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Society & Culture1 min ago
Dice
Not sure where to put this so hope this is ok here.
According to Pascal, the chance of throwing a double six with two dice is 1:36.
To my layman's thinking there are 11 possible results throwing two dice.. so why is throwing a 12 not one in eleven?
I've probably missed something that is obvious to others.
According to Pascal, the chance of throwing a double six with two dice is 1:36.
To my layman's thinking there are 11 possible results throwing two dice.. so why is throwing a 12 not one in eleven?
I've probably missed something that is obvious to others.
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The outcome with the highest probability is a total of 7 as there are six ways of achieving this. Imagine you have a blue dice (B) and a red dice(R). You can get 7 in the following ways: B1&R6, B2&R5, B3&R4, B4&R3, B5 and R2, B6 & R1. But there is only one way of getting a total of 2 (B1 and R1) or12 (B6, R6).
Not so, there are 36 possible results. If you do not believe me paste stickers over the faces of two dice you have, write unique characters on each, then see what combinations you can create.
Your example works for double six only because there is but one combination that gives you that value (12). For most other values there are multiple results that give you the same total, so for them the odds of throwing that value are higher.
Your example works for double six only because there is but one combination that gives you that value (12). For most other values there are multiple results that give you the same total, so for them the odds of throwing that value are higher.
Just a thought.
I don't think the odds are exactly 36:1
The numbers on dice are marked either by small semicircular indentations, or by paint, or (most commonly) a combination of both.
The indentations cut out of the face of the dice will reduce the weight of that face.
There are more indentations on the 6 face than on the 1 face. Whichever way up you hold the 6 and 1, the numbers around the sides will be evenly balanced from top to bottom.
But, if you hold dice perfectly horizontally with the 6 and the 1 on the sides, the 6 face weighs less than the 1 face, so the 1 is more likely to fall downwards.
So the chances of the 6 facing upwards are slightly better than 1 in 6.
So the odds of throwing a double 6 are slightly better than 1 in 36.
I don't think the odds are exactly 36:1
The numbers on dice are marked either by small semicircular indentations, or by paint, or (most commonly) a combination of both.
The indentations cut out of the face of the dice will reduce the weight of that face.
There are more indentations on the 6 face than on the 1 face. Whichever way up you hold the 6 and 1, the numbers around the sides will be evenly balanced from top to bottom.
But, if you hold dice perfectly horizontally with the 6 and the 1 on the sides, the 6 face weighs less than the 1 face, so the 1 is more likely to fall downwards.
So the chances of the 6 facing upwards are slightly better than 1 in 6.
So the odds of throwing a double 6 are slightly better than 1 in 36.
I doubt the different weights have much effect in practice, but manufacturers of precision dice recognise the need for their dice to be seen as fair, especially in casinos. Casino dice have their pips drilled, then filled flush with a paint of the same density as the material used for the dice, so that the centre of gravity of the dice is as close to the geometric centre as possible.
There's a law called "the Weak Law of Large Numbers" that basically says that the probability of some event reflects the number of times you'll see it if you have enough goes. The point then is that if you roll two dice often enough and note how many times you get a double six, you'll see whether it occurs (roughly) once every 36 rolls or (roughly) once every 12. You will get the first result, or at least something close to it, so that the layman's thinking is flawed.
At the very least, that's because there certainly aren't 11 possible results: 7 can occur as 1+6, 2+5 or 3+4; 6 as 1+5, 2+4, 3+3; and so on -- tallying all those up gives you 19 distinct sums, some of which give the same total, so that at the very least you should expect odds of 18 to 1 against a double six rather than 11 to 1. And then finally, some of these sums can occur in multiple ways: 1+6 and 6+1 are also distinct from each other. This is true whenever the two dice show different numbers, which does indeed mean that there are 36 different results that are possible -- of which only one is a double six. Hence, 35 to 1 against assuming fair dice.
At the very least, that's because there certainly aren't 11 possible results: 7 can occur as 1+6, 2+5 or 3+4; 6 as 1+5, 2+4, 3+3; and so on -- tallying all those up gives you 19 distinct sums, some of which give the same total, so that at the very least you should expect odds of 18 to 1 against a double six rather than 11 to 1. And then finally, some of these sums can occur in multiple ways: 1+6 and 6+1 are also distinct from each other. This is true whenever the two dice show different numbers, which does indeed mean that there are 36 different results that are possible -- of which only one is a double six. Hence, 35 to 1 against assuming fair dice.
Ooh! Year 8 maths again!
Perhaps a picture might help?
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Perhaps a picture might help?
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