Expected Value

[Megan315]

There are six dice. Each of the dice has five blank sides. The sixth side has a number between 1 and 6—a different number on each die. The six dice are rolled and the player wins a prize depending on the total of the numbers which turn up.
(a) Find the expected total without finding its distribution.
(b) Large prizes were given for large totals with a modest fee to play the game. Explain why this could be done.

I don't know how to do this without finding the distribution. When I calculated the expected value from the distribution, I got .014, which I'm not sure makes sense.
Any advice on how to do this problem would be greatly appreciated!

Thanks!
Megan

 

[Denis]

Well, can't make out what you're asking, but I ran a simulation and can tell you this:
you'll get zero 1/3 of the time;
the average you'll get is 3.5.

Some stuff easy enough to figure out, like you'll get maximum (21) 1/6^6 of the time.

But fear not: our expert on such Satanic Dice, Mr S.Khan, will soon ride in on Rocinante to rescue you :wink:
Note: be careful if BigGlen attempts to impress you: he hasn't been himself lately

 

[Deleted member 4993]

There are six dice. Each of the dice has five blank sides. The sixth side has a number between 1 and 6—a different number on each die. The six dice are rolled and the player wins a prize depending on the total of the numbers which turn up.
(a) Find the expected total without finding its distribution.
(b) Large prizes were given for large totals with a modest fee to play the game. Explain why this could be done.

I don't know how to do this without finding the distribution. When I calculated the expected value from the distribution, I got .014, which I'm not sure makes sense.
Any advice on how to do this problem would be greatly appreciated!

Thanks!
Megan

I had to go look for Sancho - he was as usual napping....took all my dice with him ... so I could not do any simulation....

Anyway the expected value of rolling all the 6 dice is same as one dice with six faces numbered with different numbers = 1/6*(1+2+3+4+5+6) = 3.5

since there are 30 blank faces - chances of one or many of those turning up is much higher. For example chance rolling an empty = (5/6)^6 = 15625/46656 = 0.334897977 - a little more than 1/3 - where as turning up all the numbers and hitting a 21 would be (1/6)^6 = 0.0000214335.

Hey...hey ... where did that donkey (errrr the horse - i mean the horse) go now.....

Bye for now ... the windmill is calling...