Another probablility problem.

[Guv]

I think I posted a discussion of this game in another Thread, perhaps in the Chit Chat Forum before there was a Maths Forum.

Consider three dice, each of which has the same number on opposite faces.

• Die A has 1, 6, 8 (twice each).
• Die B has 3, 5, 7 (twice each).
• Die C has 2, 4, 9 (twice each).

Now consider playing a game using the above three dice. Each of three players uses one of the above three dice. For conveneince, call the players Albert, Bob, & Charlie using die A, B, & C respectively.

A play consists of each player rolling his die once. If the number rolled by Albert is higher than that rolled by Bob, Bob pays Albert $1.00, and vice versa. Similarly: Albert pays or gets paid by Charlie; And Bob pays or gets paid by Charlie.

Note that a single roll cannot result in a tie. From the point of view of one player (say Albert), there are three possibilites.

• He loses to both Bob & Charlie and pays $2.00
• He wins from both and gets paid $2.00
• He wins from one and loses to the other player, breaking even.

I hope the above provides an adequate description of the game.

Now, calculate the following probablilites.

• P(Albert winning from Bob).
• P(Bob winning from Charlie)
• P(Charlie winning from Albert)

[thinktank2]

The greater than sysmbol is to indicate that the number on one die is greater than another die.

A > B means the value of the number on Die A is greater than the value of the number on Die B.

With that meaning,

Albert > Bob > Charlie > Albert is a wrong statement.

It can never happen.

For Each turn of 1 roll each,
The number on one die is greater than the other two
that is to say, The Possible cases are

A > B , A > C and ( B > C or C > B)
B > A , B > C and ( C > A or A > C)
C > A , C > B and ( A > B or B > A)

Never ever will this happen,
A > B , B > C and C > A

The probabilities that were calculated previously are for the
Events
1) A > B
2) B > C
3) C > A

It doesn't say anything about comparing three dice together.

The moment, A > B and B > C the probability of Event C > A becomes zero.

So ,..Can we call
the event ( A > B with B > C ) and
the event ( C > A )

Mutually Exclusive ???

[Guv]

Thinktank2: Think about the following.

Die A is likely to roll a higher number than die B.
Die B is likely to roll a higher number than die C.
Die C is likely to roll a higher number than die A.

While the overall game is fair to the three players. If one drops out, the game is not fair for the two remaining players.

As mentioned in a previous post, is likely to is not transitive like greater than, which seems counter intuitive to me. When I first read about this game, I was surprised.

Apparently, the above does not seem at all strange or counter intuitive to you.