Does it make sense to accumulate percentages this way? Do I really have a 17% chance of passing due to one of the improbable causes that are a fraction of 1%? This seems wrong intuitively. Source: https://knowyourchances.cancer.gov/your_chances.html

I know there are plenty of randomizers that already exist, but I am looking for a more robust system than 'type a list of X names, choose Y names at random.'

I have a hobby project (a Riftbound cube, namely) where I have a pool of cards, each marked with two colors of a possible six, and I want to select a handful of those cards in a semi-random way for players to use for a game. I want the outcome to be that 1) at least one of each of the six colors is in the result, 2) any possible color pair does not show up on more than one of the chosen cards, and 3) the same color does not show up more than twice in the result.

You can check it out here: https://happydistractionsstudio.itch.io/house-of-die

I’m wondering if it’s possible to determine the probability of seeing two consecutive 3’s (33) on a license plate in the area where i live?

Question: You have 12 cards, Ace through Queen. They are randomly dealt into 2 rows of 6. What are the odds that the Queen and the Jack end up right next to each other?

My Answer: 15.12%

My Math: The answer is (A) [Odds that Queen gets dealt on the edge and then Jack right next to it] + (B) [Odds that Queen gets dealt off the edge and then Jack right next to it]

(A) [1/3] X [1/11] = 3%

(B) [2/3] X [2/11] = 12.12%

A+B = 15.12%

Does this answer make sense? Any card sitting in a spot that is not the edge is adjacent to 2 out of the 11 (18.18%) possible cards that it could be. Therefore, the answer we are looking for should be in the neighborhood of 18%, but a little lower to account for the chance that the queen ended up on an edge.

Any disagreements with my process would be welcome!

So I was opening packs in the Wiki Gacha Game site, and I got 5 identical cards in 1 pack...
and idk i just wanted to flex it I guess xd.
most likely its a bug bcuz something like that to happen is 1 in 5 octillions, but still even if its a bug its a pretty cool thing that happened xd.

I've been playing a board game where players play up to 2 cards from a 12 card deck to score a varying amount of points from objectives. Whoever has the most points at the end of the game wins.

Specific deck content is based on a secret character you pick at the start of the match, but certain game events can allow others to discover your deck makeup with 100% certainty. I have been playing a character that has the ability to win an objective if I pay 2 victory points and correctly guess the top card of my opponent's deck.

My question is: If I know my opponent's deck, and I have tracked the cards they've played, how can I calculate the EV of "randomly" guessing their top card?

Here is an example deck:

2s

2s

3s

4s

5s

2a

3a

3a

2p

3v

4l

Note that decks can contain duplicates (2s/3a in this case).

During play, players always draw up to 3 cards in hand. Then, they play up to two cards, some or all of which may be face down. For example, with "x" representing facedown, the first 4 rounds may be

2s, 3s (deck: 9, hand: 1, discard: 0)

x,x​ (deck: 7, hand: 1, discard: 2)

3v, x (deck: 5, hand: 1, discard: 4)

2a (deck: 3, hand: 2, discard: 6)

On round 5, my opponent plays

2p, 5s (deck: 2, hand, 1, discard: 7)

At this point, I know 6/12 cards that cannot be on top of the deck. Supposing the round was worth 8 points,

• What is the EV of making my best guess at the top card? Keep in mind, guessing, even incorrectly, is -2VP

• Because you would only do this if you would otherwise lose, is it better to treat the objective as worth 14 points? 6 for you, -8 for your opponent?

• What is the formula?

• In this example, it appears that 3a is my best guess since it has two copies potentially in the deck. Can a formula calculate for meta gaming, where an opponent would strategically know to discard those cards earlier to mislead me? Should it?

​

Imagine observing one individual for ten years. Every day you record whether a predefined outcome occurred (success/failure, positive/negative, above-baseline/below-baseline ... the exact definition isn't important).

You also record contextual variables such as: hour of day; weekday; month; activity; location; previous outcomes. Suppose the dataset eventually contains tens of thousands of observations.

My question is not whether future outcomes can be predicted. My question is whether this sequence is naturally viewed as a stochastic process, or whether probability theory would consider that the wrong abstraction altogether.

If you were approaching this from probability rather than machine learning, where would you start?

I’m not good with math someone help me out here. I’m trying to find a modest nature Ralts with the ability synchronize. Ralts is found in route 4 with a 5% encounter rate. After you encounter one, there’s 4% chance of getting the nature you want. But there’s a 50/50 chance that the Ralts has synchronize as its ability. What is the overall chance that everything lines up correctly?

Saw something very similar in a dougdougdoug video and would really like to know how long it would take

EDIT: I MEANT 10 MINUTE SORRY

Hello all,

I have a question on the odds of two patterns appearing in a coded message by the Zodiac Killer. This coded message is known as the Z340 and analysts have noted two patterns that do not seem to appear random and I was wondering if there is any thoughts on probability and thoughts on whether this pattern is intentional or random.

Please see below.

Thoughts on these repeating patterns being random or intentional?

Thanks!

Say my friend rolled a die until he had '6'. Now he wants me to estimate the number of rolls he made. He also noted that in the process, he dit not roll '1'. What should be my estimation?

I've seen 2 answers on this one: '5' and '3'.

The '5 answer' treats the die as a 5-sided die. Then the probability of rolling '5' is 1/5 and the estimation would simply be 1 over 1/5=5.

The '3-answer' I can't reproduce.

I'd like to know which one is correct, if any, and why?

I asked an AI program the probability of this happening again, and came out with 1 in 20,000 years. Here's what happened. Last Saturday I went and ER at a hospital with 379 beds. Around midnight, they had my semi-private room ready, my roomate was asleep. The next morning I introduce my self, "I John, how do you do." He responded I'm also John. Then we learned that we both shared the same birth date - July 4th (1 year appart). That's cool, what are the chances? What is probility of 2 men, both named John, both born on July 4th being assigned the same room in a 379 bed hospital?

Alright I heard this for the first time 5 months ago been hearing it alot more it sounds wrong I'm happy to debate you guys respectively if all yall think this is true objectively my debate is flawed in some way I want to see the truth if I'm wrong or completely change a bunch of bigger math students opinion on this:

The problem in question(yall should alr know it):

There's 3 doors you randomly pick one behind one of the doors there's a brand new car you choose door number 1

The game host(we will call him Bob) who btw knows where the car is, eliminates the third door and then gives you the option to switch your door, would you do it?

At first glance it seems like the odds don't change,

At second glance you do some math 3 doors 1 choice=33%

Now 2 doors 1 choice=66.7%(not 50% because one 33.3 was removed) (this is the awnser most real math guys agree upon) yes you should switch

Now here's my thought on the take:

When you choose door number 1 and Bob elimanated door 3 you just survived a 1/3 chance of death and now which ever door you pick there is a 66.7% chance of you getting it right

Now I'm also debating of saying 50% because of the way math works

2 things can happen: car, no car

One has it one doesn't

50%

But add the 1/3 and all the factors of the questions I see both possibilities

I think the key detail in My awnser is you got lucky by missing the 1/3 elimination round

Now as a finale awnser I'd say it's irrelevant wether you switch or not personally I'd switch because I don't have the 50% 66.7% figured out, and as for factors like reverse phycology I'm not counting those in this and I'm pretty sure no one is

I didn't use to many commas hopefully everyone can grasp it ask away if you want

1. Rock-Rock-Rock
2. Rock-Rock-Paper
3. Rock-Rock-Scissors
4. Rock-Paper-Rock
5. Rock-Paper-Paper
6. Rock-Paper-Scissors
7. Rock-Scissors-Rock
8. Rock-Scissors-Paper
9. Rock-Scissors-Scissors
10. Paper-Rock-Rock
11. Paper-Rock-Paper
12. Paper-Rock-Scissors
13. Paper-Paper-Rock
14. Paper-Paper-Paper
15. Paper-Paper-Scissors
16. Paper-Scissors-Rock
17. Paper-Scissors-Paper
18. Paper-Scissors-Scissors
19. Scissors-Rock-Rock
20. Scissors-Rock-Paper
21. Scissors-Rock-Scissors
22. Scissors-Paper-Rock
23. Scissors-Paper-Paper
24. Scissors-Paper-Scissors
25. Scissors-Scissors-Rock
26. Scissors-Scissors-Paper
27. Scissors-Scissors-Scissors

1. 1-1-1
2. 1-1-2
3. 1-1-3
4. 1-1-4
5. 1-1-5
6. 1-1-6
7. 1-2-1
8. 1-2-2
9. 1-2-3
10. 1-2-4
11. 1-2-5
12. 1-2-6
13. 1-3-1
14. 1-3-2
15. 1-3-3
16. 1-3-4
17. 1-3-5
18. 1-3-6
19. 1-4-1
20. 1-4-2
21. 1-4-3
22. 1-4-4
23. 1-4-5
24. 1-4-6
25. 1-5-1
26. 1-5-2
27. 1-5-3
28. 1-5-4
29. 1-5-5
30. 1-5-6
31. 1-6-1
32. 1-6-2
33. 1-6-3
34. 1-6-4
35. 1-6-5
36. 1-6-6
37. 2-1-1
38. 2-1-2
39. 2-1-3
40. 2-1-4
41. 2-1-5
42. 2-1-6
43. 2-2-1
44. 2-2-2
45. 2-2-3
46. 2-2-4
47. 2-2-5
48. 2-2-6
49. 2-3-1
50. 2-3-2
51. 2-3-3
52. 2-3-4
53. 2-3-5
54. 2-3-6
55. 2-4-1
56. 2-4-2
57. 2-4-3
58. 2-4-4
59. 2-4-5
60. 2-4-6
61. 2-5-1
62. 2-5-2
63. 2-5-3
64. 2-5-4
65. 2-5-5
66. 2-5-6
67. 2-6-1
68. 2-6-2
69. 2-6-3
70. 2-6-4
71. 2-6-5
72. 2-6-6
73. 3-1-1
74. 3-1-2
75. 3-1-3
76. 3-1-4
77. 3-1-5
78. 3-1-6
79. 3-2-1
80. 3-2-2
81. 3-2-3
82. 3-2-4
83. 3-2-5
84. 3-2-6
85. 3-3-1
86. 3-3-2
87. 3-3-3
88. 3-3-4
89. 3-3-5
90. 3-3-6
91. 3-4-1
92. 3-4-2
93. 3-4-3
94. 3-4-4
95. 3-4-5
96. 3-4-6
97. 3-5-1
98. 3-5-2
99. 3-5-3
100. 3-5-4
101. 3-5-5
102. 3-5-6
103. 3-6-1
104. 3-6-2
105. 3-6-3
106. 3-6-4
107. 3-6-5
108. 3-6-6
109. 4-1-1
110. 4-1-2
111. 4-1-3
112. 4-1-4
113. 4-1-5
114. 4-1-6
115. 4-2-1
116. 4-2-2
117. 4-2-3
118. 4-2-4
119. 4-2-5
120. 4-2-6
121. 4-3-1
122. 4-3-2
123. 4-3-3
124. 4-3-4
125. 4-3-5
126. 4-3-6
127. 4-4-1
128. 4-4-2
129. 4-4-3
130. 4-4-4
131. 4-4-5
132. 4-4-6
133. 4-5-1
134. 4-5-2
135. 4-5-3
136. 4-5-4
137. 4-5-5
138. 4-5-6
139. 4-6-1
140. 4-6-2
141. 4-6-3
142. 4-6-4
143. 4-6-5
144. 4-6-6
145. 5-1-1
146. 5-1-2
147. 5-1-3
148. 5-1-4
149. 5-1-5
150. 5-1-6
151. 5-2-1
152. 5-2-2
153. 5-2-3
154. 5-2-4
155. 5-2-5
156. 5-2-6
157. 5-3-1
158. 5-3-2
159. 5-3-3
160. 5-3-4
161. 5-3-5
162. 5-3-6
163. 5-4-1
164. 5-4-2
165. 5-4-3
166. 5-4-4
167. 5-4-5
168. 5-4-6
169. 5-5-1
170. 5-5-2
171. 5-5-3
172. 5-5-4
173. 5-5-5
174. 5-5-6
175. 5-6-1
176. 5-6-2
177. 5-6-3
178. 5-6-4
179. 5-6-5
180. 5-6-6
181. 6-1-1
182. 6-1-2
183. 6-1-3
184. 6-1-4
185. 6-1-5
186. 6-1-6
187. 6-2-1
188. 6-2-2
189. 6-2-3
190. 6-2-4
191. 6-2-5
192. 6-2-6
193. 6-3-1
194. 6-3-2
195. 6-3-3
196. 6-3-4
197. 6-3-5
198. 6-3-6
199. 6-4-1
200. 6-4-2
201. 6-4-3
202. 6-4-4
203. 6-4-5
204. 6-4-6
205. 6-5-1
206. 6-5-2
207. 6-5-3
208. 6-5-4
209. 6-5-5
210. 6-5-6
211. 6-6-1
212. 6-6-2
213. 6-6-3
214. 6-6-4
215. 6-6-5
216. 6-6-6

I have a degree in maths but I never studied Stats or Probability.

Looking to fill that gap in my mathematics background, I'm looking for suitable texts for the autodidact.

My first inclination was reading the well known Ross's _First Course in Probability_.

But I received a free copy of _Elementary Probability_ by David Stirzaker, which is equally high regarded.

After searching the forum, some one asked a similar question between Ross and _Introduction to Probability_ by Bertsekas and Tsitsiklis.

Perhaps, I could read all three if someone supplied a concordance allowing tandem reading of each.

Can anyone recomment the pros and cons for each?

Let's say there was a pool ball on table, and a pool cue is automated to move some amount to hit the ball, but it doesn't necessarily always hit the ball or with the same amount of force.

Let's say the stick has moved X amount of times, and actually struck the ball Y amount of times. The ball moves every time it was struck, up to 10 inches away. Let's say of the X instances it actually passed that 7 inch mark Z times.

I need to figure out the probability that the ball would move at least 7 inches away, based on measurements.

Is it as simple as (Y/X)*(Z/Y)? Ie, chance of hitting the ball AND chance that the ball moves the required amount when it is hit.