Easy math puzzles

Quote: Ace2

You go to Vegas over a long weekend to play a single-deck blackjack game with a 0% edge ($500 minimum). You decide to play 1,200 hands but you will quit if you bust or double your initial bankroll before reaching 1200 hands.

What size bankroll should you bring to give yourself a 1/3 chance of busting, 1/3 chance of doubling and 1/3 chance of finishing the 1200 hands without busting or doubling?

Assume a standard deviation of 1.1547, flat betting $500 one hand at a time and perfect basic strategy to realize the 0% edge
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Quote: Ace2

5 deads

13 * combin(12,5) * 16 / combin(52,7)
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Quote: Ace2

^^^That’s exactly what I calculated, though I did edit my answer
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Quote: gordonm888

This easy math puzzle is based on the rules of Gin and Win.

Gin and Win
The object of trying to put as many cards as you can in sets, with the goal of minimizing dead wood, is also the goal of Gin and Win. It should be emphasized that pairs and flushes of at least three cards count as sets in Gin and Win. The rules start as follows:

1. Cards are ranked as in poker, except aces are low only.
2. To begin, each player must post an Ante bet.
3. The player and dealer are each dealt seven cards, face down, from a normal 52 card deck.
4. The player looks at his hand and removes any combinations of pairs, trips, and quads (rank melds), or any flushes of three cards or more cards (suit melds). One card cannot be part of more than one set.
5. The remaining cards are known as "Dead Wood" and are set apart from the cards that belong to a set.

Math puzzle:
With what probability will a player be dealt a hand with "no sets" and thus have seven Dead Wood cards?
With what probability will a player be dealt a hand with exactly five Dead Wood cards?
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Quote: DogHand

I see no mention of straights in item 4. Do they not count in Gin and Win?

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Quote: ThatDonGuy

Quote: DogHand

I see no mention of straights in item 4. Do they not count in Gin and Win?

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No, they do not. See part 5 of this Nevada Gaming Commission document about the game.
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Quote: charliepatrick

all deadwood

The first part is to get no pairs, all the ranks of the cards have to be different. At a later stage let's look at which of those could make a 3-card straight.
The second part is that to get no flushes the suits have to be divided 2-2-2-1. There are four ways to pick the singleton suit, so let's assume it's Clubs, thus the doubletons are Spades, Hearts and Diamonds. There are seven different ranks, so the ways to assign the Spades to them is 7*6/2=21. This leaves five non-Spades. The ways to assign Hearts is 5*4/2=10. Similarly Diamonds = 3*2/2=3. Leaving the Club assignment. The total number of ways for the suits is 4 * 21 * 10 * 3 * 1 = 2520.
Of the 1716 ways for seven different ranks, only 393 have no possible 3-card straights (I could only do that using a spreadsheet).
So there are 393*2520 hands = 990,360.
The total number of ways to have 7 cards from 52 is 133,784,560.
So the chances of such a hands = 990,360 / 133,784,560 which is about 0.7% (or one hand in 135).

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Quote: charliepatrick

^ I erroneously had assumed it was like regular gin. It does make the maths easier as you don’t have to check the unequal ranks.

revised numerator

So instead of 393*2520 it’s 1716*2520 giving 4,324,320 or about 3.23%.

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Quote: Ace2

One more revision for five deads

13 *6 * combin(12,5) * (4*3*10*3) / combin(52,7)

13 * 6 ways to choose the pair
C(12,5) ways to choose ranks of the five
4*3 ways to choose suits of the five (221)
10*3 ways to assign the suits to the five

Caveat: combinatorics aren’t my strongest skill
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Quote: charliepatrick

Show Spoiler

Firstly there are how many ways can you make 7 different ranks from 13, which is 13 12 11 10 9 8 7 / (7 6 5 4 3 2 1) = 1716.
The second number is how many ways the suits of seven different cards can happen so the suit distribution is 2-2-2-1.
The first part is that there can be four suits for the singleton, and once that has been determined then the other suits are fixed.
The next part is how those suits can be distributed onto the ranks.
The singleton can be put in one of 7 positions.
The highest of the other three suits can be put in 6*5/2 of the remaining 6 positions.
Similar the next suit can be put in 4*3/2 of the remaining positions.
The last suit is now set.
So you have 4 * 7 * 6*5/2 * 4*3/2 * 1 = 2520.
2520 * 1716 = 4324320.

7				4	6 864
6	1			84	144 144
5	2			252	432 432
5	1	1		504	864 864
4	3			420	720 720
4	2	1		2 520	4 324 320
4	1	1	1	840	1 441 440
3	3	1		1 680	2 882 880
3	2	2		2 520	4 324 320
3	2	1	1	5 040	8 648 640
2	2	2	1	2 520	4 324 320

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Quote: charliepatrick

^ I've now developed a spreadsheet which essentially creates every combination of how the suits are given to the various ranks.
For the 1111111 rank distribution (i.e. no pairs) its essentially like creating numbers of base 4. So there are 4⁷ lines.
For the 211111 rank distribution (i.e. one pair) the last five columns are as above, but the first two start with 12, then go 13, 14, 23, 24, 34. Thus there are 6*4⁵ lines.
For each line I then count how many 1s, 2s, 3s and 4s there are, and create four digit numbers such as 7000, 6100, 5200, 5110 etc for the suit distribution. I then count how many of these there are.
For the pairs I looked at the simple cases e.g. 6100 is where the odd suit has to be one of the pairs, and all the other cards are in the same suit. So there are 4*3 of them.
Since the numbers added up to 6144, I didn't check the more complicated ones.

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