anonymous one year ago can someone explain? two fair die are tossed 5 times, find probablity that sum of 7 will show on the first three tosses and will not show on the other two?

1. anonymous

There are 6 ways for a seven to show on one roll of two dice. 2,5 5,2 3,4 4,3 6,1 1,6 There are 36 ways for the two dice to show a total of 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12. There are 36-6 ways to roll not 7. $\left(\frac{6}{36}\right)^3 \left(\frac{30}{36}\right)^2=\left(\frac{1}{6}\right)^3 \left(\frac{5}{6}\right)^2=\frac{25}{7776}=0.00321502$

2. anonymous

oh, so 6! for total outcome

3. xapproachesinfinity

i was thinking this way there are 6^5 outcomes rolling two dice 5 times 6 of those sum up to 7 in one roll

4. xapproachesinfinity

couldn't complete my thinking here!

5. xapproachesinfinity

so i need some combinatorics to finish up this seems that way with your result

6. xapproachesinfinity

@lordofpens total outcomes is not 6! rather it is 6^5 because there are 6 possibilities for one roll and another 6 for the second roll and third and so one is total 6x6x6x6x6=6^5

7. xapproachesinfinity

@robtobey has the same thing although went a different approach, but this probabilities are like that can be done differently :)

8. anonymous

When I was much younger I calculated the player's winning percent playing craps in a Nevada gambling hall.

9. xapproachesinfinity

what was the the percentage of winning the gamble?

10. xapproachesinfinity

i need to work on this probabilities more! sometimes i get them sometimes i'm off the picture. lack of practice

11. anonymous

Close to even, but the house is still expected to win. Don't recall the exact number. I believe the players winning percentage is near 48% overall.

12. xapproachesinfinity

i see, that is not a good percentage to play the game hehehe

13. anonymous

If you never play, you will never lose. Someone is paying for the Vegas hotels, shows and so forth. I am no longer one of them.