# Solitaire problem

• Jun 25th 2007, 01:52 AM
Aliquantus
Solitaire problem
Hi,

I ran accross this problem which I'm not able to solve (unless I do really huge calculations). I haven't read any probability yet, except for what we did in upper secondary - and that wasn't much.

The solitaire is played in a way that have your deck, and then you flip card after card. While doing this, you say "one, two, three, one, two, three" and so on. Notice that you may not choose between one, two and three, they just continue looping. If you flip an ace when you are saying "one", you lose. The same goes for saying "two" - getting a two, saying "three" - getting a three. All are losing conditions. If you manage to go through the whole deck without losing, you win (haha).

1. What are the chances of winning?

2. Imagine you take out all aces, twos and threes from the deck and play the solitaire with only these 12 cards. Are chances greater or smaller you'll win compared with playing with full deck?

I've tried a bit, but I don't manage to do anything smart. First I tried only with the 12-card deck, as that was more manageable. I found that it was 4/12 to lose on the first card. Imagine we don't hit an ace on the first card. It can of course not be fair to assume that there is 4/12 to lose on the second card as well, for two reasons: We only have 11 cards to play with, and the first card we picked might have been a two. So here the chances of losing must be either 3/11 or 4/11, depending on if we picked a two or a three on the first card. However, if we create a diagram to represent all possible alternatives, it soon gets out of hands. Are there perhaps some nice formulae to use here?

However, I have a thought.. Are the chances equal (playing with 12 or 52 cards)? May we as well discard the other 40 cards, as they have nothing to do with our solitaire? When playing with 52 cards, you'll end up with saying "one" 18 times, but saying "two" and "three" only seventeen times each. I supoose this should affect the probability in some way or another?

Aliquantus
• Jul 4th 2007, 03:52 PM
rualin
Interesting problem... stupid game. I don't know either... maybe some of the better mathematicians here could help you with that.
• Sep 25th 2007, 03:45 AM
F.A.P
12 card solution
Suppose we have 3 urns...

Let every position where the player says "one" constitute urn 1, and let the "two"- and "three"-positions constitute urns 2 and 3 respectively.

For the game to be successful the following must be fulfilled

A. Urn 1 can only hold deuces and threes
B. Urn 2 can only hold aces and threes
C. Urn 3 can only hold aces and deuces

Ignoring the within order of each urn the possible ways of fullfilling this is by

Urn 1 Urn 2 Urn 3
a 2222 3333 1111
b 3333 1111 2222
c 3332 3111 2221
d 2223 1333 1112
e 2233 1133 1122

a and b contribute with one way each 1+1 = 2

c: First we have to choose three out of four 3:s and then one out of four 2:s for the first urn. Then we choose three 1:s out of four for urn 2 and the last 3also end up here. Whatever is left goes to urn 3. Number of ways:

${ 4 \choose 3 }{ 4 \choose 1 }{ 4 \choose 3 }=64$

d: Same as c due to symmetry $64$.

e: Same approach as in c yields

${ 4 \choose 2 }{ 4 \choose 2 }{ 4 \choose 2 }=216$

Total number of ways to order the numbers between the urns $2 + 64 + 64 + 216 = 346$

Now each urn can be ordered in 4! ways within. Combined with each other, the 346 ways to order the cards between the urns and a total of 12! permutations gives a probability of

$\frac{346(4!)^3}{12!}=173/17325$ .....about 0.009986

to finish the game.
• Sep 25th 2007, 03:53 AM
F.A.P
Quote:

Originally Posted by Aliquantus

"...Are the chances equal (playing with 12 or 52 cards)?..."

Aliquantus

A quick simulation of 100,000 runs indicates a lower probability of finishing the game with 52 cards... a probability of about 0.0085.
• Sep 25th 2007, 06:05 AM
Soroban
Hello, Aliquantus!

I think I have a solution . . . hope my reasoning is correct.

There are $52!$ possible permutations of the deck.

To win, the Aces must not be in positions: 1, 4, 7, 10, ... , 52
. . They must occupy four of the other 34 positions.
. . There are: . $P(34,4)$ ways.

The Deuces must not be in positions: 2, 5, 8, 11, ... , 50
. . They must occupy four of the other 35 positions.
. . There are: . $P(35,4)$ ways.

The Treys must not be in positions: 3, 6, 9, 12, ... , 51
. . They must occupy four of the other 35 positions.
. . There are: . $P(35,4)$ ways.

The other 40 cards can be order in $40!$ ways.

Hence, there are: . $P(34,4)\cdot P(35,4)\cdot P(35,4)\cdot40!$ .winning permutations.

Therefore: . $P(\text{win}) \;=\;\frac{P(34,4)\cdot P(35,4)\cdot P(35,4)\cdot40!}{52!}$

• Sep 25th 2007, 11:58 AM
F.A.P
Quote:

Originally Posted by Soroban
"..."
To win, the Aces must not be in positions: 1, 4, 7, 10, ... , 52
. . They must occupy four of the other 34 positions.
. . There are: . $P(34,4)$ ways.

The Deuces must not be in positions: 2, 5, 8, 11, ... , 50
. . They must occupy four of the other 35 positions.
. . There are: . $P(35,4)$ ways.

The Treys must not be in positions: 3, 6, 9, 12, ... , 51
. . They must occupy four of the other 35 positions.
. . There are: . $P(35,4)$ ways.
"..."

I have great confidence in my simulations and you must be wrong!!! ...:D...not really...
At first glance your reasoning seems correct. And it is up to the aces. There are indeed ${34 \choose 4}$ ways of distributing the aces.

When we come to the deuces though, we don't know whether some of the aces ended up among the 35 positions that we could choose from to start with. Thus with different probabilities we would have

${35 \choose 4},{34 \choose 4},{33 \choose 4},{32 \choose 4}$ or ${31 \choose 4}$

for the deuces.

For the threes we will have even more situations associated with different probabilities...

Conditioning... that is..
• Sep 26th 2007, 06:13 AM
Soroban
Hello, F.A.P.!

You're right . . . *blush*

I was afraid I'd missed a few cases in my reasoning.
. . Oh well, only a few million.

I'd work on it further, but my brain has turned to tapioca . . .