I was just wondering and I can't find anything about it. I guess that not every culture uses gold wedding rings, but it is common in the West. How much would it be in terms of weight and value? What percentage of the world’s gold supply is currently used in wedding rings? I suppose it's not that much...

Dear Theydidthemath

If i spend an hour in an 8nm force feedback racing sim (for example on Spa Francorchamps and assuming that each corner outputs the full 8nm of force) what would that be equivalent to in the gym world?

To preface, in case anyone is unfamiliar with magic:

To Scry X, you look at the top X cards of your library (your deck). You may then reorder them on the top, or place them on the bottom in any order. For example, if I Scry 4, I can look at the top four cards, labelled A, B, C, and D. I can put D and A on top, such that D is the top card, and put A and C on the bottom, such that C is on the bottom. Basically, look at X cards, split them into two separate piles (a pile can contain no cards), reorder those piles as you please, then place one on top and one on bottom.

The question: For what values is it possible to go from one ordering of a library to another? We will assume Y = 100 for calculations. It is trivial that to Scry 100 allows you to reorder to any arbitrary sorting, and it would seem to Scry 1 would not allow you to, no matter how many times you scry. What range of values of X allow for you to sort your library arbitrarily (given that you may scry as many times as you'd like)?

Spider-Man hits this guy with a brick pretty hard. Did he kill someone? Can someone calculate how fast the brick was going, and if it killed this dude or just sent him to the hospital with a TBI? He is unmoving for the rest of the scene, even after the battle is ended.

https://preview.redd.it/0x8qy8q6qgge1.png?width=1067&format=png&auto=webp&s=e4cb7df6ce743fefb34f4b4772d56aa79ccc8c7e

New meta for teaching ?!?

How fast do you have to be going to amputate your leg (below the knee) after hitting a tree while skiing? Weight of human ~250.

Hmmmm...

How fast do you have to be going to amputate your leg (below the knee) after hitting a tree while skiing? Weight of human ~250 lbs.

https://www.youtube.com/watch?v=o1YB-qDLsn4

Because of this sub, I now have a place to ask for answers for questions I never knew I even had. This is an iconic Brock Lesnar moment and I was wondering how in the hell he was able to take these chair shots and still stand.

Please someone, calculate how much force those chair shots are (and for bonus points: please share an equivalent example of such force)

Thank You

https://preview.redd.it/hhp0e4xnofge1.png?width=739&format=png&auto=webp&s=e93832784f2b102a9775fa723b03ed5d765f996b

For context, toki pona has a very simple syllable structure. There are 9 consonants you can have at the start of a syllable (or none at all), 5 vowels, and you can either end the syllable with an 'n' or you can not. This would make 100, but there are 8 combinations of these rules that are considered "illegal" for ease of listening, bringing the total syllable count down to 92. I think the rest of the needed context is given in the post as it's relevant, but let me know if there's anything I forgor.

Hey so I've been playing poker and for the last hand I had diamond 9 and ace preflop. The first 3 cards came to 2 aces and another 9, the 4th card was the last ace.

I believe the probability of getting quads is around 1 in 270000 so about 0.0035% (roughly, I suck at maths lol) but what would be the probability of getting quads AND a full house at the same time?

Assuming the clippers never break or wear down.

S/o u/eportillo09 for the pic.

Hey everyone! Quick questions for ya!

Scenario: two bags full of poker chips numbered 1-300 each.

What are the odds of reaching in, drawing a number. And then reaching in the second bag, and drawing the same number?

So same number, twice in a row, from two separate bags that each have 300 chips in them.

This will be an easy one for most in here, I am horrible with numbers and wouldn’t know where to start lol.

Hope everyone is having a great day!

Edit to clarify! Since I didn’t realize it mattered what number, my friend won back to back drawings with the same number: 169. So he’s holding his ticket, they draw 169, he wins the first prize. They switch bags. And draw 169 again. He wins the second prize. Does that change how the equation is done?

https://www.youtube.com/watch?v=dyC-muW7QqY

https://www.reddit.com/r/theydidthemath/comments/1i7p6mq/request_all_3_people_got_dealt_the_same_poker/

Please check my math.

First we have to get a few assumptions out of the way, as I have seen a lot of semantic arguments here and in the original reddit thread.

"3 Identical Poker Hands"

3 Hands - This means we are talking about the odds of a game with exactly 3 players, or in a game of more players, we are talking about the odds of a prespecified 3 players receiving the same hand. Otherwise as you add players the odds get easier that 3 will have equivalent hands.

Poker Hands - This establishes we are talking about a 5 card hand that follows a predefined hand order. (high card, pair, 2 pair, 3 of a kind, strait, flush, full house, 4 of a kind, strait flush)

For this problem the hands that make this interesting are the flushes and strait flushes.

Identical/Same - Technically you can not have "Identical" or "Same" poker hands as once a card is dealt, no one else can be dealt that card, there is only one ace of spades.

So we are talking about "equivalent" hands.

These are poker hands, so suits matter (flushes/strait flushes) and this is covered in the video when we acknowledge. A suited hand when compared to an off-suit hand is advantaged, so they are treated as not equivalent.

What the video and most of the discussions I have seen miss is that not all off-suit hands are equivalent when compared. (for most games)

IF you are playing some poker variant with only 3 community cards or an Omaha variant, then these hands in the video are equivalent.

However, if you are playing any poker variant such as Hold'em where you are allowed to use a single card from your hand to make your best 5 card hand, these hands stop being equivalent.

Lets look at the example provided: As8h, Ad8s, Ah8c

Ah8c is advantaged over the other 2 hands as it has flush and some unique strait flush possibilities in Hearts and Clubs.

Ad8s is the next best hand because there are 12 diamonds remaining that can appear on the board to make flush hands.

As8h only has 11 spades remaining that can appear on the board to make it's flush hands, so it is mathematically the worst hand when they are compared.

Hands like As8h, Ad8s, Ah8d are equivalent hands. In fact with 3 players, to be equivalent off-suit hands the 6 cards contained in the 3 hands must be from only 3 suits. (If you disagree, please try and find an example)

With 2 card starting hands, we have 3 types of hands a player can receive.

A pair.

A high and low card (suited)

A high and low card (off-suit)

To the Math:

Pairs:

0

High/low card suited

(52/52 * 12/51) (6/50 * 1/49) (4/48 * 1/47)

High/low card off-suit

(52/52 * 36/51) (6/50 * 2/49) (2/48 * 1/47)

So if you want the odds of 3 players receiving equivalent hands we add these 3 probabilities.

[(52/52 * 36/51) (6/50 * 2/49) (2/48 * 1/47) ] + [ (52/52 * 12/51) (6/50 * 1/49) (4/48 * 1/47) ] + [0]

The breakdown:

Pairs: There are only 4 of each rank, 3 players can not have the same pair.

Suited:

(52/52) Any first card.

(12/51) There are 12 cards left in that suit.

(6/50) 2 ranks, 3 suits left.

(1/49) Only one card this can be.

(4/48) 2 ranks, 2 suits left.

(1/47) Only one card this can be.

Off-Suit: (this one is a bit trickier)

Quick note: The only off-suit 3 hands of poker that will have the same win % against each other, are 3 hands that only use 3 suits. So these are the off-suit hands we need to calculate.

(52/52) Any first card.

(36/51) 12 ranks, 3 suits.

(6/50) 2 ranks, 3 suits.

(2/49) 1 rank, 2 suits.

Pause to talk about the this second hand (6/50)*(2/49), there are 2 types of hands to consider here, but conveniently it works out nicely. The 2 considerations are if the 6/50 card matches a suit of the first hand or does not, but in either case, there are only 2 cards the second card can be.

(2/48) 2 rank, 1 suit.

(1/47) 1 rank, 1 suit.

Total odds is the sum of the odds of equivalent pairs(0), suited hands(14976) and off-suit hands(44928).

(0 + 14976 + 44928) / 14658134400 = 0.00000408674... ~ 1 in 245k