Need a statistician or someone smarter than me...

[millstreaminn]

We are doing a charity event this summer where we will take 30 people on a wagon ride around our town. We plan to stop at 5 locations, where each person will shoot 2 dice and the highest total at the end of the ride will win $100.00. What are the chances of a tie? I'm sure it involves multiplication and a couple cans of beer, but it's too early for the beer so I need some help. Thanks- :tiphat:

 

[red angus 2010]

We are doing a charity event this summer where we will take 30 people on a wagon ride around our town. We plan to stop at 5 locations, where each person will shoot 2 dice and the highest total at the end of the ride will win $100.00. What are the chances of a tie? I'm sure it involves multiplication and a couple cans of beer, but it's too early for the beer so I need some help. Thanks- :tiphat:

Two dice can produce 21 different numerical totals with 30 people rolling this could theoretically create 630 totals at each stop. With five stops this would mean the possibility of 3,150 final totals. However there will be a probability equation that must be calculated to account for how many times people will roll the same number. Am working on that. Is there the possibility that if the five stops serve alcohol and after the fifth stop no one will remember what they have rolled and all you have to do is declare a winner?

 

[Rafter S]

I'm not a statistician and not smarter than hardly anybody, but I like working with numbers so I'll give it a shot. First, shooting two dice will give a total that can range from 2 to 12. Each person will do that 5 times, so that will give a grand total for each person that will range from a possible low of 10 to a high of 60. So unless I'm confused or missed something, which is always possible, there are 49 possible totals. However, the odds of anyone rolling snake-eyes or boxcars all 5 times has got to be pretty long, so I would think that, while I can't put a number on the odds, with 30 people I'd think you've got a fair chance of a tie.

I work with some engineers who are definitely smarter than I am, so if I get a chance I'll run this past one of them and see what they come up with.

 

[Rafter S]

With all due respect, I think you might have missed something. While I'll agree that rolling 2 dice will give 21 different possible combinations, I don't see how the total can be less than 2 or more than 12.

 

[pdfangus]

With all due respect, I think you might have missed something. While I'll agree that rolling 2 dice will give 21 different possible combinations, I don't see how the total can be less than 2 or more than 12.

yeah....dice has six sides with number ranging from one to six.....
two dice the max number is 12.....minimum is 2 so eleven variables per roll
....

 

[TennesseeTuxedo]

Just stipulate that in the unlikely event of a tie the prize money will be evenly divided and forget the math.

 

[TexasBred]

In case of a tie the winner of "Chug-a-lug a quart of beer" contest will be declared overall winner.

 

[TennesseeTuxedo]

In case of a tie the winner of "Chug-a-lug a quart of beer" contest will be declared overall winner.

Sheer genius!

 

[Alan]

Not going to try to figure this out, but you have to factor in that;

There is only one combition of the dice that rolls a two
One combo for a three
Two combos for a four
Two combos for a five
Three combos for a six
Three combos for a seven
Three combos for an eight
Two combos for a nine
Two combos for a ten
One combo for an eleven
One combo for a twelve

Too much brain work for me to try to figure the odds of a tie. I vote for the chugging contest as the tie breaker,

 

[Limomike]

I'm not a statistician and not smarter than hardly anybody, but I like working with numbers so I'll give it a shot. First, shooting two dice will give a total that can range from 2 to 12. Each person will do that 5 times, so that will give a grand total for each person that will range from a possible low of 10 to a high of 60. So unless I'm confused or missed something, which is always possible, there are 49 possible totals. However, the odds of anyone rolling snake-eyes or boxcars all 5 times has got to be pretty long, so I would think that, while I can't put a number on the odds, with 30 people I'd think you've got a fair chance of a tie.

I work with some engineers who are definitely smarter than I am, so if I get a chance I'll run this past one of them and see what they come up with.

wow. thats some good cipherin there.. :shock:

 

[pdfangus]

any single roll can only give you one of eleven totals
six seven and eight have slightly more ways to get there....but you still only get one total....

 

[millstreaminn]

Not going to try to figure this out, but you have to factor in that;

There is only one combition of the dice that rolls a two
One combo for a three
Two combos for a four
Two combos for a five
Three combos for a six
Three combos for a seven
Three combos for an eight
Two combos for a nine
Two combos for a ten
One combo for an eleven


One combo for a twelve

Too much brain work for me to try to figure the odds of a tie. I vote for the chugging contest as the tie breaker,

I saw a hamburger joint that claimed to sell over 2 million different hamburger combinations. They had something like 4 different buns, 3 sizes of beef patty and 28 toppings. I was hoping my dice question would be as easy as multiplying 30 people times 5 stops..... :nod:

 

[Dave]

In case of a tie just use the money to buy more beer and that way everyone wins.

 

[suzorse]

in case of a tie each person rolls 1 dice and the high roll wins if another tie you just roll again
simple, that is probably how they do it on a poker ride , a tie of poker hand , high card wins
Suzanne

 

[Dega Moo]

You can reduce likelihood of a tie using one or more 'twists' on the basic game:

1) multiply the the die - will get you more possibles
2) use colored die and declare one color the first number and the other color the second number. So a red and white die with red the first number and white the second red comes up 4 and white comes up 2 then the result is 42. If red is a 1 and white is a 2 then the result is 12.
3) apply a factor for each stop and multiply results by that factor at each stop
4) pick a total number they need and the closest to that number wins. Let them call a die negative or positive, before the roll of each. Multiply die results, add to total.

Really though, I like the beer solution best.

 

[melking]

I am not a math person, and not really all that bright but it would seem to me that at every stop you would have a chance of 15 ties (if you have 30 people) Seems like each stop starts a new randomness.

 

[red angus 2010]

With all due respect, I think you might have missed something. While I'll agree that rolling 2 dice will give 21 different possible combinations, I don't see how the total can be less than 2 or more than 12.

You're right I was looking at the possible combinations of dice not the numerical totals:
1 1
1 2
1 3
1 4
1 5
1 6
2 2
2 3
2 4
2 5
2 6
3 3
3 4
3 5
3 6
4 4
4 5
4 6
5 5
5 6
6 6

The numerical totals would be 2, 3, 4, 5, 6, 7, 8, 9, 10, 11 & 12

I'm sure there is a formula to determine the probability of how many time thirty people rolling five times would roll the same number total.

Thanks for catching my error.

 

[Jogeephus]

Its been a long time since I've worked with permutations but I think the answer you are looking for is 1 : 5917.

You take the number of possible outcomes, Max being 60. Subtract 1 from this and you get 59. Take 1 and divide by 59 and this is your percentage. .0169% Convert to a whole decimal number and you get .000169. Divide 1 by this and you get the 5917. Not likely but in gambling its not good odds for the house.

BTW - I reserve the right to be wrong.

 

[HOSS]

Its been a long time since I've worked with permutations but I think the answer you are looking for is 1 : 5917.

You take the number of possible outcomes, Max being 60. Subtract 1 from this and you get 59. Take 1 and divide by 59 and this is your percentage. .0169% Convert to a whole decimal number and you get .000169. Divide 1 by this and you get the 5917. Not likely but in gambling its not good odds for the house.

BTW - I reserve the right to be wrong.

I was waiting on you to jump in If anybody was going to have a shot on CT it was gonna be you. Who says an old Georgia boy ain't got brains? :mrgreen:

 

[TexasBred]

Its been a long time since I've worked with permutations but I think the answer you are looking for is 1 : 5917.

You take the number of possible outcomes, Max being 60. Subtract 1 from this and you get 59. Take 1 and divide by 59 and this is your percentage. .0169% Convert to a whole decimal number and you get .000169. Divide 1 by this and you get the 5917. Not likely but in gambling its not good odds for the house.

BTW - I reserve the right to be wrong.

I was waiting on you to jump in If anybody was going to have a shot on CT it was gonna be you. Who says an old Georgia boy ain't got brains? :mrgreen:

Hoss the impressive part is that he worked all that out on the clean side of a piece of 2x4 with a #2 pencil. :mrgreen: