You math guys out there ..

smellynuts · #1 02-23-2005, 11:02 PM

I probably could have posted this in the probability fourm.

I heard somewhere that the odds of at least one person being dealt Ax 6 handed is 33%. Can anyone confirm this?

AtticusFinch · #2 02-23-2005, 11:27 PM

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I probably could have posted this in the probability fourm.

I heard somewhere that the odds of at least one person being dealt Ax 6 handed is 33%. Can anyone confirm this?

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1) Nice handle. Real high-class [img]/images/graemlins/wink.gif[/img]

2) The easiest way to compute this is to look at the odds of this not happening. What are the odds of being dealt twelve cards in a row without an ace? For the first card, 48 cards qualify out of 52. For the second, 47 out of 51, and so on. Since these events must happen successively, you multiply the probabilities. Thus the answer is:

48/52 * 47/51 .... * 36/40 = .304

Thus the odds of none of the cards being an ace are only 30.4%. So the odds of at least one of them having an ace are actualy 69.6%.

InfernoLL · #3 02-23-2005, 11:32 PM

Now tell us how to calculate the probability that someone had an ace if one is in the flop.

SpikeSpiegel · #4 02-23-2005, 11:43 PM

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48/52 * 47/51 .... * 36/40 = .304

Thus the odds of none of the cards being an ace are only 30.4%. So the odds of at least one of them having an ace are actualy 69.6%.

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Umm, I may be wrong, but shouldnt it be

48/52 * 47/51 * ... * 37/41 = .338 ?

Because if you select a card when there are 40 left (the 36/40 in yours), you have already selected 12 = 6*2 and all the players have their cards...so, perhaps I'm wrong, but I think it's 33.8% chance of no Ace -> 66.2% chance of an Ace being somewhere

SpikeSpiegel · #5 02-23-2005, 11:52 PM

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Now tell us how to calculate the probability that someone had an ace if one is in the flop.

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Following from AtticusFinch's suggestion of computing the probability of there being no aces in people's hands, if there is exactly one ace on the flop:

46/49 * 45/48 * ... * 35/38 = .422

Start with the 46/49 I believe because we already know 3 cards so 49 = 52-3 and 46 = 49-3 because there are 3 aces left. So, that means there is a 42.2% probability of no one having an Ace -> 57.8% probability that someone has an Ace.

And just because it's something I always think about, if I don't have one, the probability of someone else having one (when there's one on the flop):

44/47 * 43/46 * ... * 35/38 = .479

So, if there is an Ace on the flop (exactly one), and you aren't holding an Ace, there is a probability of 47.9% that no one has one -> a probability of 52.1% that there is at least one somewhere in someone's hand.

AtticusFinch · #6 02-23-2005, 11:52 PM

[ QUOTE ]
[ QUOTE ]

48/52 * 47/51 .... * 36/40 = .304

Thus the odds of none of the cards being an ace are only 30.4%. So the odds of at least one of them having an ace are actualy 69.6%.

[/ QUOTE ]

Umm, I may be wrong, but shouldnt it be

48/52 * 47/51 * ... * 37/41 = .338 ?

Because if you select a card when there are 40 left (the 36/40 in yours), you have already selected 12 = 6*2 and all the players have their cards...so, perhaps I'm wrong, but I think it's 33.8% chance of no Ace -> 66.2% chance of an Ace being somewhere

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Yes, good catch. In engineering school, this sort of error was called an "off by one" error. Which was then abbreviated to O.B.1 (because they came up so often), which finally led to their nickname: Obi-Wans. Thus leading to silly sentences like, "Damn, I would have gotten an A on that exam if it weren't for my stupid Kennobi!"

But I digress.

AtticusFinch · #7 02-24-2005, 03:36 PM

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So, if there is an Ace on the flop (exactly one), and you aren't holding an Ace, there is a probability of 47.9% that no one has one -> a probability of 52.1% that there is at least one somewhere in someone's hand.

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Yes, and this probability is even higher if you have six callers in the hand, as people with aces call more often.

keikiwai · #8 02-24-2005, 03:53 PM

Another annoying one is the odds of seeing an A in the flop if one of your opponents has an A. It's lower than normal.

This came up when I went all in with AK. TT went all in, and another AK went all in. So what are our odds of beating TT?

1)If it were heads up (AK vs. TT) I would have a
3/48 + 3/47 + 3/46 + 3/45 + 3/44 = 33% chance of getting an A by the river (they are all + because this is an "or situation")

2) With my dummy opponent there are only 2 As left in the deck so odds of getting an A by river are.

2/46 + 2/45 + 2/44 + 2/43 + 2/42 = 23% (Which incidently are the odds of TT getting to TTT)

So my opponent knocked me down roughly 10%. And even if an A pops up, I will have to split with my opponent.

The dangers of playing with As.

Peter.

keikiwai · #9 02-24-2005, 04:05 PM

48/52 * 47/51 * ... * 37/41 = .338

A shortcut would be to see what the probability of all 4 Aces not being in the first 12 cards is. The probability fo 1 A not being in the first 12 cards is 40/52. This has to happen 4 times so you have (40/52)^4 = 35%

It is not exactly the same since I ignored the fact that if the 1st A is not in the top 12 it reduces the chance of the others not being in the top 12. Improved it is

(40/52)*(39/51)*(38/50)*(37/50)= .338

Exactily the same as Atticus (-Kanobe)!

Peter.

AtticusFinch · #10 02-24-2005, 04:22 PM

Ah, yes. That shortcut certainly makes it easier to avoid the dreaded Obi-Wan. (And saves a lot of typing.)