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## A Thrilling Game of Swaff

Probability puzzles require you to weigh all the possibilities and pick the most likely outcome.

 Puzzle ID: #30728 Fun: (2.48) Difficulty: (2.2) Category: Probability Submitted By: bigSWAFF_69_

In a game of Swaff, players have 1 turn to roll 2 dice, to achieve a total number that is a multiple of 3. If the combined number is NOT a multiple of three, the player is out. If it is, he/she goes on to the next round. If no players get a multiple of 3, in any round, the game is over in a tie.

Example: John rolls a 2 and a 4. Total score is 6 - a multiple of 3. John would then move on to the next round.

Lets say that John, Sally, Suzan, and Jimmy were playing a game of Swaff. What is the probability that the game ends the first round as a tie?

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 AwwwSweet May 27, 2006 Terrific job, Swaff! I like what you named the game... GebbieRose May 28, 2006 one of those classic probabilities with dice from school. always fun to keep in practice. JessicaLynn May 28, 2006 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. JessicaLynn May 28, 2006 By the way, Great teaser. One of the few probibility teasers I like. udoboy Jun 01, 2006 Silly Jessie, 7+4 =11, not a multiple of 3. It's not even odd! brainjuice Jun 05, 2006 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! ^^ sens0r108 Jun 07, 2006 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. bobthesnail Jun 09, 2006 "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. bobthesnail Jun 09, 2006 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. snowmonster Jun 09, 2006 This was great Swaff! Loved the title too! calmsavior Nov 19, 2006 That was easy, but cool. opqpop Sep 28, 2010 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.