Rabbit’s Card Puzzle – The Solution

Last night I posted a puzzle on my blog. If you haven’t read it then you can read it here.
There were quite a few answers submitted but it was quickly solved by my fellow Stationery Club member, Adam Creen.

The answer to the puzzle was 9. Surprsingly, playing the game with nine players means that it is probable that a pair will occur on the first go but the most common guesses were much higher, 26 or 27.

It is quite strange when you think about it. Nine people with a shuffled pack of cards each turn over the top card and it is likely that two or more people will turn over the same card.
How can it possible be so low? If you have nine packs of cards you can prove it for yourself through practical means. However, you can also solve the puzzle with some reasonably simple maths as long as you approach it in the right way.
I have posted the maths below so you don’t need to read it if you don’t want to but you don’t need to worry about the maths to know why the answer isn’t 27 as instinct suggests. I think this is what our instinct tells us:

I am playing in this game and I turn over the top card from my pack and I have the Ace of Spades.
Player 2 turns over their card and there is 1 chance in 52 that it is the Ace of Spades.
Player 3 turns over their card and there is 1 chance in 52 that it is the Ace of Spades.
Continue going until we have 27 players and adding it all up we have 27 “1 out of 52” chances so a pair is likely.

Where our instinct let’s us down is that we neglect to account for the fact that the other 26 players may have pairs with each other. In fact when 27 players are in the game there are many more chances of them having pairs with each other than there are of them having a pair with me.

While there are 26 opportunities for them to have a pair with me there are a total of 351 opportunities for pairs in total. The opportunity for the players to have pairs with each other, not just with me, is why you only need 9 players and not 27.

While I did come up with this puzzle, it is heavily based on the famous birthdays problem – if you take a group of 23 people, it is probable that two of them share a birthday. I think that is rather astonishing and a lovely example of when it may be better not to trust our instincts.

If you want to see the numbers then keep reading. Otherwise thanks for playing.

RedEaredRabbit


The Solution

It is easier to first look at the game being played out card by card and look at the probability of no pair being formed. Let’s go through it in order with 9 players:

Anna turns over her card first.

Now it is Belinda’s turn to turn over her card. Of her 52 cards there are 51 which will not result in a pair. Therefore the probability of no pair being formed is:

(51/52) = 98.08%

Now it is Cathy’s turn to turn over her card. Out of her 52 cards there are 50 which which will not result in a pair. Therefore the probability of no pair being formed after Cathy’s turn is:

(51/52) x (50/52) = 94.30%

Now it is Deborah’s turn to turn over her card. Out of her 52 cards there are 49 which which will not result in a pair. Therefore the probability of no pair being formed after Deborah’s turn is:

(51/52) x (50/52) x (49/52) = 88.86%

Now it is Erica’s turn to turn over her card. Out of her 52 cards there are 48 which which will not result in a pair. Therefore the probability of no pair being formed after Erica’s turn is:

(51/52) x (50/52) x (49/52) x (48/52) = 82.03%

Now it is Fanny’s (sorry) turn to turn over her card. Out of her 52 cards there are 47 which which will not result in a pair. Therefore the probability of no pair being formed after Fanny’s turn is:

(51/52) x (50/52) x (49/52) x (48/52) x (47/52) = 74.14%

Now it is Gertrude’s turn to turn over her card. Out of her 52 cards there are 46 which which will not result in a pair. Therefore the probability of no pair being formed after Gertrude’s turn is:

(51/52) x (50/52) x (49/52) x (48/52) x (47/52) x (46/52) = 65.59%

Now it is Harriet’s turn to turn over her card. Out of her 52 cards there are 45 which which will not result in a pair. Therefore the probability of no pair being formed after Harriet’s turn is:

(51/52) x (50/52) x (49/52) x (48/52) x (47/52) x (46/52) x (45/52) = 56.76%

Now it is Imogen’s turn to turn over her card. Out of her 52 cards there are 44 which which will not result in a pair. Therefore the probability of no pair being formed after Imogen’s turn is:

(51/52) x (50/52) x (49/52) x (48/52) x (47/52) x (46/52) x (45/52) x (44/52) = 48.03%

So after Imogen’s turn the probability of no pair occurring has dropped below 50%. Therefore the probability of a pair occurring has risen above 50%.

You can see how the probability of a pair increases with the number of players in the graph below. Note where the probability of a pair occurring is when you do get to 27 players – you will get a pair on the first go in about 499 out of every 500 games!

Rabbit’s Card Puzzle

Two people sit down at a table, each with a shuffled pack of standard playing cards in front of them.
They each turn over the top card and place it face up on the table. If they have the same card they shout “SNAP!”

For this to occur both number and suit must match. Four of Spades does NOT match with Four of Hearts.

The chances of this happening on the first go is 1 in 52.

Now more people want to join this fantastic game. Each time a new player joins they bring their own pack of shuffled cards.
i.e. When there are 4 players, 4 cards are turned over. If there is a pair anywhere among the 4 cards they all shout “SNAP!”

As the size of the group increases, the chances of shouting “SNAP!” increases.

At some point the group becomes so large that it is more likely than not that “SNAP!” will be shouted when they each turn their first card over.

What is the smallest number of players required to make it more likely than not that “SNAP!” will be shouted when they each turn their first card over?

I will donate £10 to the chosen charity of the person with the first correct answer.
They need a rough reason though. You can’t just guess 1,2,3,4 etc. until you get the right number.

Please put your answer in the comments on this post, not on Twitter.

Good luck!

RedEaredRabbit