Specific queston about cards an odds.!!!

oK... here is the queston:::

Usin a standard deck of well shuffeled playin cards... is a person jus as likely to be delt 2 Aces in a row... or be delt a Ace an a Doose in a row.???
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I discussed this wit somone over the weekend an i didnt convience 'em they was rong... the corect answr is of course:::

You are much mor likely to be delt a Ace an a Doose in row than be delt 2 Aces in a row.!!!

I thank when people here agree wit me... the person i argued wit will see the err of ther way.!!!

TIA

 

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Ah crap, you can do this mathematically, but right now I forget (or am too lazy or both.)

If you are dealt an ace first off. Then there is a 3/51 cahnce you'll get dealt another one but a 4/51 chance you'll get a two.

 

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Your probability of being dealt a single ace is 4/52. The probability of the second ace being dealt is 3/52 so the total is 3/676. For an ace followed by a deuce, both probabilities are 4/52 so the total probability is 1/169.

The difference between those two probabilities is only 1/169, whether that makes one scenario "much more likely" than the other is debatable.

I find your posts extremely hard to read. Could you please try and type in more coherent English in future? Modern web browsers have spell checkers so there's no excuse.

 

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I was jus tolt that my queston is beter stated this way:::

Usin a standard deck of well shuffeled playin cards... is a person jus as likely to be delt 2 Aces in a row... together or be delt a Ace an a Doose in a row... together.???
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I discussed this wit somone over the weekend an i didnt convience 'em they was rong... the corect answr is of course:::

You are much mor likely to be delt a Ace an a Doose in a row... together than be delt 2 Aces in a row... together.!!!

I thank when people here agree wit me... the person i argued wit will see the err of ther way.!!!

TIA[/QUOTE]

 

oK... did you read my restated queston... ie... the deuce dont have to foller the Ace... the order they are delt dont mater.???

 

Order doesn't matter no.

 

Prom posted his answer before you restated the question. Are you being intentionally clueless with regard to spelling, grammar, and time, cluelesshusband? Please make some kind of effort to make some amount of sense.

Anyhow, there are 1326 combinations of 2 card hands dealt from a deck of 52 cards. Of these, 16 comprise an ace-deuce combination while only 6 represent a pair of aces.

 

Hmmm. I see the reasoning, but what's wrong with:

There are 2652 ways to deal two cards from a deck. {16 of them deal you an ace and a two}. 12 of them deal you two aces.

Because it seems to me that there are a couple of different equiprobable ways to deal a given pair of aces.

PS: serious question, "what's wrong with". I know there's something wrong with it, because it conflicts too much with the intuitive implications of the fact that there are eight cards that will work on the first deal for A-2, and only 4 for A-A, 4 that will work on the second for A-2, 3 for A-A. (That gives you 32 and 12, same as the 16 and 6 but counting each one of the unordered combinations twice) But I can't spit out a clear explanation.

PPS: slaps head. There are 32 ways to deal an ace and a two. Never mind. I'll leave it as an example for the edification of the similarly confused.

edit in:

Yeah, caught it second time through. Thanks.

 

What's wrong with that is that there are another 16 of those permutations are a two and an ace. Either way you slice it, permutations or combinations, the odds of getting dealt an ace and a deuce without regard to order are more than twice the odds of getting dealt a pair of aces.

 

Originally Posted by cluelusshusbund to Prom
oK... did you read my restated queston... ie... the deuce dont have to foller the Ace... the order they are delt dont mater.???

Yes... thats why i ask if he had read my restated queston... so i coud get his answr to my restated queston.!!!

So im corect that the odds are much beter of bein delt a ace-deuce combinaton than a Ace-Ace combinaton.!!!

Thanks.!!!

The way i figered it was... for a Ace-Ace combinaton:::

The odds is 1 4 in 52 for the firs Ace... an 1 4 in 52 that the second card turned up will be an Ace.!!!
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For a Ace-deuce combinaton:::

The odds of turnin up a Ace or a deuce is 8 in 52... an the odds of the second card completin the Ace deuce combinaton is 4 in 52.!!!

 

obviously. that is if the deck contains the same number of aces and deuces.

 

This math sucks!

Once you draw a card there are not 52 cards left in the deck.

Sorry that was getting on my nerves.

 

That math is perty simple... once a card is taken away from a deck of 52... ther are 51 cards left :shrug:

Edit:::

O i see what you'r talkin about.!!!

Even if a bunch of cards have been removed from the shuffeled deck of 52... the odds of the nest card bein drawn bein a Ace stays at 4 in 52.!!!

 

delete

 

Agreed.!!!

Woudnt that also be 4/52.???

 

Neither. There are 51 cards left in the deck after dealing one card. If that first card was an ace, 3 of those 51 remaining cards are aces. The probability of the second card being an ace given that the first card dealt was an ace is thus 3/51. Putting the two together, the probability of being dealt two aces is (4*3)/(52*51) = 1/221.

 

If all 52 cards of a shuffeled deck are layin seperately an face down... any card you slide toward you has a 4/52 chanse of bein a Ace... an even tho ther is now only 51 cards left... the nest card you slide toward you still has a 4/52 chanse of bein a Ace... yes.???

Edit:::

In other words... cards are delt face down... the odds of the firs card delt bein a Ace is 4 in 52... the odds of the second card delt bein a Ace is also 4 in 52 even tho ther was only 51 cards left in the deck.!!!

 

You should be banned.

You are not even a crank, just plain dishonest.

 

More likely, yes. You got 4 deuces possible after that Ace, but only three aces left after that first ace.

 

If you don't look at either card, and they just sit in front of you, then sure this means they both have a 4 in 52 chance of being an ace. This doesn't extend to what happens when you look at them though. Once you see one card the outcome of looking at the second card is conditional on the result of the first card. Thus, on seeing an ace, the second card has it's probability of being an ace reduced from 4/52 to 3/51.
\( P(Card1=Ace,Card2=Ace) = P(Card1=Ace)P(Card2=Ace | Card1=Ace) \)
not
\( P(Card1=Ace,Card2=Ace) = P(Card1=Ace)P(Card2=Ace) \)