Hello, guys and gals!

Imagine that I have 5 index cards. On one I have written the letter A, on another is the letter B, on another is a C, on another is a D, and the final card is a E.

These cards are turned face down and shuffled into an unknown order. I hand you a pen and ask you to write the letters A, B, C, D and E so that one letter is written on the back of each card.

What is the probability that you get:

0 correct?

At least 1 correct?

At least 2 correct?

At least 3 correct?

At least 4 correct (which would be the same as getting all 5 correct)?

I am comfortable with the fact that getting 4 correct (thus 5 correct) is 1/120.

Can you help me out with the other questions?

Thanks for any insight you can offer!

None correct: 44/120 = 11/30
At least one correct: 76/120 = 19/30
At least two correct: 31/120
At least three correct: 11/120
At least four correct: 1/120

Stanalger got it! Thanks, Stanalger!!

I had to get help from PhD. Mathematician James Tanton on this. It involves something called “derangements,” and his video proved to be most helpful. (http://www.jamestanton.com/?p=828)

Here are the end results:

Probability of 0 correct is 11/30 (approximately 37%)

Probability of at least 1 correct is 19/30 (approximately 63%)
Probability of at least 2 correct is 31/120 (approximately 26%)
Probability of at least 3 correct is 11/120 (approximately 9%)
Probability of at least 4 correct is 1/120 (approximately 1%)
Probability of all 5 correct is 1/120 (approximately 1%)

Probability of exactly 1 correct is 3/8 (approximately 38%)
Probability of exactly 2 correct is 1/6 (approximately 17%)
Probability of exactly 3 correct is 1/12 (approximately 8%)
Probability of exactly 4 correct is 0/120 (0 %)
Probability of all 5 correct is 1/120 (approximately 1%)

Thanks Munsey for helping with this problem.

Excellent discussion. A pretty accurate estimate of the number of times the spectator guesses them all wrong is given by a simple formula: n!/e where n!= n(n-1)(n-2)(n-3)....(1) and e is the base for the natural log which equals about 2.71828. Rounding e to 2.718 and using n=5 in the above problem we get 5!=120 and 120/2.718=44.1 which we round to 44. If you used 2.7 the answer would still be close at 44.4. The formula essentially calculated the number of derangements (permutations with no fixed points). For numbers higher than n=5 the formula gets more accurate.

Permutations or shuffles with one fixed point in a packet of cards can produce some really nice magic. In my latest book Magica Analytica I use that concept for a few of the effects.

Larry