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## Comments

1,048Pro1,048Pro1. 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.?

6,115Administrator, LeadProhttp://en.m.wikipedia.org/wiki/Monty_Hall_problem

166Subscriber166Subscriber1,592SubscriberOk, 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%

1,592SubscriberHammah was the one that said you must of invented this all by yourself!

1,048ProANYWAY 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.

1,048Probeauregard,

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

127SubscriberWhat 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.

184Subscriber193Subscriber#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

1,048Pro193Subscriberhaha, fair enough. I was just skimming through the posts and caught this one.

428Subscriber428SubscriberIf 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.

1,573SubscriberI believe you failed the test.

794SubscriberThis 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.

2Subscriber