"Picking the highest die odds?" Topic

Roll 3d6 and pick the highest(no adding together).
Your opponent rolls 1d6.

How do I calculate the odds of me rolling higher than my opponent? I'm trying to calculate odds for close combat in Space Hulk, and other games.

Given that the highest you can roll on 1d6 is a six, and 3d6 produces a bell-shaped curve with 216 possible combinations, you have a 196/216 chance of rolling 7-18, or roughly a 91% chance of rolling higer than a 6.

higher.

Virtualscratchbuilder, please check out the original description – the OP was not actually rolling 3d6 ("no adding together), but rolling three different d6 and picking the highest number amongst them.

Manual (slow) solution for such a case is to go through all the 216 combinations, and then count the frequencies of the cases where the highest number is 1, 2, 3, 4, 5 or 6.

For starters, the chance of the highest number being 1 is 1/216 (as only the combo 1-1-1 gives it).

Once you know the distribution, you can compare it with the regular d6 distribution (count the probability of rolling a given number, and then multiply that with the probability of the other guy rolling higher).

Griefbringer

Duh… duh duh… I kin read. I kin spel too.

"For starters, the chance of the highest number being 1 is 1/216 (as only the combo 1-1-1 gives it)."

Not sure this works. If you have 3 dice, the chance of the highest number being 1 is 3/18.

87.5/100 if you start that 1d6 to 1d6 is 50/50 and 2d6 to 1d6 is 75/100 I did not count in the chance of ties.
that drips in a rock that I dont have the math to work that in.

I could be totally wrong

Not sure this works. If you have 3 dice, the chance of the highest number being 1 is 3/18.

Nope – it's 1/6 x 1/6 x 1/6, which is indeed 1/216. 3/18 is 1 in 6….

Dom, now trying to remember if there's an easy way of doing this….

OK, thought had. 'Scuse working out; I'm figuring this out as I type….

1/6 chance of opponent getting a 1. If he does the odds of failing to beat it are 1/6 x 1/6 x 1/6. Gives 1/1296 total.

1/6 chance of opponent getting a 2. If he does the odds of failing to beat it are 2/6 x 2/6 x 2/6. Gives 8/1296 total.

3 gets you 27/1296

4 gets you 64/1296

5 gets you 125/1296

6 gets you 216/1296 (since you can't beat his 6 however many you roll….)

Total probability 441/1296 that you fail to beat him, so 66% chance of you beating him, 34% chance of you not. (Ties are "not" since the question was rolling higher than him.)

Dom.

PS – Quester; your principle falls apart because of the ties issue; effectively a tie is a "win" for the 1d6, since the 3d6 haven't rolled higher than it.

Final PS – to get the odds of being beaten by him, we can work it the other way:

1 – 0/6 x 0/6 x 0/6 = 0/1296 'cos his 1 don't beat nuffink….
2 – 1/6 x 1/6 x 1/6 = 1/1296
3 – 2/6 x 2/6 x 2/6 = 8/1296
4 – 3/6 x 3/6 x 3/6 = 27/1296
5 – 4/6 x 4/6 x 4/6 = 64/1296
6 – 5/6 x 5/6 x 5/6 = 125/1296

This gives a total chance of 225/1296 that the 1d6 beats the 3d6, or 17%. Combining this with the original figure of 441/1296 that the 3d6 fails to win, we get 17% chance of 1d6 being higher, 17% chance of the draw, and 66% chance of 3d6 being higher.

Alright, thanks for the calculations :) I'll try running some numbers now to figure out how many dice to give what units.

Silly me for not thinking of the way of calculating that Dom presented – that is the way to go.

Griefbringer

Dunadan – drop me an email on dom <at> domsdecals <dot> com – just had a very minor eureka moment and found a p*ss-easy way of calculating the probabilities for any number of dice – good old Excel makes it not much more than a copy and paste job…. Drop me a line and I'll email you the spreadsheet.

Dom.

Dom Skelton's numbers look good to me. If you're still foggy on why, this explanation might help.

First, the chance of 1D6 beating a value ranging from 1-6 versus the chance of failing:

target value: success chance vs failure chance
6: 0 vs 6/6
5: 1/6 vs 5/6
4: 2/6 vs 4/6
3: 3/6 vs 3/6
2: 4/6 vs 2/6
1: 5/6 vs 1/6

On a random roll though we don't know the number we need to beat, so we average (add the chances of success up and divide by the number of cases, 6); this makes the denominator 6 to the power of the number of dice (2 in this case).

So … D6 to beat D6 = (1+2+3+4+5)/6^2 = 15/36 = 41.67%.

What happens when we throw another D6?

We can't improve on success, only on failure. So to our chance of success, the second die would add the same success value but since it only applies when the first die fails, we have to multiply it by the chance of failure:
6: 0 + 0
5: 1/6 + (5/6*1/6) = 11/36
4: 2/6 + (4/6*2/6) = 20/36
3: 3/6 + (3/6*3/6) = 27/36
2: 4/6 + (2/6*4/6) = 32/36
1: 5/6 + (1/6*5/6) = 35/36

Again, we have to average the 6 cases, which makes our denominator 6^3 or 216.

So … Two D6 to beat D6 = (11+20+27+32+35)/6^3 = 58.33%

For a third D6, it gets hairier because now we have to add the chance of success multiplied by the chance of failure on two dice, or in other words, our chance of failure to the power of 2. But the process is the same: add them up and divide by the number of cases … our denominator is now 6^4 or 1296.
6: 0 + 0 + 0
5: 1/6 + (5/6*1/6) + ((5/6)^2)*(1/6) = 91/216
4: 2/6 + (4/6*2/6) + ((4/6)^2)*(2/6) = 152/216
3: 3/6 + (3/6*3/6) + ((3/6)^2)*(3/6) = 189/216
2: 4/6 + (2/6*4/6) + ((2/6)^2)*(4/6) = 200/216
1: 5/6 + (1/6*5/6) + ((1/6)^2)*(5/6) = 215/216

So, three D6 to beat D6 = (91+152+189+200+215)/6^4 = 65.35%