BrainDen.com - Brain Teasers

## Question

I shuffle a deck of cards and deal them into two piles.

.

1. The first card I place in front of me

2. If the second card is lower in rank, I start a pile to the left of the first; otherwise to the right.
Now I have two piles [one card in each pile, with the lower ranked card on the left.]
I continue, until they are all dealt, as follows:

3. If a card is lower in rank than the previous card, I place it on the left pile.

4. If a card is higher in rank than the previous card, I place it on the right pile.
.

When I'm done, the left pile will tend to have lower ranked cards, and the right pile will tend to have higher ranked cards.

But not completely: a J that follows a K will be in the left pile and a 5 that follows a 3 will be in the right pile.

What is the expected highest ranking card in the left pile? Of the lowest ranking card in the right pile?

To make the rankings clear, I numbered the cards: from 1 [Club deuce] to 52 [spade Ace]. You can give the answer as a number or card, as you like.

## Recommended Posts

• 0

What if you deal 2 cards of the same rank back to back - i.e., J of hearts followed by J of spades. Would they be in the same pile or different piles? Does one suit outrank another?

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• 0

What if you deal 2 cards of the same rank back to back - i.e., J of hearts followed by J of spades. Would they be in the same pile or different piles? Does one suit outrank another?

Clubs [lowest] Diamonds, Hearts, Spades [highest].

Or just assume they're numbered 1-52.

No ties. My bad: rank was the wrong term to use.

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• 0 card 26 on the left pile and card 25 on the right pile?

the probability of card 52 to be on the left pile is 0

the probability of card 51 to be on the left pile is 1/51 because card 52 is the only card that can appear next for card 51 to be placed on the left pile

the probability of card 50 to be on the left pile is 2/51 because card 52 or card 51 are the only cards that can appear next for card 51 to be placed on the left pile

going in this fashion the probability of card 26 to be on the left pile is 26/51 which is around 50% probability. so card 26 is the highest card that is expected on the left pile.

following the same logic, card 25 has a 50% probability of appearing on the right pile.

the probability of card

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• 0

Doing a simulation of 10,000,000 plays, I get 44.55 as the largest on the left

and 9.47 as the smallest on the right. That translates to the Queen of Spades

on the left and the 4 of Clubs on the right. I have no idea as to how one can

compute this analytically!

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• 0 Great approach, but why isn't the left and right pile expected differential the same (instead of 7.5 vs. 8.5

Doing a simulation of 10,000,000 plays, I get 44.55 as the largest on the left

and 9.47 as the smallest on the right. That translates to the Queen of Spades

on the left and the 4 of Clubs on the right. I have no idea as to how one can

compute this analytically!

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• 0

Great approach, but why isn't the left and right pile expected differential the same (instead of 7.5 vs. 8.5

I don't know. I don't even know how the can opener works!

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• 0 Calculating the highest expected value on the left:

Define pn as the probability that card i is the highest value (for 1 <= n <=52).

Clearly p52 = 0;

We derive pn = (1 - Σ52i=n+1 (pi))*(52-n)/52

That is, the probability that card 48 the highest card is the product of the probability no higher card is the highest card (49, 50, 51 or 52) and the odds that the card preceding it is higher (49, 50, 51 or 52 = 4/52).

I'm not sure how to convert that to a direct formula (remove the series). Plugging this into an excel spreadsheet:

n p

50 3.77%

49 5.44%

48 6.84%

47 7.89%

46 8.55%

45 8.83%

44 8.73%

43 8.31%

So card 45 has the highest probability and, thus, is the highest expected value for the left side.

Calculating the lowest value on the right side should be a identical mirrored calculation.

Edited by vinays84
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• 0 I don't know. I don't even know how the can opener works!

LOL

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• 0 right pile cannot contain anything below 2

left pile cannot contain anything above 51

so left pile shows the linearly decreasing probability of having cards 1 through 51 in it

similarly right pile shows the linearly increasing probability of having cards 2 through 52 in it

thus highest(left)=26 and lowest(right)=27 seems to be the fair choice for the answer...

Edited to make the answer easier to comprehend

Edited by KlueMaster
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• 0

Great approach, but why isn't the left and right pile expected differential the same (instead of 7.5 vs. 8.5

I transcribed the numbers from my simulation incorrectly. The expected highest on the left is 43.55 and the expected lowest on the right is 9.45. I must have been sleepy!

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• 0

In order for a card to be the highest rank in the left pile, two things have to happen:

1. The card has to be in the left pile

2. All cards with a higher rank have to be in the right pile.

So, assuming that cards are randomly ordered in the deck, the probability that a card is in the right pile is:

p(R,right) = (R - 1) / 51 where R is the rank of the card in question

p(R,left) = 1 - p(R,right)

Then it's just a matter of combining probabilities of event occurring simultaneously.

So,

p(1,highest in the left) = p(1,left) * p(2,right) * p(3,right) * ... * p(50,right)

p(2,highest in the left) = p(2,left) * p(3,right) * p(4,right) * ... * p(50,right)

and so on. Similarly

p(50,lowest in the right) = p(50,right) * p(49,left) * p(48,left) * ... * p(1,left)

p(49,lowest in the right) = p(49,right) * p(48,left) * p(47,left) * ... * p(1,left)

Tabulating those products, you will see that they reach their maximums at:

p(8,lowest in the right) = 0.0892

p(45,highest in the left) = 0.0892

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