O/T Stumped on this one HELP

[CASS]

Any one have a clue ?

"You've scattered a deck of cards and one joker face down on table so that you dont know which card is where.
Next you start turning the cards over one by one.
Assuming that you cant flip the same card twice, what are the percentage odds that you will turn over all four aces before you turn over the joker."

From Professor Layton and the Curious Village.
This has got us by the short an curlies,I is off to the Sausage for a glass of inspiration.

[Gnasher]

20%

[CASS]

20%

Well that's what we thought,but I guess we we rather complicated the situation...20% seems logical.

[Gnasher]

Talking about a pack of cards makes you think the rest of the cards are important, they aren't. You've got 5 cards and the chance of turning a specific one of them over last is 1 in 5, 20%

[Alan]

Talking about a pack of cards makes you think the rest of the cards are important, they aren't. You've got 5 cards and the chance of turning a specific one of them over last is 1 in 5, 20%

Guessing on this one, but with 5 cards, 4 of them the aces, surely the odds on not picking the joker are 80% only on the first pick. If an ace is picked, then the odds drop to 75% on the second pick, to 66.6% on the third go and finally 50% with 2 cards left.

Some clever dick somewhere wil have a formula for these sort of computations...but like CASS
says...maybe we are making it over complicated.

[Keith]

Talking about a pack of cards makes you think the rest of the cards are important, they aren't. You've got 5 cards and the chance of turning a specific one of them over last is 1 in 5, 20%

but like CASS says...maybe we are making it over complicated.

Yup, I agree with Gnasher and therefore Alan!

Of the five cards, the only point of interest is which one comes last. Therefore each of the Aces and the one Jack all have the same chance of coming out last, 1 in 5, or 20%

[Alan]

Guessing on this one, but with 5 cards, 4 of them the aces, surely the odds on not picking the joker are 80% only on the first pick. If an ace is picked, then the odds drop to 75% on the second pick, to 66.6% on the third go and finally 50% with 2 cards left.

Some clever dick somewhere wil have a formula for these sort of computations...but like CASS
says...maybe we are making it over complicated.

As this is my theory ('and this is what it is'... Ann Elk) I claim full marks for my proof..which gives the answer...

80% x 75% x 66.66% x 50% = 20%

o.k. I'll get my anorak.....