unusual combinatronic puzzle

Dunno about the others, but I'm going to have to go with 1 on this one.

I think this depends on the drawing procedure, about which the O.P.'s description wasn't entirely clear. After all the names are placed in the hat, the protocol might be either of

1: Albert draws and returns names until encountering a non-Albert, then Betty draws and returns until encountering a non-Betty, and so on through Yousef and Zorah. (We assume a forty-letter alphabet. ObWhimsy: What are the fourteen new letters? Seuss' letters come *after* Z, but this puzzle's new letters obviously must lie between B and Y.)

2: Each player draws a name, then all who drew their own names re-deposit them and re-draw, then all who drew their own names a second time re-deposit and re-draw, then ...

The 1/40 answer is correct for case 2, but the question is only of interest in case 1

Assuming you meant to say, '... for case 1 also,' I'd like to see that demonstration, as well! Seems to me that if players 1..39 follow the rules, player 40 has 0 probability of drawing his own name.

Have you looked at how close .63 is to 1 - 1/e ?

Bob H

See the value of 1-1/e, sequence A068996 in Neil Sloane's Online Encyclopedia of Integer Sequences (OEIS), for relevant comments and links about problems with 'derangements':
http://www.research.att.com/cgi-bin/access.cgi/as/njas/ sequences/eisA... um=A068996

(You may have to paste the above link back together.).

Rick Shepherd

Um, er, yes.

No, the probability must be positive. For example, if Albert draws Betty, Betty draws Charles, Xena draws Yousef, and Yousef draws Albert, the first 39 players (still waiting for those fourteen new letters) have followed the rules and each holds a slip with someone else's name. But Zorah has not yet drawn, and the only slip still in the hat says 'Zorah.' Hence, the probability of the last player being stuck with his or her own name is greater than zero.

Specifically, Zorah will get stuck with her own name iff nobody else draws it first. I don't know how to compute that probability because the number of draws is variable (for example, if Quincy draws the 'Quincy' slip, puts it back and then draws 'Renata,' he's had two chances rather than just one on which he might have drawn 'Zorah'. This seems to make things a bit more complicated than just totting up the derangements on 39 objects

Ugh, sounds painful!

Thanks guys, I get it, but will need more reading to really get my head around it!

Remind me to introduce you to my wife some time.

No, each player only gets one chance to draw somebody else's name.

The probability that Albert draws Zorah is 1/39, no matter how many draws Albert takes...you can ignore all but one.

The problem I see, however, is that everybody after Alfred has a different chance of drawing Zorah depending on whether or not their name has already been picked by somebody else. If Alfred drew Betty's name, Betty has a 1/39 chance of drawing Zorah. If Alfred did not draw Betty's name, Betty has a 1/38 chance of drawing of drawing Zorah.

Remind me to introduce you to mine, she's an aussie windsurfer with a particular like of Perth (won a couple of races in Lancelin a while back.)

Ciao,

Hmmm... I'm not sure that's the same problem. If you shuffle together all the clubs and hearts in one pile, and all the spades and diamonds in another, then deal pairs one from each pile, you'd have the (approximately) 1-1/e result.

What you wrote in the paragraph above seems different. But is it?

Bob H