# 4 dices with different numbers, which has more chance of getting a higher number

• Aug 10th 2009, 09:56 PM
b0mb3rz
4 dices with different numbers, which has more chance of getting a higher number
There are 4 die, one with numbers 4-4-4-4-0-0, one with 3-3-3-3-3-3, another with 2-2-2-2-6-6 and the last with 1-1-1-5-5-5.

Person A selects a die, then person B selects one from the remaining 3. Then they roll the die and the person with the higher number wins.

Which person A or B has a better chance of winning? Can you calculate these chances?
• Aug 10th 2009, 10:53 PM
CaptainBlack
Quote:

Originally Posted by b0mb3rz
There are 4 die, one with numbers 4-4-4-4-0-0, one with 3-3-3-3-3-3, another with 2-2-2-2-6-6 and the last with 1-1-1-5-5-5.

Person A selects a die, then person B selects one from the remaining 3. Then they roll the die and the person with the higher number wins.

Which person A or B has a better chance of winning? Can you calculate these chances?

Since all 12 possible choices of die can occur and they all have the same probability, choosing first (at random) gives no advantage, so the probability of A winning is 0.5 and of B winning is 0.5 (a draw is not possible).

Now if A can deliberately choose which die they will use and B make their choice knowing what A has choosen we have a different problem. However whatever A chooses B has a choice that restricts the probability of A winning to 1/3 (assuming I have done the sums correctly).

CB
• Aug 11th 2009, 10:18 AM
Soroban
Hello, b0mb3rz!

This is a classic problem . . . "The Non-transitive Dice."
It takes a bit to show our work, but the answer is surprising.

Quote:

There are 4 dice, one with numbers 0-0-4-4-4-4, one with 3-3-3-3-3-3,
another with 2-2-2-2-6-6 and the last with 1-1-1-5-5-5.

Person $A$ selects a die, then person $B$ selects one from the remaining three.
Then they roll the die and the person with the higher number wins.

Which person, $A$ or $B$, has a better chance of winning?
Can you calculate these chances?

It turns out that $B$ has a distinct advantage.

No matter what die $A$ chooses,
. . $B$ can choose a die which gives him a $\tfrac{2}{3}$ probability of winning.

Call the dice: . $\begin{array}{cc}W\!: & \text{0-0-4-4-4-4} \\ X\!: & \text{3-3-3-3-3-3} \\ Y\!: & \text{2-2-2-2-6-6} \\ Z\!: & \text{1-1-1-5-5-5} \end{array}$

There are four cases to consider . . .

[1] $A$ chooses die $W\!:\;\;P(0) = \tfrac{1}{3},\;P(4) = \tfrac{2}{3}$
. . Then $B$ chooses die $Z\!:\;\;P(1) \:=\:\tfrac{1}{2},\;P(5) = \tfrac{1}{2}$

. . $\begin{array}{c||c|c|}
& 1 & 5 \\ \hline\hline \\[-4mm]
0 & \frac{1}{3}\cdot\frac{1}{2} & \frac{1}{3}\cdot\frac{1}{2} \\ \\[-4mm]\hline \\[-4mm]
4 & \frac{2}{3}\cdot\frac{1}{2} & \frac{2}{3}\cdot\frac{1}{2} \\
\end{array}$

. . $P(\text{B wins}) \:=\:\tfrac{1}{3}\cdot\tfrac{1}{2} + \tfrac{1}{3}\cdot\tfrac{1}{2} + \tfrac{2}{3}\cdot\tfrac{1}{2} \;=\;\frac{2}{3}$

[2] $A$ chooses die $X\!:\;\;P(3) = 1$
. . Then $B$ chooses die $W\!:\;\;P(0) = \tfrac{1}{3},\;P(4) = \tfrac{2}{3}$

. . $\begin{array}{c||c|c|}
& 0 & 4 \\ \hline\hline \\[-4mm]
3 & 1\cdot\frac{1}{3} & 1\cdot\frac{2}{3}
\end{array}$

. . $P(\text{B wins}) \:=\:1\cdot\tfrac{2}{3} \:=\:\frac{2}{3}$

[3] $A$ chooses die $Y\!:\;\;P(2) = \tfrac{2}{3},\;P(6)= \tfrac{1}{3}$
. . Then $B$ chooses die $X\!:\;\;P(3) = 1$

. . $\begin{array}{c||c|}
& 3 \\ \hline\hline \\[-4mm]
2 & \frac{2}{3}\cdot1 \\ \\[-4mm]\hline \\[-4mm]
6 & \frac{1}{3}\cdot1 \end{array}$

. . $P(\text{B wins}) \:=\:\tfrac{2}{3}\cdot1 \:=\:\frac{2}{3}$

[4] $A$ chooses die $Z\!:\;\;P(1) = \tfrac{1}{2},\;P(5) = \tfrac{1}{2}$
. . Then $B$ chooses die $Y\!:\;\;P(2) = \tfrac{2}{3},\;P(6) = \tfrac{1}{3}$

. . $\begin{array}{c||c|c|}
& 2 & 6 \\ \hline\hline \\[-4mm]
1 & \frac{1}{2}\cdot\frac{2}{3} & \frac{1}{2}\cdot\frac{1}{3} \\ \\[-4mm]\hline \\[-4mm]
5 & \frac{1}{2}\cdot\frac{2}{3} & \frac{1}{2}\cdot\frac{1}{3} \\
\end{array}$

. . $P(\text{B wins}) \:=\:\tfrac{1}{2}\cdot\tfrac{2}{3} + \tfrac{1}{2}\cdot\tfrac{1}{3} + \tfrac{1}{2}\cdot\tfrac{1}{3} \;=\;\frac{2}{3}$

~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

Isn't that mind-boggling?

. . $\begin{array}{c}
\text{Die }W\text{ beats die }X\\
\text{Die }X\text{ beats die }Y \\
\text{Die }Y\text{ beats die }Z \\
\text{Die }Z \text{ beats die }W \end{array}$

FYI: There is a set of three non-transitive dice.

. . . . . $\begin{array}{c}
(2,2,4,4,9,9) \\
(1,1,6,6,8,8)\\
(3,3,5,5,7,7) \end{array}$

• Aug 17th 2009, 03:38 AM
b0mb3rz
thanks for the replies, im just wandering how it is you came about with the chance of 1/2 that they win if they select the die randomly
• Aug 17th 2009, 04:14 AM
CaptainBlack
Quote:

Originally Posted by b0mb3rz
thanks for the replies, im just wandering how it is you came about with the chance of 1/2 that they win if they select the die randomly

If the die are selected randomly the game is symmetric between the players, and as there can be no draw each must have a win probability of 1/2.

CB