do you have a calculator ready??

Question

here is the problem:

you have 25 cards lying face down, consisting of 12 paired cards plus one odd card, all placed randomly in a five by five configuration.

your job is to pick out the pairs by a process of memory and elimination, until you are left with with the odd card face down. you may only turn over one card at a time, and if you fail to make a pair then both cards must be turned back over- like the kids game "husker du?"

question is, what is the probability that by pure chance, every pair of cards you turn over in sequence makes a matching pair until you are left with only the odd card i.e. you complete the 'game' in the fewest moves possible?

Answer 1

Ok, the best case scenario is to pick a pair after pair until you are left with the odd card.

The odds go as follows. Since you win if the odd card is left last, that means the odds of randomly picking a card which is not the odd one are 24/25. Next, you need to pick a card which will match the one you picked first, and the odds for that are 1/24 (since you are looking for an exact one among 24 that are left). That leaves you with 23 cards. The logic stays the same - 22/23 (for picking any except the odd one out) and 1/22 (for picking it's pair), 20/21 and 1/20, and so on. So, by calculating

24/25 * 1/24 * 22/23 * 1/22 * ... * 2/3 * 1/2

you get the odds of winning the game in the smallest number of moves possible. That's 1 / 7,905,853,580,625 !

Answer 2

Turn over one card. The probability of this not being the odd card is 24/25. Turn over a second card. The probability of this being the matching card is 1/24. So the probability of getting a match right on the first two card turns is (24/25)(1/24) = 1/25.

Discard the matching pair. This leaves 23 cards unturned. The probability that the next card isn't the odd card is 22/23. The probability of the next card after this being the match is 1/22. The product together is 1/23. There are 21 cards remaining, so the same process goes on. Eventually you end up with 3 cards. The chance of not picking the odd card is 2/3, and the chance of finding the match is 1/2, giving the probabilty of making the final move right 1/3.

So the total probabiltiy of getting a match every time until the only card remaining is the odd card is:
(1/25)(1/23)(1/21) * ... * (1/5)(1/3)

This gives you 1 in 7,905,853,580,625. That's about 8 million times worse than trying to get a royal straight flush in a poker hand.

Answer 3

Answer :

1.26489E-13

Solution :

Probability of first card having a pair = 24/25
Probabiliy of next card matching it = 1/24

Therefore probibility of first two cards being a pair = 24/25 * 1/24 = 1/25

Similarly probibility of third and fourth cards being a pair = 22/23 * 1/22 = 1/23

Probability of solving game with no wrong moves = product from n=1 to 12 (1/((2 * n)+1))

Answer 4

First choose any card. As long as you don't choose the odd card, any one of the other 24 is OK. There are then 24 remaining, exactly one of which is a match.

Once those 2 have been removed, choose another card. As long as you don't choose the odd card, any one of the other 22 is OK. There are then 22 remaining, exactly one of which is a match.

And so on, so...

(24/25*1/24)*(22/23*1/22)*...*(2/3*1/2)
= 1/25*1/23*...*1

Answer 5

I'll solve any math problem for ¢20, at least 5 problems I assume. Quickly and guaranteed quality.

Answer 6

is this your homework?

Answer 7

a 6 out of 81 chance??
I dunno, just a guess