I forget the equation for this problem... some help?
Join Date: Jan 2008
Posts: 4
I think this problem is pretty simple, just haven't done this kind of math in a while. Anyway, here's the problem.
You get 6 dice. Each dice has 6 sides. Numbered 1-2-3-4-5-6. The dice are put in a cup, shaken, and then dumped onto a table. What are the odds/ % chance that all the die fall on the same number. Let's say it costs $1 for 2 rolls, how much money would you have to win on your $1 wager for this to be profitable.
Thanks in advance to any reply's if you could show the equation i'd appreciate it.
Join Date: Jul 2004
Posts: 483
It's late, but I'll take a stab.
(1/number of sides)^(number of dice) would give you chances of say 111111.
(1/6)^6 = 0.000021433 or about 46656-1. But adding 222222, 333333, etc....to your slim chances would be like 7776-1. For $0.50/roll, they should offer close to $3900 for any six dice matching.
Let's check with with two dice, chance to throw: 11, 22, 33, etc...
(1/6)^2 = 0.0277778 for say 11. Times 6 for 22,33,44...would be 0.166667 or 1/6, which would make sense. For two dice that match, they should pay $3 for a $0.50 shot that matches two dice (or $2.50 + return your $0.50).
Last edited by bbarker703; 01-25-2008 at 02:30 AM.
Join Date: Jul 2007
Posts: 7,474
Each die has 6 possible values.
There are 6 die.
The total number of outcomes is 6^6.
Only one of those shows a '1' on all the dice.
So the probability of all dice showing a '1' is 1 out of (6^6).
The probability that all dice show the same number (where that number can be 1,2,3,4,5, or 6) must be 6 x (1 out of (6^6))
which = 1 out (6^5) = 1 out of 7776
You'd need to be getting odds of 7775 - 1 to breakeven.
At $0.50 a roll, the prize would have to be at least $3888.
I think that's right.
Join Date: Jan 2008
Posts: 4
looks good to me as well, thanks a lot guys.
