Math Fun: With Dice

I thought it would be fun to share a little math with you,
so I wrote up a page how dice A can beat dice B, while B beats C, and C beats A.

 

You walk into a bar with three (three sided) dice A, B, C, and find a man to gamble with.
 For concreteness,

 

1st side

2nd side

3rd side

Dice A

10

21

32

Dice B

11

22

30

Dice C

12

20

31

 

You will each roll one dice. The one with the lower number pays the other $10.

To be fair, you let him reach roll choose his favourite dice.

If he chooses A, you choose B.

If he chooses B, you choose C.

If he chooses C, you choose A.

Is it possible that you always expect to win 5 out of every 9 rolls?

So the expected/average  gain is $10´5/9 -$10´4/9 = $1.11 each roll. Not a big profit.

 

If he has A and you have B        

If he has B and you have C        

If he has C and you have A        

A\B

11

22

30

10              

Win

Win

Win

21

Lose

Win

Win

 32

Lose

Lose

Lose

B \C

12

20

31

11              

Win

Win

Win

22

Lose

Lose

Win

 30

Lose

Lose

Win

C\ A

10

21

32

12              

Lose

Win

Win

20

Lose

Win

Win

 31

Lose

Lose

Win

 

An explanation is the following.

·         We each roll a dice. Let “Blue” be the event that your 10’s digit is different than his.
See all the blue Lose/Win.
This happens 2/3 = 6/9 of the time.
 Because whatever his first digit is, you have 2 out of 3 digits that are different.

·         Conditioned on “Blue”, the probability of you winning is ½.
Note in 3 out of the 6 blue cases, you win.
The reason is that there is symmetry between you and him winning.
We write Pr[You win|Blue] = ½.

·         Conditioned on “Red”, the probability of you winning is 2/3.
Note in 2 out of the 3 red cases, you win.
The reason is that if you count around 0,1,2,0,1,2,0,....
your second digit (in the red case) is always one ahead of his.
Eg He gets 10 on dice A, you get 11 on dice B. 0 vs 1
Eg He gets 21 on dice A, you get 22 on dice B. 1 vs 2
Eg He gets 32 on dice A, you get 30 on dice B. 2 vs 0
And you win 2 out of three of these.

We write Pr[You win|Red] =2/3.

·         Combining these we write
Pr[You win] = Pr[Blue] ´ Pr[You win|Blue] + Pr[Red] ´ Pr[You win|Red]
                      =     2/3     ´           ½                   +      1/3    ´   2/3
                      =      3/9  (note 3 blue wins)     +      2/9     (note 2 red wins)
                      =      5/9 (note 5 wins out of 9)

A general principle is that in most fair games,
if your opponent chooses his strategy first and reveals it to you,
then you can respond with a strategy that beats his.

This is why to win you need secrecy.