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Mike Haven
10-05-2002, 05:53 AM
this is a great one that i found in a probability book

if lolly deals out a hand of 13 cards and offers to pay you \$10 to your \$10 bet if the hand contains at least two aces you would of course refuse to bet - in fact only 26% of such hands contain two or more aces

but if she keeps dealing out one 13 card hand at a time from a well-shuffled deck until she can tell you honestly that "this hand contains at least one ace" would you then be prepared to bet that the hand contains at least two aces, at even money?

or if, instead, in the latter case, she could say honestly "this hand contains the ace of spades" would you then be prepared to bet that the hand contains at least two aces, at even money?

BruceZ
10-05-2002, 06:54 AM
but if she keeps dealing out one 13 card hand at a time from a well-shuffled deck until she can tell you honestly that "this hand contains at least one ace" would you then be prepared to bet that the hand contains at least two aces, at even money?[

No, your odds just got even worse.

Odds of 2 or more aces is 4.4%. 4.4%/26% = 17%.

or if, instead, in the latter case, she could say honestly "this hand contains the ace of spades" would you then be prepared to bet that the hand contains at least two aces, at even money?

Yes, now you are a favorite.

P(2+ aces|As) = [P(As|2 aces)P(2 aces) + P(As|3 aces)P(3 aces) + P(As|4 aces)P(4 aces)]/25%

= [.5(.216) + .75(.041) + 1(.0026)]/.25 = 56.5%

BruceZ
10-05-2002, 07:45 AM
In part 1 I computed the probability of getting 3 or more aces given that you got 2 aces. We want the probability that we have 2 aces given 1 ace which is just P(2 ace)/P(1 ace) = 26%/48% = 54%, so you are a favorite here too.

BruceZ
10-05-2002, 08:25 AM
I can't seem to add the right 4 numbers together. P(1+ aces) = 69.6%. 26%/69.6% = 37.4%, so you are not a favorite as I said in the first place.

My original post just had answers with no math. I should have left it that way...

lorinda
10-05-2002, 03:33 PM
If i offer anyone \$10 if they win a bet with my current form, they had better factor in the probability of getting paid if they win!!!

Bozeman
10-05-2002, 09:24 PM
Don't take the bet. But if you replaced "aces" with "aces or kings", then you win.

As for the As bet, then it is like 1 ace out of 12, which I believe is slightly better than 50%.

Craig

Bozeman
10-05-2002, 09:27 PM
Also, she could only tell you the hand contains at least one ace when it contains exactly one ace, and then you EV goes way down /forums/images/icons/smile.gif

Mike Haven
10-08-2002, 01:59 PM
before you calculate an exact answer it's good to make a rough estimate first - this gives an indicator that things haven't gone astray during the number-crunching

as a 13 card hand will have on average one ace in it hands with no aces in them have to be more common than hands with two or more aces in them

you can expect to see one ace, no ace, two aces, three aces ( - seldom), or four aces (-rare), in that order of probability - this infers that you will see a lot more hands with one ace than with two aces in them - maybe twice as many? - if this is true then when lolly ignores the hands with no aces there might be only one in three hands containing two or more aces - after the arithmetic the exact answer turns out to be 37% - a very bad bet to take at even money!

in the second case, the hand consists of the ace of spades plus 12 of the other remaing cards - there are 3 aces in these 51 cards so any card dealt has a 3/51 chance of being an ace - with 12 cards the average number of aces will be
12 x 3/51 = 2/3

obviously there will actually be zero, one, two, or three aces in the 12, with no or one ace being the most common - to make up the average of 2/3 of an ace then one ace will have to be more likely than none - the chance of there being at least one more ace in the 13 cards therefore seems to be between 1/2 and 2/3

in fact, it turns out to be 56% - so you should take the offered evens bet and slowly remove all of lolly's money