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A paradox of changing odds I read about - doing my head in, but it must be easy to explain why.

Three sets of two playing cards: AA KK AK

With the cards turned face down, you task is to pick the AK pair. Odds are 3 to 1 you pick the correct pair. That is not the problem.

You pick a pair, and one card is turned over - it's a K. That means now, the odds of having the AK pair are 2 to 1.

How? Nothing changed, no magic, yet just by seeing one card of the pair you chose the odds change from 3 to 1 -> 2 to 1.

I have read the solution, but still don't understand this simple logic.

Nick

Linicks
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    "Odds are 3 to 1 you pick the correct pair" is "the problem", since the odds then are 2 to 1, corresponding to a probability of 1/3. ​ ​ –  Aug 28 '16 at 17:05
  • OK, reading the answer below, and then re-reading this, you are right - it is not 3 to 1 getting the correct pile, but in fact 2 wrong 1 right ~ 2 to 1 against. – Linicks Aug 28 '16 at 18:52

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Now you know that the pair is not AA. That new information could change the odds. Usually the original odds would be quoted as 2 to 1 against. In fact if you turn just one card at random and find a K, the odds you have AK are still 2 to 1 against. You now have $2/3$ chance of having the KK and $1/3$ chance of having AK because it is twice as likely you found a K from KK. To see this in detail, list the six possibilities of which pair you pick and which card you pick from the pair. Initially two of the six have you picking AK. When you find a K, three of the possibilities are ruled out, but only one of the remaining three has you choosing AK.

Ross Millikan
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  • Thanks, the solution I read was if you number the K's as K1, K2, K3 (it can apply to Aces also), then when the card is turned over you don't know what Kn it is... thus still a 3 to 1 chance. But... by seeing a king after your choice, then you either have KK or AK? – Linicks Aug 28 '16 at 17:18
  • @Linicks : ​ ​ ​ ​ ​ ​ ​ Yes. ​ The answer will depend on how the "one card" to be "turned over" was chosen. ​ ​ ​ If it was chosen in a way that will reveal a K whenever there is at least one, then K_is_revealed will have the same probability (namely, 1) of happening for each of ​ KK,AK , ​ so since they were equally likely before, they will still be equally likely after. ​ ​ ​ (continued ...) ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ –  Aug 28 '16 at 17:48
  • (... continued) ​ If it was chosen at random, then [the probability of K_is_revealed given that the pair is KK] will be twice [the probability of K_is_revealed given that the pair is AK], so since KK,AK had equal likelihoods before, [[pair is KK] and K_is_revealed] will be twice as likely as [[pair is AK] and K_is_revealed]. ​ ​ ​ ​ –  Aug 28 '16 at 17:48
  • OK, I still been thinking about this since. I still do not get it. Suppose you choose a set of two cards and neither are shown. Touch one card and you either have an A or a K. That means the other is either a A or a K - back to 2 to 1 again. This is doing my head in. – Linicks Sep 17 '16 at 16:03
  • @Linicks : ​ ​ ​ Take 3 cards - 2 As and 3 Ks. ​ Shuffle them and touch one of them. ​ You either have an A or a K. ​ In summary, different cases are not necessarily equally likely. ​ ​ ​ ​ ​ ​ ​ ​ –  Sep 24 '16 at 11:52
  • @Ricky Demer.

    OK, I got to the bottom of this, which the first response correctly stated the original odds are 2-1 against.

    I showed a seasoned gambler, and straight away on being shown the 3 sets of two cards he said "2 to 1", Why, I asked. Well, the reason is with three selections to pick 1, it works out 3 [to pick] minus 1 [to get] = 2 to 1. Now I understand.

    Thanks all,

    Nick

     – Linicks Sep 26 '16 at 18:11