Hi folks

First post. Long time BG player. Love the game. Mr Robertie, thanks for your years of contributions to it.

Now a I have simple conundrum to pose to all here which is counter-intuitive and I would value all opinions.

Let me first set out a scenario.

You roll A SINGLE DIE

What are the odds that you will roll a given number, say a 6?


The answer is surely 1 in 6 (16.67%)

I doubt anyone here would dispute this.


Now, what if I allow you to roll that same die a second time?

So you get 2 bites at this cherry, 2 chances to roll a 6

What are your odds now?

I would say that 1 in 6 plus 1 in 6 equates to 1 in 3 or (33.3%)


If you agree with this so far then read on.



Now I would contend that:

1. The odds of rolling a 6 on a die will always be 1 in 6 and this will never ever change even if you get to roll the die a million times. Every roll is a 1 in 6 chance.

2. If I make the first roll with a blue die and make the second roll with a different red die, then the odds are still the same. It doesn't matter which die is used at a given time, the odds will always be 1 in 6 to roll a 6 for each individual die. The odds will be 1 in 3 if I get to roll the same die twice or if I first roll a red die and then a blue die.

Are you still nodding and agreeing? Read on

I further contend that if I am rolling a blue die and then a red die, this is absolutely no different to rolling both dice together. Both dice are still being rolled. It does not make a difference whether the red or blue is rolled first, nor does it matter if they are rolled together. The odds of rolling a 6 are still 1 in 3 because I have 2 chances at a 1 in 6 outcome.


It is at this point that things go awry for some people. That's because the moment you roll both dice together rather than one then the other, they start to get lost in the world of BG 2-dice combinations which most of the game is about.

At this point people say, well there are indisputably only 36 possible outcomes with 2 dice. Of those there are 11 throws that contain a 6.

(1-6, 6-1, 2-6, 6-2, 3-6, 6-3, 4-6, 6-4, 5-6, 6-5, 6-6)

Thus the odds of getting a 6 MUST be 11 in 36 or just 30.5%

And this is indeed the odds you see quoted all over the internet for producing a specific SINGLE DIGIT number.


As much as I have thought about this and wrangled with it I just do not believe it is true.

The consideration of throwing a required single digit number, say a 6, has no concept or meaning in terms of dice combinations. For this specific requirement/outcome the dice are independent and have no connection with each other whatsoever.

Elsewhere in the game this is not true. Elsewhere there is always a connection because BG concerns itself with the TOTAL SUMs of both dice and with results greater than a single digit.

But in this specific instance, which incidentally is all about the odds of getting a blot off the bar in a home board that has 5 primes made, is all about producing ONE SINGLE NUMBER.

As such at the moment I can only contend that the odds of throwing a 6 to get on the board is still 1 in 3 or 33% and can not be 11 in 36 (30.5%)


The act of throwing both dice together doesn't alter the fact that each die has that 1 in 6 chance to produce a 6. They are in effect 2 separate dice throws, the red and the blue that happen to be being made at the same time. If the red die came up 6 then it doesn't matter one iota what the other die comes up because we have at that point already satisfied the requirement of throwing a 6.

This is twisting my melon

Trust me I clearly understand the notion of 36 possible combinations and 11 of them containing a 6. Yet something there just doesn't sit right for me.

Any thoughts?

Do not agree with this.

The chances you will roll one 6 in two rolls is chance you roll a 6 in first roll (1/6) + chance you didn't in the first roll * chance you will in the second roll (5/6)*(1/6).

6/36 + 5/36 = 11/36

But this statement suggests that the 2nd roll is somehow dependent on the first roll, which it isn't. The 2nd roll is a purely independent action.

Remember that if the first roll comes up 6 then the 2nd roll has no significance because we already have the 6 needed. Yet we can still roll that 2nd die anyway for the hell of it and just ignore whatever comes up.

If the first roll of 1 in 6 chance fails, that in no way changes the odds of the 2nd roll coming up a 6. Any way you cut it the odds of any single rolled die coming up a 6 is 1 in 6.

Both dice have an equal 1 in 6 chance of coming up a 6

They are 2 distinct and separate events, throwing the dice together doesn't change that

Nope, it's 11/36. You're looking at it all wrong. Here are your possible outcomes when you roll one die and then a second die looking for a six:

1. You roll the first die and get a six. There is no need to roll the second die. This happens 6 out of 36 times.

2. You roll the first die and do not get a six. Therefore, you need to roll the second die. You roll the second die and get a six. This happens 5 of 36 times (remember, you don't even need to roll the second die six times, leaving 30 rolls of which 5 will be a six).

3. Neither roll is a six.

If you add up numbers 1 and 2, you get 11 out of 36 times. 6 times on the first die and 5 times on the second die. The only difference is, in backgammon, we are rolling the second die even when we get a 6, but it doesn't change the odds since we already have our six.

If you want to do it the hard way, simply count all the possible rolls and then count which ones have 1 or more sixes:

11 21 31 41 51 61
12 22 32 42 52 62
13 23 33 43 53 63
14 24 34 44 54 64
15 25 35 45 55 65
16 26 36 46 56 66

That's every possible roll (36 of them) and they all have an equal chance of appearing. Only 11 of them have at least 1 six. You are getting hung up on double six and counting it as two outcomes instead of one.

Hi

No not remotely getting hung up on double 6 and as stated in the OP I don't think dice combos apply here because we're looking for a single number so there is no connection between the dice or their rolls.

In your first post there you say:
"Therefore, you need to roll the second die. You roll the second die and get a six. This happens 5 of 36 times "

This can not possibly be true. The chances of the 2nd die rolling a 6 is and always will be 1 in 6, just as it is for any die. What happened before you roll it is of no consequence. The odds of rolling a 6 are 1 in 6

I'll make a point and then I'll show you the way it's actually supposed to be done mathematically:

If you roll 6 dice, do you think you are 100% to get some number? If not, then the original premise MUST be wrong.

The right way to do the math is to take the compound probability of neither die being a 6 and subtract it from 1. Each die is 5/6 NOT to be a 6 so we do (5/6)*5/6 and get 25/36. 1-(25/36) is 11/36.

If you didn't understand that, let me try to explain this more simply for you then. Let's look at the manual way first:

Manual Way

First, look up above. Count the number of possible dice rolls. It's 36, right?
Second, look up above. Count the number of possible dice rolls that have one or more sixes in them? It's 11, right?
Third, do we agree that every combination has the same probability of being rolled? I'm going to guess you agree.
Therefore, it's obviously 11 out of 36. That's the "hard" way. Manually count all possibilities and manually count all that contain the attribute for which we are looking.

Mathematical Way

Now, let me take a shot at showing you why you are making a mathematical mistake, again.

Define your goal: To roll a 6.
Define your parameters: You have two dice. A roll of a 6 on either die meets the goal. It doesn't matter if you roll one 6 or two 6s.

I think we all agree so far.

Just for ease, let's say you roll the dice one at a time. We agree that on the first die, there are six possibilities and one of those possibilities is rolling a 6. So, the odds of the first die containing a six is one out of six. We agree on that, right?

So, to say it another way, six out of 36 times, we do not care what is on the second die. We've met our goal. 30 out of 36 times, we care what is on the second die. Still with me?

Now, for the 30 times that we care about the second die, we agree that there is a one out of six chance that it will be a 6 for the times we care what it is, right? Here is where you get hung up, though. You just say there is a one out of six chance of rolling a 6 and leaving it hang there. That's incomplete because there is a seventh variable for the second die and that is, we don't care what is on it. So, the odds of rolling a 6 with the second die is 5 out of 36, not 6 out of 36. To spell it out:

• 5x - Roll a 1
• 5x - Roll a 2
• 5x - Roll a 3
• 5x - Roll a 4
• 5x - Roll a 5
• 5x - Roll a 6
• 6x - We Don't Care

Therefore, it's not 6/36 + 6/36 = 12/36 (or 1/3). It is 6/36 + 5/36 = 11/36.

Just to prove this, let's say you have six dice and roll them all needing just one 6. By your math, 1/6 + 1/6 + 1/6 +1/6 + 1/6 +1/6 = 6/6 or 1. By your math, it's impossible to roll six dice and not roll a 6. We know that is not true, though. The true answer would be 6/36 + 5/36 + 4/36 + 3/36 + 2/36 + 1/36 or 21/36.

Also the post directly above this one is absolutely correct, but I thought this might be easier for you to understand.

21/36 is not right for rolling a 6 in six rolls. Easiest way to calculate it is to calculate the chance of not rolling a 6 as (5/6)^6 and then subtract that from 1.

Just like you can calculate the 2 roll case by 1 - (5/6)^2 (as another poster mentioned).

Yes, you are absolutely right. Not sure what I was thinking and I realized it last night - I was trying to oversimplify it so that he could understand. He doesn't seem to understand your way (the best way). Thanks for the fix!!!

Puzzled, you should try extending your logic to prove that it’s wrong before you start arguing with others who, based on their posts, actually understand the math. You say that the odds of rolling a six in two attempts is 1/6 + 1/6 = 1/3. Then, by that logic, the odds in THREE attempts should be 1/2 (3 x 1/6). Extending to 6 attempts, your logic says the probability is 1. Thus if you roll a die six times you claim it’s certain you’ll roll a six. Obviously this isn’t true (try it yourself!) so your logic is flawed.