Easy math puzzles
Poll
 | 20 votes (46.51%) | ||
| | 14 votes (32.55%) | ||
| | 6 votes (13.95%) | ||
| | 2 votes (4.65%) | ||
| | 12 votes (27.9%) | ||
| | 3 votes (6.97%) | ||
| | 6 votes (13.95%) | ||
| | 5 votes (11.62%) | ||
| | 12 votes (27.9%) | ||
| | 9 votes (20.93%) |
43 members have voted
April 6th, 2022 at 6:39:30 PM
permalink
I'll let the Solitaire puzzle go another day in case someone is working on it. Admittedly it's a long slog which is unusual for a Riddler puzzle. One fanatic there went on to solve for a "draw one card at a time" scenario. Offhand, I didn't even think that was mathematically possible.
Have you tried 22 tonight? I said 22.
April 7th, 2022 at 2:31:19 AM
permalink
It would be possible if say all the up-cards left at least 24 cards that didn't go e.g. KKKKJJJ would leave 33 cards (J plus all 9s thru 2s couldn't go).
April 8th, 2022 at 9:50:04 PM
permalink
Quote: GialmereIt's Toughie Tuesday. Let's waste time playing...
While killing some time at your desk one afternoon, you fire up a new game of Solitaire on your computer (specifically the version where you deal out three cards from the deck at a time). But your boredom quickly turns to rage because your game is unplayable — there's nothing you can do on the board and you can flip through your deck, but you never have any legal moves!
Rounded to the nearest quarter percent, what is the probability of being dealt such a nightmare scenario?In America, Klondike is so popular that if you simply say you're playing Solitaire it's assumed that's the game you're playing. This is somewhat baffling since Klondike has a very low win ratio compared to most other solitaire card games. What are the odds that you'll win? Well, according to Wikipedia...Quote: WikipediaThe probability of being able to win a game of Klondike with best-possible play is not known, although Hoyle's Rules of Games suggests the chances of winning as being 1 in 30 games. The inability of theoreticians to precisely calculate these odds has been referred to by mathematician Persi Diaconis as "one of the embarrassments of applied probability".
The game enjoyed a surge in popularity when it was included as part of the Windows operating package. The idea was to get users comfortable using a mouse by dragging and dropping red 6s onto black 7s. The result was workers wasting endless hours playing the game. But hey, it was worth all that losing to see those cards fly around the screen when you finally won right? According to Microsoft, Solitaire was its most popular program for many years.
Klondike was also called Canfield in America, perhaps because it was once a casino game at the Canfield Casino in Saratoga Springs, New York. I've heard of Solitaire in casinos but have never seen it myself. I do think that the reason "Las Vegas Scoring" is so popular on computer versions is that you can lose the game but still claim a minor victory if you got enough cards up top to end with a positive cash amount.
Anyways, although we may never know the true odds of winning Klondike, the above puzzle (from the Riddler) is solvable (but not a cakewalk).
link to original post
They are about 0.25 percent, or about 1 in 400.
Let’s lay out the things we need to consider to tackle this problem. The first thing is the board of cards we’ve arranged on the table to begin the game. That board includes seven face-up cards (and 21 face-down cards). The second thing is the deck that we’ll deal from during the game. That deck includes the 24 remaining cards, but because we’re dealing three cards at a time, we only care about eight of them (the ones that will appear at the top of the three-card draws). These two sets — the seven face-up cards and eight in the deck — will determine whether we’ll play a nightmare game of Solitaire in which we have no legal moves.
What’s left to do is an intricate counting problem involving some very big numbers. In what follows, I’ve adapted the approach of solver Jacob Kes, who was kind enough to provide the code he used for counting card combinations.
Brute force is always the most elegant solution! The key criteria in this problem are a card’s rank (that is, whether it’s a number or a face card) and its color (red or black). You can use a computer script to find all the possible combinations for the seven face-up cards by rank and color only — for example, a red five, a black seven, a red jack, etc.3 But we can’t forget the suits!
For each set of seven number-and-color combinations, there’s a corresponding number M of possible combinations of seven cards (now taking suit into account). For each of these combinations, you can calculate how many of the remaining cards (of which there are 45 and might end up in the deck) wouldn’t allow us any legal Solitaire moves — let’s call this number N. That’s how many are neither aces, which can be moved to a designated area on the Solitaire table, nor are cards that can be moved onto a face-up card.
This means that, for a given seven cards, our chance of a nightmare deal is of P = (N choose 8)/(45 choose 8). Each face-up card combination has a 1/(52 choose 7) chance of occurring, so for a total probability, you just need to sum M*P/(52 choose 7) over all of the face-up number-and-color combinations. This winds up being 643,746,385,468/257,479,369,193,475, or about 0.0025, or about 1/400.
In addition to walking through the intricate math to arrive at this number, Laurent Lessard calculated the probability of a nightmare game in the version where you deal one card at a time. It’s significantly lower: about 1.8⋅10^−7
Good luck avoiding a nightmare game, you brave, solitary warriors.
Let’s lay out the things we need to consider to tackle this problem. The first thing is the board of cards we’ve arranged on the table to begin the game. That board includes seven face-up cards (and 21 face-down cards). The second thing is the deck that we’ll deal from during the game. That deck includes the 24 remaining cards, but because we’re dealing three cards at a time, we only care about eight of them (the ones that will appear at the top of the three-card draws). These two sets — the seven face-up cards and eight in the deck — will determine whether we’ll play a nightmare game of Solitaire in which we have no legal moves.
What’s left to do is an intricate counting problem involving some very big numbers. In what follows, I’ve adapted the approach of solver Jacob Kes, who was kind enough to provide the code he used for counting card combinations.
Brute force is always the most elegant solution! The key criteria in this problem are a card’s rank (that is, whether it’s a number or a face card) and its color (red or black). You can use a computer script to find all the possible combinations for the seven face-up cards by rank and color only — for example, a red five, a black seven, a red jack, etc.3 But we can’t forget the suits!
For each set of seven number-and-color combinations, there’s a corresponding number M of possible combinations of seven cards (now taking suit into account). For each of these combinations, you can calculate how many of the remaining cards (of which there are 45 and might end up in the deck) wouldn’t allow us any legal Solitaire moves — let’s call this number N. That’s how many are neither aces, which can be moved to a designated area on the Solitaire table, nor are cards that can be moved onto a face-up card.
This means that, for a given seven cards, our chance of a nightmare deal is of P = (N choose 8)/(45 choose 8). Each face-up card combination has a 1/(52 choose 7) chance of occurring, so for a total probability, you just need to sum M*P/(52 choose 7) over all of the face-up number-and-color combinations. This winds up being 643,746,385,468/257,479,369,193,475, or about 0.0025, or about 1/400.
In addition to walking through the intricate math to arrive at this number, Laurent Lessard calculated the probability of a nightmare game in the version where you deal one card at a time. It’s significantly lower: about 1.8⋅10^−7
Good luck avoiding a nightmare game, you brave, solitary warriors.
----------------------------------------------------------------
...when you’re cheating at solitaire and a fight breaks out.
Have you tried 22 tonight? I said 22.
April 18th, 2022 at 3:14:31 PM
permalink
Slight variation on a previously posted problem:
Henry and Tom decide to bet on a coin flip. Henry wins on heads, Tom wins on tails.
It’s $1 per flip and they are really bored, so they decide to do one million flips. After each flip, the loser pays the winner $1. The players may bring any bankroll they want, but if a player goes bankrupt he automatically loses the game. Tom is very wealthy and tells Henry he's bringing a million dollars to the game. Henry is not rich and decides to bring an amount that will give him 50% confidence of not going bankrupt. How much should Henry bring?
Henry and Tom decide to bet on a coin flip. Henry wins on heads, Tom wins on tails.
It’s $1 per flip and they are really bored, so they decide to do one million flips. After each flip, the loser pays the winner $1. The players may bring any bankroll they want, but if a player goes bankrupt he automatically loses the game. Tom is very wealthy and tells Henry he's bringing a million dollars to the game. Henry is not rich and decides to bring an amount that will give him 50% confidence of not going bankrupt. How much should Henry bring?
It’s all about making that GTA
April 18th, 2022 at 3:44:40 PM
permalink
I've been meaning to do something with this kind of problem, but haven't found the time. For now, my answer, which is probably wrong, is:
$337
“Extraordinary claims require extraordinary evidence.” -- Carl Sagan
April 19th, 2022 at 12:56:35 AM
permalink
I get the same answer*** as the Wizard above
***: $337 when rounded
Formula used:
~0.67449σ = "50% of proportion within"
n = "number of coin flips" = 1 million
~0.67449 x0.5 x (n^0.5) = 0.337245 x (1,000,000^0.5) = 0.337245 x 1000 = 337.245
Note: I don't use this (or a variation of this formula) very often, so I could have used it incorrectly / applied it wrong.
Last edited by: ksdjdj on Apr 19, 2022
April 19th, 2022 at 7:48:53 AM
permalink
I respectfully disagree and suggest that you double (cough, cough) check your answers
Last edited by: Ace2 on Apr 19, 2022
It’s all about making that GTA
April 19th, 2022 at 1:56:55 PM
permalink
Quote: Ace2I respectfully disagree and suggest that you double (cough, cough) check your answers
link to original post
Because I am a relative beginner at this, can you give me a hint? (in spoilers if you like)
Something like, "the formula you used was correct in this situation, but the numbers are wrong", thanks
April 19th, 2022 at 2:42:42 PM
permalink
This hint was in my last reply!
It’s all about making that GTA
April 19th, 2022 at 4:35:53 PM
permalink
Quote: Ace2This hint was in my last reply!
link to original post
Thanks.
~674.
But when I looked up the SD for a single coin flip online, I found it was 0.5 from multiple sources (the above figure assumes 1 for the SD).
If 674 is closer to correct, please explain why my original POSTED answer was wrong.
But when I looked up the SD for a single coin flip online, I found it was 0.5 from multiple sources (the above figure assumes 1 for the SD).
If 674 is closer to correct, please explain why my original POSTED answer was wrong.
Edited (about 455pm, Pac time): In the spoiler see word in capitals, for the edit.
Last edited by: ksdjdj on Apr 19, 2022
