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| Posted by AwwwSweet | 05/27/06 |
| Terrific job, Swaff! I like what you named the game... :lol: |
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| Posted by GebbieRose | 05/28/06 |
| one of those classic probabilities with dice from school. always fun to keep in practice. |
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| Posted by JessicaLynn | 05/28/06 |
| I feel stupid. When I got 2/3 to the power of 4, I multiplied 2/3 x 2/3. I got 4/9 x 2/3 = 8/27. When I multiplied the 27 by 3, somehow I ended up with 74. So I got 16/74. |
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| Posted by JessicaLynn | 05/28/06 |
| By the way, Great teaser. One of the few probibility teasers I like. :D |
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| Posted by udoboy | 06/01/06 |
| Silly Jessie, 7+4 =11, not a multiple of 3. It's not even odd! |
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| Posted by brainjuice | 06/05/06 |
| easy and nice teaser :) but i'd rather find combination of dice which is multiple of three than find combination of dice which is not multiple of three. never mind, nice one! ^^ |
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| Posted by sens0r108 | 06/07/06 |
| Wouldn't the total possible combinations be 21?
1,1 2,2 3,3 4,4 5,5 6,6
1,2 2,3 3,4 4,5 5,6
1,3 2,4 3,5 4,6
1,4 2,5 3,6
1,5 2,6
1,6
Since 1,6 and 6,1 would equal the same number either way, those shouldn't be counted I don't think. I'm not real sure, but that was the way I was thinking. |
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| Posted by bobthesnail | 06/09/06 |
| "Since 1,6 and 6,1 would equal the same number either way, those shouldn't be counted I don't think. I'm not real sure, but that was the way I was thinking."
Good question, but they should be counted seperately.
Here's why:
There are two ways to get a combination of 1 and 6. Lets make the dice red and blue to differentiate them.
You can have a Blue 1 and a Red 6.
You can have a Blue 6 and a Red 1.
But there is only one way to get a 3 and 3 (or any other pair). That is a Blue 3 and a Red 3.
Counting 1-6 without counting 6-1 discounts the fact that rolling a 1 and 6 is twice as likely as rolling a 3-3. Your counting scheme would count them equally likely.
Another way to look at it. You are told that someone has two kids. What is the likelyhood that they are a boy and a girl?
The way you calculated, you would say that there are three possibilities. 2 boys, 2 girls, 1 of each. In actuality, the 1 of each has twice the probability of being true, since it can be satisfied with an older boy, younger girl, or an older girl, younger boy, while the other two (2 boys and 2 girls) have only one way to be true - older boy and younger boy (or girl, for the 2 girl scenario).
So the answer would be that there is a 1/2 chance of a boy and girl combo, but if you don't count the two possible permutations of BG, then you might mistakenly call it a 1/3 possibility. |
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| Posted by bobthesnail | 06/09/06 |
| oooh. I just thought of a different way to look at this problem.
Roll one die. Regardless of the roll on the first die, the sum will be a multiple of three if one of two numbers turns up on the other die.
In other words, you roll a 1 on the first roll. Your sum is a multiple of three if you roll a 2 or 5 on the second.
You roll a 2 on the first roll. Your sum is a mult. of 3 if you roll a 1 or 4 on the second.
You roll a 3 on the first roll. Your sum is a mult of 3 if you roll a 3 or 6.
In other words, the first roll just dictates what modulo of 3 on the second roll wins. Since there's two numbers each of mod 0, 1, and 2 on a die, there is always a 1/3 chance the second roll will make you a winner. |
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| Posted by snowmonster | 06/09/06 |
| This was great Swaff! Loved the title too! |
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| Posted by calmsavior | 11/19/06 |
| That was easy, but cool. |
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| Posted by opqpop | 09/28/10 |
| Let p be the probability you don't get multiple of 3, which is 1 - (2 + 5 + 4 + 1) / 36 = 2/3.
(2/3)^4 = 16/81. |
