a problem
 

One day, a friend approaches you with 2 decks of standard 52 cards. He tells you, "I want to make you a bet. I'm going to shuffle both deck of cards thoroughly. Then, I'm going to turn over the cards from the decks simultaneously, one by one. In other words, I'll turn over the first cards of each deck simultaneously. Then, I'll turn over the next cards simultaneously. I'll continue doing this, until I've finished turning over all the cards in the decks."


"Now, my offer is this. I'll give you 3:2 odds that at no time while I'm revealing the cards, will 2 of the EXACT same card fall down simultaneously. In other words, you give me $2. If I get through the whole deck without any two cards falling identically, then I'll pay you $3. But, if 2 cards fall at the same time, and they're identical, I win and keep your money."


Should you accept this bet?

Re: a problem
 

Well, if I've absorbed the lesson of the inclusion-exclusion principle, the probability of a match is:


52*51!/52! - C(52,2)*50!/52! + C(52,3)*49!/52! - C(52,4)*48!/52! +...


= 63%.


So yes you should take 3-2 odds since you are better than a 2-1 favorite. I like this principle.

Re: a problem
 

Except 63% isn't better than a 2-1 favorite...

oops
 

Ok, let's get the easy part of this problem right. You LOSE 63% of the time when there is a match, so you are more than a 3-2 dog, so you should NOT take the bet.

Re: a problem
 

I approached this problem two ways:


The simple way, giving each trial a 51/52 chance of failure with 52 trials giving (51/52)^52 =36.43% which means I'm under the 40% chance required to take your bet.


I then needed to confirm to myself that each trial was independent becuase it felt wrong, so i assumed the first deck came through in order As.2s.3s etc

Obviously the chances of first card matching is 1/52.

The 2s will have a 1/51 chance of SUCCESS every time it wasn't the first card which is 51/52 of the time so we get 1.51/51.52 or 1/52.

The 3s will have a 1/50 chance of SUCCESS every time it's not one of the first two cards (50/52) so again we get 1.50/50/52 or 1/52.


So these trials can be treated as independent i believe.


My maths is rusty so please check this for yourselves and flame me if approppriate.


Lori

Sorry Bruce, you posted whilst I was typing mine *NM*
 

Re: a problem
 

I read in "Gambling Scams" that for such a wager, the odds are in favor of the two cards being in identical locations is better than 5:3, or 10:6. Since you are being paid 3:2, or 9:6, you should not accept the bet (my math is completely off base, huh?).

solution
 

No, you're absolutely correct. You should not accept the bet. The reason is inclusion-exclusion. What we're really doing is looking at the second deck as a permutation (shuffling) of the first, and asking if there are any fixed points. Let N(i) be the number of shufflings in which the i'th cards are identical. We want to know how many shufflings there are where this never happens for any i.


Suppose our deck has n cards, so here n = 52. Let N(i,j,k) be the number of shufflings where the i'th, j'th, and k'th cards are identical, and so on.


N(i) = (n-1)!, because there are (n-1)! ways to permute the remaining cards.


N(i,j) = (n-2)!, because there are (n-2)! ways to permute the remaining cards.


And so on.


Also, when we apply inclusion-exclusion, there are C(n,1) terms of type N(i), C(n,2) terms of type N(i,j), etc. In general, there are C(n,k) terms corresponding to shufflings which have k identical cards.


Thus, the number of ways that none of the cards can fall identically is


n! - (n-1)!*C(n,1) + (n-2)!*C(n,2) - (n-3)!*C(n,3) + ... + (-1)^n,


the final term corresponding to k = n.


Now, if we divide this by n!, we get that


1 - 1/(1!) + 1/(2!) - 1/(3!) + 1/(4!) - ... + (-1)^n/(n!)


is the probability that none of the cards will fall identically.


This is the series expansion of e^x, where x = -1. So, the above expansion is just a partial sum approximation for 1/e. This means the probability that at least one of the cards WILL fall identically is (e-1)/e, and so the odds for the bet should be set at (e-1):1, which is about 1.718:1. This is why both 3:2 and 5:3 are losing bets.

That\'s not 3:2 odds...
 

If I start with $2 then I should finish with $5 if I win my bet and $0 if I lose.

Re: solution
 

[(n-1)/n]^n also equals 1/e for large n, so in this problem the cards are independent, and we can just do (51/52)^52 = 36.4% without the inclusion-exclusion principle.


If you do something n times where the probability of success each time is 1/n, and the tries are independent, then the probability of at least one success is approximately 1-1/e = 63%, the approximation becoming better as n becomes sufficiently large.