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## The Red Baron – Fighter Ace

Posted by Karl Sharman on April 21, 2014 – 3:41 am

Baron von Richthofen died on this day in 1918 – He was a fighter Ace in WW1 – so, on that tenuous link, here’s an Ace question by Martin Gardner 1914-2010 – After going down in flames for my last question, this is a probability question by someone else…

Someone deals you a bridge hand (13 cards from a regular deck of 52 cards). You look at the hand and notice you have an Ace and say “I have an Ace”. What is the probability that you have another Ace?

The cards are collected and different hand is dealt. This time you look at your hand and state “I have the Ace of Spades” (which is true), what is the probability, this time, that you have another Ace?

Question: Is the probability in the second case the same as before, a lower probability, or a higher probability?

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This post is under “Tom” and has 9 respond so far.

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1. 1. Chris Said：

Hi Karl. Finding the probability of having two or more aces is a pretty hard challenge. However, for the second part, the (relative) probabilities are the same (unless I’ve gone daft).

At this point, I’ll simply state that the probability of having at least one ace is higher that the probability of having at least the ace of spades.

I’ll be back with the ratio of the latter two probabilities later (prolly today). I don’t know if I can face doing the absolute probability of either case though.

2. 2. Chris Said：

I’ve just realised this problem is easier than I had first thought. I’ll be back.

3. 3. Wizard of Oz Said：

For the first part of the problem, the probability of another ace in your hand is 1 – (probability of no other aces).

There are 12 cards remaining in your hand and 51 cards from which these 12 can be chosen. 3 of these are aces so 48 are not. So the chances of the rest of your hand NOT having an ace are:

48/51 * 47/50 * …. * 37/40 = 48!39!/36!51! = about 39%

So there is about a 61% chance of at least one more ace in your hand.

This is true whichever is your first ace, so nothing changes when it is the ace of spades.

4. 4. Chris Said：

Hi Wiz. I agree, except you have made a computational error: 39% should be 43.9% and so it’s about 56.1% probability of having at least one more ace.

5. 5. Chris Said：

Something’s not right. When I try to calculate other probabilities, I get ridiculous results. Probably a brain fart though.

6. 6. Karl Sharman Said：

As I am off to Grimsby tomorrow, I shall post Mr Gardeners answer now:-

The surprising answer is that, if you call “I have the Ace of Spades”, it is more likely that you have another Ace than if you just state “I have an Ace”!

If you state you have the Ace of Spades there is an 11686 / 20825 chance you have (at least) another Ace, which is 56.12%. If you state you have An Ace then the chance you have (at least) another Ace is 5359 / 14498 (which is 36.27%). It’s over 50% less likely!

How come? Well, this puzzle is more about a confidence problem than a probability problem. It’s about the disclosure of information.

Thinking about it another way. What this puzzle is really about is comparing the probabilities of:

# hands with at least two Aces / # hands with at least one Ace

and

# hands with at least two Aces (one of which is the Ace of Spades) / # hands with the Ace of Spades

OK, hold onto your hats, here we go with the math. To keep the equations of manageable size, I’m going to use the abbreviation b{n,r} to represent the binomial expansion function n!/(r!(n-r)!) (often referred to as n choose r).

First a few basics:

Total_Number_Hands— All possible ways to combine the cards to make a Bridge hand. There are 52 cards in a deck, and 13 cards in a hand = b{52,13}

No_Aces— The number of hands that have no Aces. There are 48 non-Ace cards, and 13 in a hand = b{48,13}

At_Least_One_Ace— The number of hands that have at least one Ace = Total_Number_Hands – No_Aces

Excatly_One_Ace— The number of hands that have just one Ace. (Chose any Ace, then choose 12 non-aces) = b{4,1} x b{48,12}

At_Least_Two_Aces— Start with the total number of hands, then subtract hands with either no Aces, or one Ace. You are left with the number of hands with two or more Aces = Total_Number_Hands – (No_Aces + Exactly_One_Ace)

We now have enough to calculate our first ratio (chances of having a second Ace if you state you have an Ace):

At_Least_Two_Aces / At_Least_One_Ace = 36.27%

To calculate the ratio where the person claims to have the Ace of Spades, we need some additional terms:

With_Ace_Spades— Number of hands with Ace of Spades. (We choose the Ace of Spades, and then have to choose an additional 12 cards from the remaining 51) = b{51,12}

No_Other_Aces— Number of hands with no other Aces = b{48,12}

At_Least_Two_Aces_(Including_Ace_Spade)— Number of hands with at least two Aces, one of which is the Ace of Spades = With_Ace_Spades – No_Other_Aces

We now have enough to calculate our second ratio (chances of having a second Ace if you state you have the Ace of Spades):

There you have it!

7. 7. Wizard of Oz Said：

I do have a problem with this solution.

Presumably the same result as for the ace of spades would coma about if I nominated the ace of hearts or diamonds or clubs, i.e. 56% chance of another ace in my hand.

What you seem to be saying is that if I have an ace and I DON’T nominate what it is then my chances of another ace in my hand of 13 will be 36%. Now, if nothing else changes but I DO nominate the suit of my ace then my chance of another ace in my hand, dealt from the same pack in the same order, immediately jumps to 56%.

All this just from stating the suit of my initial ace. I do find this hard to believe!

8. 8. Chris Said：

I agree with Wiz. But I only have a vague understanding of where the reasoning is faulty. I’ll definitely be looking more closely at this problem.

9. 9. Wizard of Oz Said：

With all due respect to Martin Gardner I can’t agree with his answer. It makes no sense. I haven’t yet got my head around his reasoning (or your presentation of it). I’ll let you know if I do.

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