# Rolling Dice Problem

1. Sep 14, 2011

### dspampi

Consider the experiment of rolling 5 fair dice. You “win” if all the dice show different numbers. I have to decide in advance how many times to repeat the experiment.
I will be very happy if I win at least once. What is the least number of times I need to plan to repeat the experiment so that my probability of winning at least once is more than 99%?

So there is an equal chance that one of six possibilities are possible for for each dice roll.
The first roll will have a different number = 1; 2nd is different 5/6, 3rd is 4/6, 4th is 3/6, 5th is 2/6 and the 6th is 1/6....so is the probability just the multiplication of all the chances and how do I get to 99% confidence?

2. Sep 14, 2011

### Ray Vickson

Are you sure you are setting up the problem correctly? To me, it reads that in each round of play you roll all 5 dice at once, and win if all 5 numbers are different. If so, then each round of play is independent of the previous, unless you have already stopped because you have already won. If my interpretation is correct the problem is just like that of tossing a biased coin and winning as soon as you get 'heads'.

RGV

3. Sep 14, 2011

### PeterO

As RGV says, you are rolling 5 dice at a time. The chance that all 5 are different can thus be calculated - lets assume it is 20%.
That means there is an 80% chance you have not won.

You now work on the probability of NOT winning.
If you roll again there is now a 64% chance you have not won [80% of 80%]
You want to calculate how many rolls until the chance you have not won is less than 1%
If there is less than 1% chance of NOT winning, there must be a 99% chance you have won.

Don't forget, I "made up" the 20% probability for explanation purposes; you need to use the real answer.

4. Sep 14, 2011

### dspampi

Ok well I agree that all 5 dice are being rolled at the same time so I think regardless that
P(win) = 6/6 * 5/6 * ... * 2/6 = 6!/6^5
and P(lose) = 1 - P(win) = 0.907407407

So to calculate the rolls that I have lost down to 1%,
is that just (P(lose))^n = 1?

5. Sep 15, 2011

### PeterO

That is correct, except it should be <= 1 , not just =1 , with n an integer.

6. Sep 15, 2011

### willem2

And 1% is a probability of 0.01, not 1

7. Sep 15, 2011

Indeed!!!!