Peeping not allowed
- Do you have time?
- I feel time has me, but shoot.
- I want you to play a solitaire game several times in a row.
- What could me more fun?
- I will shuffle this deck of cards and put the pile face down on the table.
- I can dig that.
- You will turn one of the cards at the time starting from the top.
- OK.
- Before you do I want you to guess when you will draw the first red ace and when you will draw the second red ace.
- I have reason to believe that I will draw the first before the second.
- Brilliant! But can you be more precise? Where in the stack will the first red ace most likely be and where will the second be?
- I may have to peep?
Problem source: Martin Gardner, 1970, via Wordplay.
June 21st, 2010 at 2:57 am
It’s easier than we could assume:
The first red ace is most likely to be the first card, because that’s where there’s the most possible other places for the second red ace to be.
There is C(52,2) = 1326 ways to choose to places for the red aces.
There is 51 ways to have the first red ace in the first place.
Probability is 51/1326 = ~3,85%.
Similarly, the second red ace would appear most likely as the last card (just start discolsing the cards from the last one and you get the same problem than with the first red ace.)
Even though you should bet on the first card for the first red ace and on the last card for the second one, if you bet on where would be the two red aces at the same time (thus winning only if you get both on the right place), betting on the first and last cards doesn’t give you any better chance to win the bet than any other choice!
It may be counterintuitive, but that’s because if the first red ace shows up in the first place, (with odd 3,85%), it decrease the odds to find the other at the end (odds are then 1/51 =~ 1,96%, but they would have been 100% if first red ace was the penultimate one…).
A mathematician would simply says that the place of the first red ace and the place of the second one are two dependent variables: that means that the result of one has consequences on the result of the other and that implies that the odds of finding the two red aces on places A and B is not the product of the odds of finding the first red ace on place A with the odds of finding the second one on place B.