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  • AesahAesah Posts: 1,048Pro
    yes but unless you agree with me your answer will be incorrect ;)
    Thanked by 1beauregard
  • AesahAesah Posts: 1,048Pro
    edited December 2014
    bearugeard let me ask you this. Texas Hold'em:

    1. You look at your first card and it's an ace, what is the probability the other card is an ace

    2. What is the probability that you're dealt at least 1 ace

    3. What is the probability you're dealt pocket aces

    Are you implying that the answer to 1. should equal 3. divided 2.?
  • Bart HansonBart Hanson Posts: 6,115Administrator, LeadPro
    Beaugard -- I didn't come up with this. It's the "Monty Hall Problem"

    http://en.m.wikipedia.org/wiki/Monty_Hall_problem
  • doubletapdoubletap Posts: 166Subscriber
    25%
  • doubletapdoubletap Posts: 166Subscriber
    what weighs nothing but if you put it in a barrel, it makes the barrel lighter
    Thanked by 1neverlearn2
  • beauregardbeauregard Posts: 1,592Subscriber
    Aesah wrote:
    bearugeard let me ask you this. Texas Hold'em:

    1. You look at your first card and it's an ace, what is the probability the other card is an ace

    2. What is the probability that you're dealt at least 1 ace

    3. What is the probability you're dealt pocket aces

    Are you implying that the answer to 1. should equal 3. divided 2.?

    Ok, Mr. Wizard... let's have a go!

    1. 52 cards, right?
    you have one card (Ace), so 51 cards remaining.
    3 Aces left.
    odds are 3/51 that the next one will be an Ace - 1/5.8

    2. 52*51 = 2652 ways to get two cards
    2652/2 = 1326 possible outcomes
    12*16 = 192 of those have an A
    192/1326 = 1/14.5

    3. 52*51 = 2652 ways to get two cards
    4 ways to get one A
    3 ways to get another A
    3*4/2652= 1/221

    or something like that.
    at least that's theoretical.
    in reality, the odds are very different.
    for example, the way I'm running -
    1.) 0%
    2.) 5%
    3.) -4%
  • beauregardbeauregard Posts: 1,592Subscriber
    Bart wrote:
    Beaugard -- I didn't come up with this. It's the "Monty Hall Problem"

    http://en.m.wikipedia.org/wiki/Monty_Hall_problem
    I didn't say you did.
    Hammah was the one that said you must of invented this all by yourself!
    :lol:
  • AesahAesah Posts: 1,048Pro
    edited December 2014
    ok well your answers to 1 and 3 are correct. however you can trivially see that 1/14.5 is lower than 1/13 (the chance you'll have an ace with 1 card), which can't possibly be right.

    ANYWAY disregarding the exact math, the point is that the "implication" i suggested is NOT true, and it's not true in parallel question about coins and children either. The probabilities are different under different conditions.
  • AesahAesah Posts: 1,048Pro
    edited December 2014
    ACK wrote:
    1) If I flip two coins and one of them is heads, what is the probability that they are both heads?

    beauregard,

    If you think the probability 50/50 how about we prop bet on it. We flip two coins and if one is heads, I'll give you $600 if they're both heads and you give me $500 otherwise ;)
  • sleepyboysleepyboy Posts: 127Subscriber
    Let me see if I understand what Aesah is saying about the coin flips. The way I first interpreted it was that hes going to flip a coin, if its heads hes going to flip a second coin and the odds of that one being heads is not 50/50, but that's not what hes saying, the odds are 50/50.

    What I believe he is saying is that hes going to flip two coins and tell you if one is heads. In this case you can see that 75% of the time he will tell you one is heads. (HT TH HH TT). Then you have a 67% chance that the remaining coin is tails.
  • AJD804AJD804 Posts: 184Subscriber
    To piggyback off of sleepyboy, i think i truly understand what is going on here. When looking at the original post, some of us were looking at it differently than others. I and and the majority were looking at it from the vantage point of one coin has already flipped and now we are ready to flip the second one. This is why we all believed the answer to be 50% or 1/2 or whatever. ACK and aesah were looking at it from the view point that all of the coin flipping was done in the past and they were just looking at the data. From that POV, then yes the answer is 1/3 or 13/27 for Q3
    Thanked by 1beauregard
  • JT00JT00 Posts: 193Subscriber
    Aesah wrote:
    ok well your answers to 1 and 3 are correct. however you can trivially see that 1/14.5 is lower than 1/13 (the chance you'll have an ace with 1 card), which can't possibly be right.

    ANYWAY disregarding the exact math, the point is that the "implication" i suggested is NOT true, and it's not true in parallel question about coins and children either. The probabilities are different under different conditions.

    #2 is astonishingly simple if you just think in reverse. What are the odds that neither of my two cards contains an Ace? A lot of probability and/or combinatorics problems are much easier solved by trying to solve for the opposing outcome.

    48 out of 52 cards are not an Ace

    48/52 - First card is not an Ace
    48/51 - Second card is not an Ace, if First card is not an Ace
    48/52 * 48/51 - First card is not an Ace AND Second Card is not an Ace

    48/52 * 48/51 = 2304/2652 = approx. 86.9%

    Simply subtract 0.869 from 1, and you have your answer: 13.1% OR 1/7.6
  • AesahAesah Posts: 1,048Pro
    Um yea literally the very first sentence I posted in the thread explained that, and I was responding to someone who quoted it so I figured we were with that
  • JT00JT00 Posts: 193Subscriber
    Aesah wrote:
    Um yea literally the very first sentence I posted in the thread explained that, and I was responding to someone who quoted it so I figured we were with that

    haha, fair enough. I was just skimming through the posts and caught this one.
  • ACKACK Posts: 428Subscriber
    edited December 2014
    I made typo in my earlier post, I meant the answers are 1/3, 1/3 and 13/27
  • ACKACK Posts: 428Subscriber
    Similarly you could phrase a question:

    If you deal me 5 cards and I look at them and tell you that "I have an ace". What is the probability that I have more than one ace?

    If you deal me 5 cards and I look at them and tell you that "I have the ace of spades". What is the probability that I have more than one ace?

    Again, the answer to the second question is higher than the answer to the first question.
  • workinghardworkinghard Posts: 1,573Subscriber
    ACK wrote:
    I made typo in my earlier post, I meant the answers are 1/3, 1/3 and 13/27

    I believe you failed the test.
    Thanked by 1beauregard
  • RogerHardyRogerHardy Posts: 794Subscriber
    AJD804 wrote:
    To piggyback off of sleepyboy, i think i truly understand what is going on here. When looking at the original post, some of us were looking at it differently than others. I and and the majority were looking at it from the vantage point of one coin has already flipped and now we are ready to flip the second one. This is why we all believed the answer to be 50% or 1/2 or whatever. ACK and aesah were looking at it from the view point that all of the coin flipping was done in the past and they were just looking at the data. From that POV, then yes the answer is 1/3 or 13/27 for Q3

    This is the correct answer. The first thing you learn in probability is that you have to define exactly what you're trying to solve for. This whole thread was a lesson in semantics more than it was probability.
    Thanked by 1beauregard
  • Limpy808Limpy808 Posts: 2Subscriber
    Whats the probability I'll get the 10 wasted minutes of my life back after reading this thread.....
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