\G/

09-15-2009, 07:38 AM

So a couple of months ago I was at a party in Surfers Paradise (not my home town) and a dice game broke out. The rules were local to the reigon and I had never heard them before, but being a bit of a gambler I thought I would learn and have a punt. I came out behind about $160 but that doesn't bother me at all. What bothers me is that the game seems fundamentally flawed (not that it is rigged, because all players play by the same rules - but you will see what I mean).

The rules were simple.

You play with three dice.

All three are rolled at once.

If your roll is a scoring roll, you pass to the player to your left.

If your roll is not a scoring roll, you roll again.

The dice are passed round the circle until every player has a scoring roll.

Highest scoring roll wins the buy ins (equal for every player - in this case $20 per game).

If the highest score is equalled, the game resets from the start and all scores are cancelled.

Scoring rolls go like this (from lowest to highest):

1,2,3

A double. (doubles are ranked by the 3rd dice - ie: the dice which is different to the other two. EG: double one with a six, is better than double six with a two)

A triple.

4,5,6

Now I am not a great statistician but appear to me that:

The highest roll and the lowest roll have the exact same probability of occuring.

Worse still the odds of rolling the second highest roll (triple 6) or any other triple are substancially lower than rolling the highest scoring roll (4,5,6).

Here is my logic:

The odds of rolling triple 6 are:

6x6x6 = 216 (6x6x6 because only one number may appear on each die)

216 to 1.

The odds of rolling 4,5,6 are:

2x3x6 = 36 (as the first die can be any one of three numbers, the second can be either one of the remaining numbrs, while the third

can only be one).

This would appear to mean you are 6 times more likely to roll the best score than the second best.

Is my logic correct or flawed, please tell me!

Cliffs:

- Trying to work out is my odds calculations are correct.

The rules were simple.

You play with three dice.

All three are rolled at once.

If your roll is a scoring roll, you pass to the player to your left.

If your roll is not a scoring roll, you roll again.

The dice are passed round the circle until every player has a scoring roll.

Highest scoring roll wins the buy ins (equal for every player - in this case $20 per game).

If the highest score is equalled, the game resets from the start and all scores are cancelled.

Scoring rolls go like this (from lowest to highest):

1,2,3

A double. (doubles are ranked by the 3rd dice - ie: the dice which is different to the other two. EG: double one with a six, is better than double six with a two)

A triple.

4,5,6

Now I am not a great statistician but appear to me that:

The highest roll and the lowest roll have the exact same probability of occuring.

Worse still the odds of rolling the second highest roll (triple 6) or any other triple are substancially lower than rolling the highest scoring roll (4,5,6).

Here is my logic:

The odds of rolling triple 6 are:

6x6x6 = 216 (6x6x6 because only one number may appear on each die)

216 to 1.

The odds of rolling 4,5,6 are:

2x3x6 = 36 (as the first die can be any one of three numbers, the second can be either one of the remaining numbrs, while the third

can only be one).

This would appear to mean you are 6 times more likely to roll the best score than the second best.

Is my logic correct or flawed, please tell me!

Cliffs:

- Trying to work out is my odds calculations are correct.