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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Applying my conjecture that the probability of reaching a point at any time during a session is double the probability of ending the session at/beyond that point , we need the z-score corresponding to the probability of 1/3 * 1/2, which is 0.967. Then take 1200^.5 * 1.1547 * 0.967 to get the answer of 38.7 units * $500 = $19,350.

Verification: Knowing the standard deviation and edge, you can easily calculate that this game is statistically equivalent to a bet with a 3/7 probability of winning 7 for 3. Markoving 1200 bets shows that a bankroll of 38 units * $500 = $19,000 gives bust/double/finish probabilities of 33.4%/33.4%/33.2%. I believe this is the closest you can get to 1/3.

So I'd bring $20,000

You don't need that much computing power. You do, however, need to know how to code around the limits of floating-point numbers.

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?

5 deads

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

Quote: Ace2

5 deads

13 * combin(12,5) * 16 / combin(52,7)
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In order to have exactly 5 dead cards, two of your cards must be a pair, and the other 5 cards must have no pairs and your seven cards must have no more that 2 cards in any given suit.

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

Quote: Ace2

^^^That’s exactly what I calculated, though I did edit my answer
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Well you got the denominator right, lol. I am struggling with your numerator.

Try answering the question about 7 deads, which may illuminate the path to wisdom on 5 Deads.

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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gordonm888,

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

Dog Hand

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.

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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That is correct.

I asked about straights because I noticed charliepatrick mentioned them in his answer.

Dog Hand

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

^ 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.

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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To confirm the comments of others, in this puzzle (and in Gin and Go) one cannot meld straights.

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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I get a slightly higher answer. I am having difficulty critiquing your answer because you have used numbers rather than formulas such as c(n,m) for combinations.