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Doubling Dice Game

Submitted By:javaguru
Fun:*** (2.4)
Difficulty:**** (3.22)

You are playing a game of dice with two friends, Tom and Harry. Each person has a cup with six dice. Each round the players ante $1 each and then all slam their dice cup open-end down on the table. Each player lifts the edge of their cup to look at the roll of their six dice, but keeps their roll concealed from the other players.

On their turn each player may either pass or double the antes for all the players. After the third player's turn all the dice are revealed to determine the winner.

The winner is the person with the most dice with the same value. If two or more players have the same number of matching dice, then the higher value of the dice wins. For example, four ones beat three sixes, but three sixes beat three fives. If two or more players have the same number and value for the winning roll, then they split the antes. The starting person rotates so that the person who went first in a round goes last in the next round.

On the first round you roll a pair of sixes. Harry passed on his turn and now it's your turn. You know that Harry would have doubled the ante if he had three of a kind or better, but that he wouldn't double the ante with a pair when having to go first. So Harry could have the same roll as you, but you're not worried about Harry beating you. You also know that Tom will also double the ante on his turn if he has three of a kind or better, but that he won't double it with a pair.

Should you pass or double the antes?

Bonus question: Does your action change if instead of splitting the antes on a tie, the antes instead stay in the pot for the next round?

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Comments on this teaser

Posted by dreamndesire02/11/09
i understood half of the question and very little of the answer, but nice! :D

Posted by Zag2402/14/09
It wasn't quite clear to me what you meant by "doubling the antes." I assume the other players have to match your increase (or it would just be stupid) but do they have to match it or can they fold?

Posted by javaguru02/14/09
I tried to make it clear: "double the antes for all the players". I called them antes instead of bets, since that's what you have to put in just to participate in the wager, whereas a bet implies something you can either accept or not. So no, they can't drop. All players will have to double the money they have at risk.

Posted by javaguru05/12/09
Nate, the problem with C(6,3) * 5 * 6 is that you count each combination of triples twice: once as three ones plus three twos, and then again as three twos plus three ones. :o Here are all the combinations of two different numbers from 1 to 6: (1,2), (1,3), (1,4), (1,5), (1,6), (2,3), (2,4), (2,5), (2,6), (3,4), (3,5), (3,6), (4,5), (4,6), (5,6). Your approach also counts (2,1), (3,1), (4,1), ...

Posted by azbycx06/09/10
That was a long expanation. I got part of it, but not all. :-?

Posted by masutra9505/27/15
Good problem. Only the analysis of having a pot in case of ties seems faulty. On the next turn each of the players has the same chance of winning the pot regardless of who goes first, since it is impossible to fold. So your expected winnings do not change at all with a pot.

Posted by javaguru05/27/15
The odds are not the same when the money stays in the pot. When the money stays in the pot, you have zero advantage. If you tie with one person, you win one and a half times what you risked. It's only in the case that there is a three-way tie that the odds are the same.

Posted by Robot08/29/15
A simpler way of looking at this is: 1)You're not worried about Harry beating you 2)You want to beat Tom 3)Tom will also double the ante on his turn if he has three of a kind or better, but that he won't double it with a pair 4)Note the rules for winning => You should pass if Tom has a better chance of combination for winning than yours, lest double up. Ignoring the bluffing chances. The only set of possibilities of having a better combination of the dice for winning than yours (pair of sixes) are: 3 common, 4 common, 5 common, and all the same = Total no. of possibilities - 1 common, 2 common Total number of possibilities = 6^6 => P(better combination for Tom)= 1- {# 1 common + # 2 common}/6^6 = 1- {6! + 6!*40}/6^6 =119/324 < 1/2 => Double up. The bonus question was answered in the most shortest way.


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