chances of throwing a seven on the next roll are 6/36. Thus the probability of sevening out within 2 rolls is 24/36 * 6/36 = 2/3 * 1/6 = 2/18 = 1/9 = 0.111111111. This explains the first 2 entries which are pretty easy to see from simple probabilities. It gets more complicated to calculate the other numbers. I would have to explain conditional probabilities and how they apply to the various situations. You don't need to understand the math to understand the chart. If you go down the chart you will see that approximately 50% of random shooters will seven out by the 6th roll and
57.1% will seven out by the 7th roll.
We know from another of Dr. Catlin's studies
57% of random shooters do not make it past the 5-count. From this analysis of his we see that 57% of random shooters will also seven out by the 7th roll.
This means if you place your first bet on the 8th roll of every random roller you can expect the same results on average as those who use the 5-count on random rollers. It does not mean that you will both bet on the same random rollers. But it does mean you will both be eliminating 57% of the random rollers. It is just not the same 57%. I don't remember if you like to put your first bet down after the 6th or 7th roll. But if you are doing it after the 7th roll, counting all the rolls, you can expect the same results as those using the 5-count.
I remember you said Frank told you that the two methods were almost identical. From our experience we have seen both techniques seemed to work out about the same. I am happy that I now have seen a mathematical explanation of why they are in fact the same.
Another interesting thing I came across was that the probability of sevening out on or before the 49th roll is 0.99913902. Subtracting this number from 1 will give us the probability that the shooter makes it to 50 rolls; the number is 0.00086098. This means that only 8.6 random shooters out of 10,000 can expect to have a roll of 50 or longer. Another way of looking at this would be that a random roller can expect to have 8 or 9 rolls of 50 or longer for every 10,000 times he shoots the dice. Frank, Dominator, Stickman and Rock and Roller have 13, 11, 10 and 9 fifty plus rolls respectively, that have been observed by another person in the last 5 years. There is no way they have shot the dice 10,000 times in the past 5 years. Thus their results are significantly better than what can be expected by a random shooter. But we all know that already. I just like playing with the numbers to keep demonstrating it mathematically.
