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BGonline.org Forums
dice probabilities and shot counting
Posted By: Daniel Murphy In Response To: dice probabilities and shot counting (Bill Riles)
Date: Thursday, 1 September 2011, at 9:52 p.m.
Yes -- and if you can remember the sequence 11-9-7-5-3-1, you can remember the sequence 11-20-27-32-35-36 since 11+9=20, 20+7=27, 27+5=32, 32+3=35, 35+1=36. Either sequence answers all questions about how often you enter a checker on an X-point board or how many direct shots you have if you hit with X direct numbers.
How often do you enter one checker on a 3 point board? Three points are vacant, so 11+9+7=27. How often do you enter one checker on a 4 point board? Two points are vacant, so 11+9=20.
You can also calculate those examples another way. Count the occupied points, square that number, and subtract from 36. That's how many numbers enter. How often do you enter 1 checker on a 3 point board? 3 points are occupied. 3 squared is 9. 36-9=27. How often do enter a single checker on a 4 point board? 4 points are occupied. 4 squared is 16. 36-16=20.
That last method is longer. You don't need to use it. But I think knowing multiple ways to find an answer helps your memory and facility in counting. Squaring is also helpful if the question is how often do you enter 2 checkers on an X point board? How often do you enter 2 checkers on a 3 point board? 3 points are vacant. 3 squared is 9. You enter both checkers 9 times in 36. How often do enter a single checker on a 4 point board? 2 points are vacant. 2 squared is 4. You enter both checkers 4 times in 36.
Similarly with direct shots. If two direct numbers hit, 11+9=20 rolls hit. How many rolls don't hit? 4 squared is 16. If three direct numbers hit, 11+9+7=27 rolls hit. How many rolls don't hit? 3 squared is 9. If four direct numbers hit, 11+9+7+5=32 rolls hit. How many rolls don't hit? 2 squared is 4.
For counting both direct and indirect (combination) shots, a sequence that should be memorized is 11-12-14-15-15-17. That's how many direct and indirect numbers hit when your checker is 1, 2, 3, 4, 5 and 6 pips away.
To take Bill's example, where you hit with both direct and indirect 6's and 5's. You can use the 11-9-7-5-3-1 method: 11 sixes plus 9 more fives is 20, to which you must add the indirects 4-2, 3-3, 2-2, 3-2 and 4-1, for a total of 28 rolls in 36. But if you've memorized that 6's give 17 direct and indirect hits and 5's give 15, then 17+15=32, from which you must subtract the numbers you've counted twice, 5-1 and 6-5, for the same total of 28.
It's faster, of course, to eventually memorize that 6's and 5's hitting means 28 hitting numbers, including all combination numbers.
Bill P asked: how often does a six appear?
In one roll? There are 6 numbers. 5 of them are not sixes. 5 squared is 25. 36-25=11. So 11 times in 36, you will roll at least 1 six in 1 roll. That is where the 11 in Bill R.'s 11-9-7-5-3-1 sequence comes from.
Might you instead need to roll at least 1 six in 2 rolls? From above, you will fail to roll a 6 in 1 roll 25/36 of the time. You will fail on your second roll 25/36 of the time. You will fail on both rolls 25/36 times 25/36 of the time. Which is 625/1296. The rest of the time -- 671/1296 -- you don't fail. So you are a slight favorite to roll at least one six in two rolls.
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