• TalkBass has been independent since 1998. Add your voice.
    Create a free account to reply to discussions, view embedded media, and browse with fewer display ads.
    Join free Log in
    Want zero display ads or expanded classifieds tools? Compare plans.

Help me out with a Dice Probability Question

Johnny Fila

Johnny Fila

Formerly "The Crusader"
Nov 21, 2004
3,460
27
59
Macungie, PA
#1
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.
 
#2
The Crusader said:
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.
Click to expand...

1295 to 1 to be specific.
But 1 in 1296 is also correct :)
 
#3
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
 
#7
PSPookie said:
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
Click to expand...

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.
 
#8
The Crusader said:
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.
Click to expand...

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.
 
#10
PSPookie said:
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.
Click to expand...

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:o, 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.:D
 
#12
The Crusader said:
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:o, 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.:D
Click to expand...

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?

RWP said:
Along the same lines, how many 3 letter combination can there be? 26x26x26=17,576?
Click to expand...

Right, and if the order is unimportant then there are fewer combinations but the same number of permutations.
 
#13
PSPookie said:
Yup, you're overthinking this ;)

Probability = (Number of Favorable Outcomes) / (Total Number of Possible Outcomes)
Click to expand...
Thats what I did the first time.

PSPookie said:
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?
Click to expand...

I guess not, but the fact that there are only 252 non-distinct combinations screwed me up.
 
#14
PSPookie said:
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?
Click to expand...

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.

Invalid Link Removed
 
#17
PSPookie said:
^ 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.
Click to expand...

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.
 
#18
The Crusader said:
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.
Click to expand...

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.
 
#19
i_got_a_mohawk said:
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.
Click to expand...

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.
 

Latest posts

Top Bottom
Back