Card Game: Less, between, greater

[BMAD]

If you give me $10, I give you three cards (from an ordinary playing deck). I then deal me 1 card. You get to look at your hand and then must claim whether my card is lesser in value than your cards, greater in value of your cards, or between your least & greatest valued cards. If you guess correctly you will receive $20 (your 10 plus my 10), otherwise i keep your $10. Then we play again (if you like).

Who is advantaged in this game?

[bonanova]

You have the advantage.

[bonanova]

After my 3 cards are dealt, there are 49 cards that you might hold.

On average my three cards will divide those remaining 49 cards into equal intervals.

The interval below my lowest card will be on average 12.25 cards

The interval above my highest card will be on average 12.25 cards.

The interval between my lowest and highest card will be on average 24.5 cards.

That interval size is the break-even point.

The "between" interval, being in each case an integer, will never be 24.5 cards.

Rather, it will be an integer n between 0 and 49, and n will be symmetrically distributed.

The number of hands with interval = n will equal the number of hands with interval = 49-n.

This means that if I mindlessly bet between on every hand, my winnings and losses will cancel.

That is, I will break even. That strategy alone keeps you from winning.

But I can do better, of course.

One eighth of the time, the interval less than my lowest card will exceed 24.5 cards,

One eighth of the time, the interval greater than my highest card will exceed 24.5 cards.

Three-fourths of the time, my hand will contain an interval that exceeds 24.5 cards.

Three-fourths of the time, I will be favored to win my bet.

By simulation, the average number of cards in the largest interval in my hand is 30.36 cards.

My average gross winnings for each hand is thus $20 x (30.36/49) = $12.39+
My average net winnings (after paying you $10) will thus be $2.39+ per hand.

When do we play?

[BMAD]

After my 3 cards are dealt, there are 49 cards that you might hold.

On average my three cards will divide those remaining 49 cards into equal intervals.

The interval below my lowest card will be on average 12.25 cards

The interval above my highest card will be on average 12.25 cards.

The interval between my lowest and highest card will be on average 24.5 cards.

That interval size is the break-even point.

The "between" interval, being in each case an integer, will never be 24.5 cards.

Rather, it will be an integer n between 0 and 49, and n will be symmetrically distributed.

The number of hands with interval = n will equal the number of hands with interval = 49-n.

This means that if I mindlessly bet between on every hand, my winnings and losses will cancel.

That is, I will break even. That strategy alone keeps you from winning.

But I can do better, of course.

One eighth of the time, the interval less than my lowest card will exceed 24.5 cards,

One eighth of the time, the interval greater than my highest card will exceed 24.5 cards.

Three-fourths of the time, my hand will contain an interval that exceeds 24.5 cards.

Three-fourths of the time, I will be favored to win my bet.

By simulation, the average number of cards in the largest interval in my hand is 30.36 cards.

My average gross winnings for each hand is thus $20 x (30.36/49) = $12.39+

My average net winnings (after paying you $10) will thus be $2.39+ per hand.

When do we play?

Shouldn't your average intervals be of a size more like 16 1/3?

[bonanova]

Shouldn't your average intervals be of a size more like 16 1/3?

Sure ... if you deal me only TWO cards.

But you deal me THREE cards.

|----12.25----card1----12.25----card2----12.25----card3----12.25----|

|----12.25----locard------------24.50------------hicard----12.25----|

|----LOWER----locard-----------BETWEEN-----------hicard---HIGHER----|

|--------------------------------49---------------------------------|

[BMAD]

Ahh. I stand corrected