A game with two dices

lgon

New Member
#1
Imagine a game with two players and referee.
The referee rolls two unbiased dices: one of the dices outcome is a score and the other dice outcome is the number of times that the first dice should be rolled.
For instance, if the second dice outcome is 4, then the referee will roll 4 times the first dice, and the result will be the sum of the 4 outcomes.
The players never know the second dice outcome; theu are only informed of the first dice outcome.
In my example, (first dice will be rolled 4 times) the players would only be informed of each outcome of the first dice. Imagine that the outcomes will be 2, 6, 3, 3 (total 14).
You must put your bets before the result of each outcome.
In the begining, without any information, you would place a bet for a total outcome.
After you know the first outcome (in our example, 2) you can revise your bet.
Then, after knowing the second outcome (6), you can revise your bet again. And so on.

What is the best strategy for this game (not using brut force...)?
 

Englund

TS Contributor
#2
A tip is to calculate the expected value,

\(E[T=total]=E[XY]=E[X]E[Y]=\sum_{x=1}^6{xP(Y=y)}\sum_{y=1}^6{yP(X=x)}=(7/2)^2\),

where y is the second die and x is the first. If you get to roll a second time, you know that \(y>1\), so the new expected of Y is conditioned on that \(y>1\). I hope I've pointed you in the right direction.
 
Last edited:

Englund

TS Contributor
#3
I'll give you the correct expected values and then you can try getting to the same conclusion.

\(E[T|y>1]=x_1+10.5\)
\(E[T|y>2]=\sum_{i=1}^2{x_i}+8.75\)
\(E[T|y>3]=\sum_{i=1}^3{x_i}+7\)
\(E[T|y>4]=\sum_{i=1}^4{x_i}+5.25\)
\(E[T|y>5]=\sum_{i=1}^5{x_i}+3.5\)

I just figured that it is impossible to play according to this strategy since the value \(x_1+10.5\), and so on, is impossible to attain in a game. The score can only be integers. But these values can serve as hints to how to place your bets.
 

lgon

New Member
#4
Thanks for the answer. But it seems to me that knowing the expected value doesn't help much about the strategy. For instance, before any information is given, EV is indeed 12.25, but the most probable outcome is 6 (with 6% prob), the second is 5 (with 5.15% prob), and the third is 11 (with 4.88% prob).
Is there a set of formula to compute the probabilities for each outcome, in each phase of the game?
 
Last edited: