Jump to content

Wyrd Chronicles #22


Lucidicide

Recommended Posts

By my math, for a given numbered scheme the % chance is (4/52 + 4/51) * 100 = 15.6% (edit: I had it right the first time)

4/52 because you have 4 cards of a given number, then add 4/51 because you have a second chance to flip that number. 

 

I haven't done stats for about 15 years though, so if there's a better way of doing it I'd like to know! 

Link to comment
Share on other sites

7 minutes ago, Rathnard said:

By my math, for a given numbered scheme the % chance is (4/52 + 4/51) * 100 = 15.6% (edit: I had it right the first time)

4/52 because you have 4 cards of a given number, then add 4/51 because you have a second chance to flip that number. 

 

I haven't done stats for about 15 years though, so if there's a better way of doing it I'd like to know! 

You can't just add up the probabilities because they aren't independent. The correct way to calculate this would be 1-(48/52 * 47/51) = 14.9%

(Also you really shouldn't multiply by 100. You can multiply by 100% if you want, because that's just multiplying by 1.)

Link to comment
Share on other sites

6 minutes ago, Bengt said:

According to gimp the colours are 238-130-232 (ee82e8) and 192-199-203 (c0c7cb).

Or a screenshot.

pink_on_light.png.0b79ff77264af16930d5ac

I used Paint since I don't have anything fancier on this machine. But interesting that the numbers are that different. OTOH, that screenshot looks quite different from what I see, so there's that as well.

 

screenshot.png

Link to comment
Share on other sites

Okay, let me see if I understand the math here.

For doubles, you flip a card and out of the 51 remaining cards there are three that match the value and twelve that match the suit, so you get 15/51 (or 780/2652) odds, or about 29%.

For the odds of flipping a given numeric value, doing it the case by case manner (a.k.a the hard way):

Card 1 but not card 2:  4/52 * 48/51

Card 2 but not card 1:  4/52 * 48/51

Card 1 and Card 2:  4/52 * 3/51

Sum of cases = 396/2652 or about 14.9%.  In other words, figuring out the odds of not getting the number of either card and then subtracting from it from one is simpler.  1- (48/52 * 47/51).

You can't do (4/52 + 4/51) because succeeding on the first flip changes the odds for the second flip, so you have use dependent probabilities (multiplication) instead of independent probabilities (addition).

 

  • Like 1
Link to comment
Share on other sites

13 minutes ago, solkan said:

Okay, let me see if I understand the math here.

For doubles, you flip a card and out of the 51 remaining cards there are three that match the value and twelve that match the suit, so you get 15/51 (or 780/2652) odds, or about 29%.

For the odds of flipping a given numeric value, doing it the case by case manner (a.k.a the hard way):

Card 1 but not card 2:  4/52 * 48/51

Card 2 but not card 1:  4/52 * 48/51

Card 1 and Card 2:  4/52 * 3/51

Sum of cases = 396/2652 or about 14.9%.  In other words, figuring out the odds of not getting the number of either card and then subtracting from it from one is simpler.  1- (48/52 * 47/51).

You can't do (4/52 + 4/51) because succeeding on the first flip changes the odds for the second flip, so you have use dependent probabilities (multiplication) instead of independent probabilities (addition).

 

Your math is correct. But that last paragraph doesn't make much sense.

A simpler way to explain it might be:

One of four things can happen when you flip two cards:

  • Card 1 is an ace and card 2 is not 4/52*48/51
  • Card 1 is not an ace and card 2 is 48/52*4/51
  • Both cards are aces 4/52*3/52
  • Neither of the cards is an ace 48/52*47/51

We can calculate the probability of getting at least one ace by adding up all the cases in which we flip an ace, but there is also an easier way. We know that one of the four cases must happen, so the probability of at least one of them happening is 1 (or 100%). The cases are separate, so no two of them can happen at the same time, (because card 1 either is an ace or it isn't), so the probabilities are also separate, meaning that they must all add up to 1. That means that we can very easily calculate the probability of getting at least one ace by subtracting the probability neither of the cards is an ace from 1, i.e. 1 - (48/52 * 47/51).

Here's the proof for those who doubt:

1 = 4/52 + 48/52 = 4/52*1 +  48/52*1 = 4/52*(3/51 + 48/51) + 48/52*(4/51 + 47/51) = 4/52*3/51 +  4/52*48/51 + 48/52*4/51 + 48/52*47/51

<=> 1 - 48/52*47/51 = 4/52*3/51 +  4/52*48/51 + 48/52*4/51

 

It is also possible to calculate the probability of getting at least one ace another way by using conditional probability.

The probabilities of getting an ace from two different cards is not independent, so you can't simply add up the two probabilities. What you can do however is take the probability of getting an ace when you flip one card 4/52 and add to that the probability of getting an ace on your second flip on the condition that the first card you flipped was not an ace. That gives us 4/52 + 4/51*48/52.

That formula can also be written in several different ways:

4/52 + 4/51*48/52 = 4/52 + 4/51 - 4/52*4/51 = 1 - (1 - 4/52)*(1 - 4/51) = 1 - 48/52*47/51 :huh:

 

  • Like 1
Link to comment
Share on other sites

Join the conversation

You can post now and register later. If you have an account, sign in now to post with your account.

Guest
Reply to this topic...

×   Pasted as rich text.   Paste as plain text instead

  Only 75 emoji are allowed.

×   Your link has been automatically embedded.   Display as a link instead

×   Your previous content has been restored.   Clear editor

×   You cannot paste images directly. Upload or insert images from URL.

×
×
  • Create New...

Important Information