[bonanova]

May and Moe want to see the numbers 1-6, using dice.

They adopt different plans: May uses six dice; Moe uses only one.

They roll their dice at the same time: May rolling her 6 dice, Moe rolling his single die.

They repeat this process until one of two things happen:

1. May sees all six numbers on her dice at once.
   
2. Moe keeps track of his numbers, and finally his sixth number appears.
   

Who gets there first?

[Guest]

At each roll, May has a chance of 6!/6⁶ to see all numbers.

After 5th roll, Moe also has the same chance to see a track of all numbers.

So, at average, May will see them at 6⁶/6!.th roll but Moe see them at (6⁶/6!+5).th roll.

Result: May has a (5x6!/6⁶ + 0.5)/1 chance to see her numbers before Moe.

[bonanova]

At each roll, May has a chance of 6!/6⁶ to see all numbers.

After 5th roll, Moe also has the same chance to see a track of all numbers.

So, at average, May will see them at 6⁶/6!.th roll but Moe see them at (6⁶/6!+5).th roll.

Result: May has a (5x6!/6⁶ + 0.5)/1 chance to see her numbers before Moe.

Are you sure about Moe's expected wait?

[Guest]

Are you sure about Moe's expected wait?

To clarify the question:

May rolls all 6 dice at once, while Moe rolls one dice. Then again May rolls 6 dice at once and Moe rolls one more???

or

May rolls 6 dice at once, waits till Moe rolls 6 dice one by one. Then again May rolls 6 dice and waits till Moe rolls 6 dice one by one.

I understood the question as second choise.

If May rolls 6 when Moe rolls one, obviously May has a more greater chance. (1/6) But I dont expect this problem to be so easy.

[Guest]

Does Moe have to see all of the numbers in a row, or can they be separated by other numbers?

Then I think Moe would have a better chance, although I don't have the math right now to prove it.

[Prime]

I am not sure I understand the condition.

Bonanova has already posted same problem before. At least it is the same as Moe's probability. Which was found to be 14.7 rolls of die on average needed.

For May the probability is as Nobody has calculated:

As Nobody calculated, Moe's chance to see 6 different numbers on 6 dice is 6!/6⁶. Therefore, the average number of rolls she needs is

6⁶/6! = 64.8.

On the other hand, if OP meant Moe must see 6 different numbers in a row, then we have a problem...

At first glance it appears, that in this case May takes 6 turns while Moe takes one. But it is not so. Moe does not have a fixed position from which to start counting his 6 consecutive rolls. Let's say, the rolls 1 through 6 did not reach the objective, then it is still possibe for the rolls 2 through 7, or 3 through 8, and so on.

[bonanova]

I am not sure I understand the condition.

Bonanova has already posted same problem before. At least it is the same as Moe's probability. Which was found to be 14.7 rolls of die on average needed.

For May the probability is as Nobody has calculated:

As Nobody calculated, Moe's chance to see 6 different numbers on 6 dice is 6!/6⁶. Therefore, the average number of rolls she needs is

6⁶/6! = 64.8.

On the other hand, if OP meant Moe must see 6 different numbers in a row, then we have a problem...

At first glance it appears, that in this case May takes 6 turns while Moe takes one. But it is not so. Moe does not have a fixed position from which to start counting his 6 consecutive rolls. Let's say, the rolls 1 through 6 did not reach the objective, then it is still possibe for the rolls 2 through 7, or 3 through 8, and so on.

Prime has it. Moe sees his numbers first.