[bonanova]

You and Joe are partners at Bridge. You, having no aces, are told by a reliable informant that Joe holds an Ace. You quickly calculate the probability that Joe holds at least two Aces, which would be enough to bid game. Wait. Your informant now tells you Joe holds the Ace of Hearts. Are Joe's chances worse, better, or the same?

[Guest]

Assuming you don't mean "Hearts" value differ from "Diamond"..etc.

It seems same but

If we know that one of Joe's Aces is Heart, we can get the odds of one of the three other Aces to be at Joe: There is (2/3)³ = 8/27 odds that Joe has no other Aces, than Joe's chance = 1-8/27=19/27.

Without knowing that it is Hearts, odds for four Aces to be zero at Joe=16/81, 1 at Joe= 32/81..etc.

Since we know that Joe has at least one Aces, we must exclude "zero at Joe" projection(16 of 81 cases).

Now Joe's chance is 1-(32/65) = 33/65.

I solved it back to front so Joe's chance is lowered from 33/65 to 19/27 when you heard "Hearts".

It's very different from 19/27, I didn't expect that, so this solution should be wrong!!!

[Guest]

You and Joe are partners at Bridge. You, having no aces, are told by a reliable informant that Joe holds an Ace. You quickly calculate the probability that Joe holds at least two Aces, which would be enough to bid game. Wait. Your informant now tells you Joe holds the Ace of Hearts. Are Joe's chances worse, better, or the same?

I don't know from Bridge. I just showed up for the snacks...

[bonanova]

It seems same but ...

Joe's chance is lowered from 33/65 [0.507...] to 19/27 [0.703...] when you heard "Hearts".

Confused. Lower? or higher?

[Guest]

so far....for the probability of at least 2 aces....

Probability at least 2 aces = 1 - Prob(0 aces) - Prob(exactly 1 ace)

= 1 - [ ₄₈C₅/₅₂C₁₃ ] - [₄₈C₁₂/₅₂C₁₃]

= 1 - 0.303 - 0.439

= 0.257 or 25.7%

this is certainly shaking some cobwebs out of the noggin....haven't had a class in probability since I was in college, and that was in upstate NY, so skiing, hockey and drinking all came before studying probability.....translation...hope I did this correctly....

[Guest]

Confused. Lower? or higher?

Sorry, since I explained my solution from back to front, I silly confused.

Joe's chance was raised from 33/65 to 19/27 when you heard that his aces was hearts.

And I give my regards to all cristians for Noel, from world of muslims.

[bonanova]

Sorry, since I explained my solution from back to front, I silly confused.

Joe's chance was raised from 33/65 to 19/27 when you heard that his aces was hearts.

And I give my regards to all cristians for Noel, from world of muslims.

Nice job. Not a simple problem.

I thought about pointing out that it could have been spades or diamonds or clubs.

I any case the probability increases; then asked why?

But the increase in odds in this case is difficult to see, and one gets lost in the calculations.

So I did it in terms of dice, where one's intuition leads to problems much more quickly.

Thank you for the seasonal wishes ... most gracious.

[Guest]

So I did it in terms of dice, where one's intuition leads to problems much more quickly.

I assume that the answer I gave was correct?

In dice problem, as I've posted, there is a definite clue that helps subject 1-6 to have a raised chance.

But in this aces problem I can't see a logical clue.

Having the information that one of the aces is hearts shouldn't help somebody to have a more chance.

I'm conflicted about my answer, and suppose it was wrong.

Bonanova, can you help us to understand the solution? Is being one of the Aces, Hearts is a clue?

[bonanova]

Deck has 8 cards:

Aces or spades, clubs and diamonds, call them AS AC AD,

and five non-aces, call them just 1 2 3 4 5.

Deal 4 cards, just to you and your partner.

Never mind the restriction that you have no aces for now.

The hands your partner can have, that have at least one A are:

1 Ace: [15 hands]

AS + 1234 1235 1245 1345 2345 [5 hands]

AC + ditto

AD + ditto

2 Aces: [30 hands]

AS AC + 123 124 125 134 135 145 234 235 245 345 [10 hands]

AS AD + ditto

AD AC + ditto

3 Aces: [10 hands]

AS AC AD 12 13 14 15 23 24 25 34 35 45 [10 hands]

One ace - 15 hands; 5 of them have AS.

Two aces - 30 hands; 20 of them have the AS.

Three aces - 10 hands; 10 of them have the AS.

Given at least one ace -> 55 cases, 40 of which have two aces. p[2 or 3 aces] = 8/11.

Given the Spade ace -> 35 cases, 30 of which have two aces. p[2 or 3 aces] = 6/7.

[Guest]

Deck has 8 cards:

One ace - 15 hands; 5 of them have AS.

Two aces - 30 hands; 20 of them have the AS.

Three aces - 10 hands; 10 of them have the AS.

Given at least one ace -> 55 cases, 40 of which have two aces. p[2 or 3 aces] = 8/11.

Given the Spade ace -> 35 cases, 30 of which have two aces. p[2 or 3 aces] = 6/7.

Thank you for your work. It proves the raising chance, without any word.

This is an interesting case. As Monty Hall problem, it is hard to believe. Even more interesting than it, because I could easily understand it. But, although you've proved Aces problem, I still hardly understand that how it happened that my chance is raised when I heard "hearts". It must be magic

[bonanova]

Thank you for your work. It proves the raising chance, without any word.

This is an interesting case. As Monty Hall problem, it is hard to believe. Even more interesting than it, because I could easily understand it. But, although you've proved Aces problem, I still hardly understand that how it happened that my chance is raised when I heard "hearts". It must be magic

When I think of increased chances [probabilities] I tend to think it's because the favorable cases increased.

Here, that isn't what happens - the total cases shrink a lot [all the at least one ace cases, where the ace is not ace of hearts]

The favorable cases shrink, too, but not proportionately as much.

So the probability increases.