Help me out with a Dice Probability Question

[The Crusader]

Good morning fellow TBers. Lately at my local watering hole, we've been throwing a little dice for 2 bucks a roll (maximum one roll per session!) Now, I am sure that many if not most of you are familiar with it. It's 5 dice, die, (whatever) and whoever rolls all 1s or 2s or 3s or 4s or 5s or 6s wins the pot.

So my question to you guys is, what is the probability of doing that?

Is it 6x6x6x6x6/6=1296 or 1 in 1296? Or some other method? Please explain. Thanks.

[Ívar Þórólfsson]

Originally Posted by The Crusader

Good morning fellow TBers. Lately at my local watering hole, we've been throwing a little dice for 2 bucks a roll (maximum one roll per session!) Now, I am sure that many if not most of you are familiar with it. It's 5 dice, die, (whatever) and whoever rolls all 1s or 2s or 3s or 4s or 5s or 6s wins the pot.

So my question to you guys is, what is the probability of doing that?

Is it 6x6x6x6x6/6=1296 or 1 in 1296? Or some other method? Please explain. Thanks.

1295 to 1 to be specific.
But 1 in 1296 is also correct

[PSPookie]

I get 1/1296 as well.

1 (1st die can be any number) x 1/6 (probability of rolling the same number on the 2nd die) x 1/6 (probability of rolling the same number on the 3rd die) x 1/6 (probability of rolling the same number on the 4th die) x 1/6 (probability of rolling the same number on the 5th die) = 1/(6^4) = 1/1296

[need4mospd]

The total number of possible outcomes is...

6^5 = 7776

Since there are 6 specific combos out of that 7776...

7776/6 = 1296

[DeluxeRed]

Almost as good as the odds of getting busted...

[PSPookie]

Originally Posted by Ívar Þórólfsson

1295 to 1 to be specific.
But 1 in 1296 is also correct

Probability, not odds . . . there is a tiny symantic difference, in this case amounting to 5.958e-7

[The Crusader]

Originally Posted by PSPookie

I get 1/1296 as well.

1 (1st die can be any number) x 1/6 (probability of rolling the same number on the 2nd die) x 1/6 (probability of rolling the same number on the 3rd die) x 1/6 (probability of rolling the same number on the 4th die) x 1/6 (probability of rolling the same number on the 5th die) = 1/(6^4) = 1/1296

I get that also, but doing it that way makes the dice "distinct", I think. These dice are all being rolled at once and from what I just read, with 5 dice there are 252 "non-distinct" combinations. For example 1-2-3-4-5 is no different than 1-3-2-4-5. We don't care what the order is. For a small example, with 2 dice, there are 21 non-distinct combinations, not 36 as one might assume.

[PSPookie]

Originally Posted by The Crusader

I get that also, but doing it that way makes the dice "distinct", I think. These dice are all being rolled at once and from what I just read, with 5 dice there are 252 "distinct" combinations. For example 1-2-3-4-5 is no different than 1-3-2-4-5. We don't care what the order is. For a small example, with 2 dice, there are 21 distinct combinations, not 36 as one might assume.

That's why it's 1/(6^4) and not 1/(6^5).

The dice may be rolled together but each die represents an independent probabilistic event. That is: the number showing on the 1st die does not effect the probability of what will show on the 2nd, etc. As such, one can simply multiply the probabilities of each event to arrive at a solution. There may only be 21 distinct combinations but there are many more permutations.

[Truktek2]

my brain hurts.

[The Crusader]

Originally Posted by PSPookie

That's why it's 1/(6^4) and not 1/(6^5).

The dice may be rolled together but each die represents an independent probabilistic event. That is: the number showing on the 1st die does not effect the probability of what will show on the 2nd, etc. As such, one can simply multiply the probabilities of each event to arrive at a solution. There may only be 21 distinct combinations but there are many more permutations.

But since there are only 252 non-distinct combinations (my apologies, I meant to type non-distinct earlier) by rolling all 5 at once, could it be 6 in 252? i mean people are winning the pot when the pot is still small, not like $1296.

the way I see it, and I probably am 100% wrong knowing me, if we were to roll one die at a time, then we have given the dice individuality and we would be looking for a certain combination/order. But by rolling all at once, they are non-distinct.

It's probably 1 in 1296. I think I'm probably overthinking this.

[RWP]

Along the same lines, how many 3 letter combination can there be? 26x26x26=17,576?

[PSPookie]

Originally Posted by The Crusader

But since there are only 252 non-distinct combinations (my apologies, I meant to type non-distinct earlier) by rolling all 5 at once, could it be 6 in 252? i mean people are winning the pot when the pot is still small, not like $1296.

the way I see it, and I probably am 100% wrong knowing me, if we were to roll one die at a time, then we have given the dice individuality and we would be looking for a certain combination/order. But by rolling all at once, they are non-distinct.

It's probably 1 in 1296. I think I'm probably overthinking this.

Yup, you're overthinking this

Probability = (Number of Favorable Outcomes) / (Total Number of Possible Outcomes)

There are only 6 possible premutations that are favorable and there are 7776 total permutations. In this case the favorable combinations only have one corresponding permutation.

Let me ask you this question: If you were to roll each die individually, would this impact your chances of winning?

Originally Posted by RWP

Along the same lines, how many 3 letter combination can there be? 26x26x26=17,576?

Right, and if the order is unimportant then there are fewer combinations but the same number of permutations.

[The Crusader]

Originally Posted by PSPookie

Yup, you're overthinking this

Probability = (Number of Favorable Outcomes) / (Total Number of Possible Outcomes)

Thats what I did the first time.

Originally Posted by PSPookie

There are only 6 possible premutations that are favorable and there are 7776 total permutations. In this case the favorable combinations only have one corresponding permutation.

Let me ask you this question: If you were to roll each die individually, would this impact your chances of winning?

I guess not, but the fact that there are only 252 non-distinct combinations screwed me up.

[The Crusader]

Originally Posted by PSPookie

Yup, you're overthinking this

Probability = (Number of Favorable Outcomes) / (Total Number of Possible Outcomes)

There are only 6 possible permutations that are favorable and there are 7776 total permutations. In this case the favorable combinations only have one corresponding permutation.

Let me ask you this question: If you were to roll each die individually, would this impact your chances of winning?

I thought about this some more, as you may have figured.
There are indeed 7776 permutations. However, many of them repeat as combinations, which is to say that when the dice are rolled they may read 1-2-3-4-5 OR to you, you may read them in the order of 1-2-3-5-4 or to someone else 5-4-3-2-1. On the surface, they appear different, but for the sake of what we are doing, it is repetitious. How many different ways can 1-2-3-4-5 be arranged, I'm not gonna figure it out.

BUT, since we are not looking for the dice to roll in a certain order, the duplications must be discarded, which increases the probabilities.

The same goes for 1-1-1-1-1. We don't care what way the dice tumble out die number 3 could come out first and be a 1, die number 2 could be next and be a 1, etc.

Or on a smaller scale: 2 dice. 36 permutations, but only 21 combinations:

1 2 3 4 5 6
1 11 12 13 14 15 16
2 21 22 23 24 25 26
3 31 32 33 34 35 36
4 41 42 43 44 45 46
5 51 52 53 54 55 56
6 61 62 63 64 65 66

12 and 21 are the same, so I delete 21 because they are being rolled at the same time. If we were to assign name to the dice, A and B, THEN, 12 and 21 would be individual.

[PSPookie]

^ Try testing your hypothesis experimentally with just 2 dice. If you roll doubles more frequently than 1 in 6 times over a reasonable statistical sampling then you can conclude you are correct.

[RWP]

Originally Posted by PSPookie

........Right and if the order is unimportant then there are fewer combinations but the same number of permutations.

Thanks!

[The Crusader]

Originally Posted by PSPookie

^ Try testing your hypothesis experimentally with just 2 dice. If you roll doubles more frequently than 1 in 6 times over a reasonable statistical sampling then you can conclude you are correct.

I'm not sure if you're being facetious or experimental, but it's good either way. One of the reasons I think I'm correct is because people have been winning more, actually much more, than 1 in 1296. The pot starts at $100 automatically, the biggest pot has been $250, which would signify 75 rolls.

[i_got_a_mohawk]

Originally Posted by The Crusader

I get that also, but doing it that way makes the dice "distinct", I think. These dice are all being rolled at once and from what I just read, with 5 dice there are 252 "non-distinct" combinations. For example 1-2-3-4-5 is no different than 1-3-2-4-5. We don't care what the order is. For a small example, with 2 dice, there are 21 non-distinct combinations, not 36 as one might assume.

1-2-3-4-5 and 1-3-2-4-5 may not be any different when you are looking for non-distinct combinations. But, even though they are non-distinct combinations they are still two seperate possibilities. (If that makes sense? The result of each may be the same in the manner you are looking at, but the paths are different, if you catch my drift)

Rolling the die one at a time or all 5 at once has no impact on the chances.

[The Crusader]

Originally Posted by i_got_a_mohawk

1-2-3-4-5 and 1-3-2-4-5 may not be any different when you are looking for non-distinct combinations. But, even though they are non-distinct combinations they are still two seperate possibilities. (If that makes sense? The result of each may be the same in the manner you are looking at, but the paths are different, if you catch my drift)

Rolling the die one at a time or all 5 at once has no impact on the chances.

I understand what you are saying, but let's give the dice names, and they tumble out like this:

A)1 B)2 C)3 D)4 E)5 or A)1 C)3 B)2 D)4 E)5 Same thing, right?

OR

A)1 B)1 C)1 D)1 E)1 or A)1 C)1 B)1 D)1 E)1. So how many different ways can the dice be arranged for all ones?

abcde
abced
acbde
acdeb
acebd
adbce
adceb
adebc etc, etc.

If it helps, picture the dice as each one having a different color.

[clmayhew]

I like to think everything in life is a 50/50 chance, but that could just be the tiny optimist in me.