Math problem - Dice outcomes

[mrsix]

I've done the math, but I'm no mathematician. I need verification and I'm counting on TERBites to check my work...

You have 3 dice that are to be thrown.

Each number has a 1 in 6 chance on showing up on any of the three dice, and choosing just one number to come up would have a 50% of showing up at least once among the three dice (16.7% *3 = 50%).

If I pick 2 numbers, eg. 4 and 6, what are the odds that they will both show up in one throw of the three dice?

 

[Perry Mason]

There are 6 x 6 x 6 possible combinations: 216

There are 2 x 2 x 2 possible throws that will show 2's and 4's. Two of those are all 2's or all 4's.

What is left?

Do your own maths and figure it out.

Rather easy, really...

Perry

 

[Yoga Face]

Each dice has a 1/3 chance of being 2 or 4

meaning 1/ 27 chance of being all 2 and/or 4



total number of combinations are 216

OF THOSE 216 divided by 27 are 2 or 4




OR 8/216 OF BEING EIHTER A 2 OR 4

2 OF THOSE ARE ALL 2 OR ALL 4


6/216

2.7 % ???????


no wait

every 27 throws you got all 2 and/or all 4 on each die

every 216 divided by 2 or 108 throws you 1 chance of all 2 or all 4

every 108 throws you got 4 all 2 and/or 4

4-1 =3

3 out of 108 is the answer or


2.7%

 

[fuji]

There are 6 x 6 x 6 possible combinations: 216

There are 2 x 2 x 2 possible throws that will show 2's and 4's. Two of those are all 2's or all 4's

The various throws are, with x being 1/3/5/7 or four ways;

24x
2x4
X24
42x
4x2
X42
242
224
422
244
424
442

So 30 combinations. 30/216 is the answer.

 

[Yoga Face]

The various throws are, with x being 1/3/5/7 or four ways;

24x
2x4
X24
42x
4x2
X42
242
224
422
244
424
442

So 30 combinations. 30/216 is the answer.

no

1/3 chance of each die being 2 or 4

so


1/27 all three die are 2 or 4

now imagine throwing the dice 216 times a few million times or until all possibilities occur as chance would make it happen


of these throws 1/27 are a combination of 2 or 4 in every 216 throws

odds are all three will be either 2 or 4 but not all 2 or all 4 is 216 divided by 27 or 8 out of 216

2 of these throws out of 216 are all 2 or all 4

216 divided by (8 -2) is the answer

or

2.777777777777 %

verify my methodology


WTF happens when we use all six numbers


6/6*6/6*6/6


is 216 out of 216


216/(216-6)=97.2 %


that seems about right

WTF are the odds the dice roll all 1,2,3,4,5 or 6

should be 100-97.2 = 2.8%


1/6*1/6*1/6

1/216 for each number

or .46% * 6 = 2.777777%




my methodology is sound

 

[Protoss]

(4 OR 6 on die #1) AND (( the other on #2 ) OR (the other on #3))

( 1/6+1/6 ) x ((1/6) + (1/6))

1/3 X ( 1/3 )

1/9

 

[red]

i can't see why casinos are so successful

 

[RemyMartin]

By the way, the chance of any one number showing up at least once on three dice is not 50%, it's 91/216. The math in the first post fails, since that would extrapolate to a 100% chance of rolling any particular number if six dice were rolled (1/6 * 6 = 100%). Clearly, that's not the case - it's not a sure thing.

If you pick one number the chance of it show up is 50%, he was right

 

[Yoga Face]

Two thirds of the time (4/6), you don't roll either number on the first die, meaning that you need to roll both numbers on the second two dice, in any order: 4/6 x 2/6 x 1/6 = 8/216.
One third of the time (2/6), you roll one of the numbers on the first die: 2/6. After that, 5/6 of the time you fail to roll it on the second die, then have a 1/6 chance of rolling it on the third die (2/6 x 5/6 x 1/6), while 1/6 of the time you roll it on the second die, then it doesn't matter what the third die is (2/6 x 1/6 x 6/6). Adding those two gives another 22/216.
30/216
Fuji's post shows the brute force method.
Protoss fails to account for NOT rolling the number on the first die. Yoga Face's math is a mess, sorry to say. Perry didn't even bother, and gets an incomplete.

I say it is 2.7%

I verified my math by figuring out what the odds are of all 6 numbers being shown on any of the dice but no one number being on all of them


add to this the odds of all the dice being one number should give 100%

My mathematical answer was 100% verifying my methodology


in post 5 my math is as simple as it can be made
So WTF do you say it is if it is not 2.7%

 

[fuji]

Yoga, look at my post. Are there any combinations I left out? Any I should have included? Name the combination I got wrong.

There are 30 ways to get your result, out of 216 total possible combinations. I listed all thirty possibilities.

If your math produces a different result, it's wrong.

 

[Yoga Face]

ok


i misread the problem

 

[RemyMartin]

No, he wasn't. If so, tell me what the odds are of throwing any one particular number when throwing six dice. ... If you say it's not 100%, that's why it's not 50% when throwing three dice.

Casinos are successful because people are bad at math.

if throwing six dice, in a long run is 100%.

 

[RemyMartin]

if you pick a number say 2, throw 10,000 times, the 2 should come 10,000 times or very close to it.
impossible closer to 2/3

 

[RemyMartin]

you should take each dice as seperate chance.

 

[gar]

I used a probability tree to map out my answer. There are 3 paths (scenarios) that result in a win.

Path A) hit one number on first die, hit other number on second die, third die is irrelevant
33.3% times 16.6% which will happen 5.5 %

Path B) hit one number on first die, miss second die, hit third die
33.3% times 83.6% times 16.6%, which happens 4.6 %

Path C) miss both on first die, hit one on second die, hit other on third die
66.6% times 33.3% times 16.6%, which happens 3.7%

Add up the 3 paths 5.5 +4.6 +3.7 =13.8% of the time you will be successful

 

[Born2Star]

The various throws are, with x being 1/3/5/7 or four ways;

24x
2x4
X24
42x
4x2
X42
242
224
422
244
424
442

So 30 combinations. 30/216 is the answer.

Fuji is correct... except there's no 7 on the dice.... LOL

 

[Capital Amatuer]

It's a statistics question...
There's a proper formula, pretty sure it involves factorials and applying multiplication and division. I think I need to understand the question better. Rolling three dice at once resulting in a 4 or a 6 showing on each of the three dice ? Or rolling three die and keep the one(s) showing a 4 or a 6 and then rolling the remaining die to show a 4 or a 6, and so on. As in Yatzee.
That's why statisticians are banned from casinos, they know how to calculate probabilities !

 

[Protoss]

Protoss fails to account for NOT rolling the number on the first die. Face's math is a mess, sorry to say. Perry didn't even bother, and gets an incomplete.

Aaaah yes Mr, Fonzerelli so I did. <smacks his noodle> Very astute of you. I reckon then that if i had accounted for that, my first line would remain as is and then OR-ed with (NO 4 or 6 on the first roll AND a 4 or 6 on #2 AND the other on #3) (your 4/6x2/6x1/6=8/216)

Now doing that I would arrive with 1/9 + 8/216 or 24/216 + 8/216 = 32/216 which differs from your 30/216. Do you see the reason for the discrepancy? This has been fun. Used neural pathways that haven't been active in *quite* a while :thumb:

 

[Perry Mason]

It's not statistics, it's simple mathematics.

It is permutations and combinations question... and I was just too lazy, have no incentive, to do the mathematics.

Like Protoss, that would involve accessing neural pathways I haven't used in decades... and I would assume they are still there even if calcified!

Now if it were a pretty lady asking and there are prospects of a certain kind of reward, I would be happy to punch the figures into a calculator and work them out.

Perry

 

[fuji]

Fuji is correct... except there's no 7 on the dice.... LOL

LOL. Just draw another dot.