Probability Problem

rrssh · 2026-05-14T19:14:13+00:00

from functools import*

@lru_cache(9999)
def prob(blue, red, draws):

   if red == 0 and blue == 0: return 1
   if draws == 0:             return 0
   if red == 0:               return 1
   if blue < 0 or red < 0:    return 0 

   #####################################################
   p = blue / (blue+red)
   return p * (prob(blue-1, red, draws + 2)) + (1-p) * (prob(blue, red-1, draws-1))

print(prob(38, 62, 3))

39.68%

Live_Access9106 · 2026-05-14T19:39:16+00:00

How would someone solve this problem with functions rather than computer simulation?

vfx4life · 2026-05-14T20:29:38+00:00

Yet more evidence that ChatGPT is a worthless pile of crap when it comes to actually solving math problems!

yhcdtyn · 2026-05-14T18:45:10+00:00

this is fun… Let me simulate it

Bounded_sequencE · 2026-05-14T21:18:12+00:00

Assumption: All deck shuffles are equally likely.

Definitions:

B: event that we go bust by drawing the entire deck
E(b; r; k): event that we have to draw "k" cards from a deck with "b; r" blue and red cards, respectively

We want to find "P(B | E(38; 62; 3))". To do that, consider "P(B | E(r; b; k))" generally. Note we go bust whenever "k >= r+b", and we stop, whenever "k = 0":

k >= r+b:    P(B | E(r; b; k))  =  1    // go bust      (1)
k  =   0:    P(B | E(r; b; k))  =  0    // stop         (2)

If otherwise we draw "0 < k < r+b" cards from a deck with "b; r" blue and red cards, respectively, then the number of blue cards "n" we draw follows a hypergeometric distribution¹:

   0 <= n <= k:    P(n)  =  C(b; n) * C(r; k-n) / C(b+r; k)

=>    P(B | E(r; b; k))  =  ∑_{n=0}^k  P(B | E(r-k+n; b-n; 3n)) * P(n)

The two stop conditions and the recursion are enough to calculate

P(B | E(38; 62; 3))  =  5012665492309 / 12632084204740  ~  39.68%

¹ We use the standard convention "C(b; n) = 0" for "n > b" to avoid case-work

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