The "Deal or no Deal" dillemma

[Simon]

The problem below combines "Deal or no Deal" with Monty Hall.

 

There are 24 shuffled cards laid face down by a dealer. 12 are red and 12 are black. The dealer knows which are which.

 

You choose a card for yourself, hoping it is black.

 

You randomly start to turn over cards. To your good fortune, the first nine you reveal turn out to be red. The dealer interrupts the game.

 

He complains that your chance of having a black card has increased too much. Ideally he'd like you to eliminate black cards from now on to even things out.

 

"Alright", you say "Why not perform the eliminations yourself and make sure those three reds are not turned over.

 

The dealer agrees and eliminates 11 cards, all black.

 

There are 3 red cards and 1 black card unrevealed . One of them is yours.

 

What is the probability your card is black?

 

Simon

[snark1100]

lets go for 4/5.

[CraigD]

[math]\frac12[/math]. I have not been offered the opportunity to change my choice, so the probability of having chosen one of 12 out of 24 cards remains unchanged.

[Boerseun]

Well, seeing as there's only four cards, three of them red and one black, and I have to pick one of them, chances are 25% that the first card I pick up will be black.

[snark1100]

[math]\frac12[/math]. I have not been offered the opportunity to change my choice, so the probability of having chosen one of 12 out of 24 cards remains unchanged.

 

yes, but suppose that after you had made your choice, all 12 black cards had been faced. Then you would know that you did not hold a black card.

[Tormod]

yes, but suppose that after you had made your choice, all 12 black cards had been faced. Then you would know that you did not hold a black card.

 

AFAIK it doesn't matter what happened before - the odds for the game on the whole have not changed but the game has. It's now 1 black card and 3 red cards on the table.

 

I suspect the probability is the chance of picking the right card (1/4) times that of picking the wrong card (3/4) which is .187 or just about 1/5.

[Jay-qu]

I have heard many variations of this question and they always get me thinking!

 

In this case we have what is the chance of picking 1 of 12 from 24 so initially from your choice you have a 50:50 chance of getting either colour. This probability is the same as other cards are turned over as they havent affected your choice.

 

But since you ask what the probability is of your card been black we are now given conditions. So what is the chance of one card of 24 been black knowing that 11/12 blacks arent your card and 8/12 reds arent your card. This would make it 1/4.

 

An interesting question is knowing this, if given the offer of changing your choice will your chances be better or worse? :)

[Qfwfq]

I vote that the chance of the card you chose first being black is 1 in 37 while, for each of the other three, the chance is 12 out of 37.

 

Edit: No, wait, :( I got something backwards, it's 4 out of 5 for the first choice and 1 out of 15 for each of the others.

[Simon]

A fascinating range of solutions. :(

 

I have considered this carefully. From what I can tell, only one among you has nailed it.

 

Lets see how the discussion plays out before I give you mine.

 

But there's one point I will make that shouldn't give anything away.

 

The chance that your card is black = the chance that one of the other cards is red. Once you know the former, you'll know exactly what advantage or disadvantage, if any, there is in swapping.

 

Ok I'm going to rule out one answer. The probability of your card being black is not 1/4.

 

Simon

[CraigD]

I want to change my answer to the one snark1100 gave, 4/5. ;)

 

What this question is really asking by “What is the probability your card is black?” is, “what is the probability your card is black given that you have successfully picked 9 other cards which were red?”

 

The actions of the dealer, who knows which cards are which, are of no consequence.

 

Here’s why I think the answer is 4/5:

After you have chose your card, there are equal probabilities (12/24 = 1/2) that you have chose a winning black card, or a losing red one.

If you have chose a winning black card, the probability A of then choosing 9 consecutive red cards is [math]\frac{12}{23} \cdot \frac{11}{23} . . . \cdot \frac{4}{15} [/math].

If you have chose a losing red card, the probability B of then choosing 9 consecutive red cards is [math]\frac{11}{23} \cdot \frac{10}{23} . . . \cdot \frac{3}{15} [/math].

So, given that you have chosen 9 consecutive red cards, the probability that you’ve won is [math]\frac{A}{A+B} = \frac{4}{5}[/math] (sparing the tedious but convenient factoring and canceling steps)

 

It’s a pretty cool illustration of conditional probability. Up to drawing 12 cards after your initial choice, the more unlikely the probability of having drawn those cards, the more additional information about your probability of winning you gain. Drawing 9 red cards is pretty (about 0.03%) unlikely, so you gain a lot of information. Drawing 10 or 11 red cards gain even more, telling you your probability of winning is 6/7 or 12/13. Drawing 12 red cards, of course, tells you your probability of winning is 1.

 

If the game’s rule were expanded a bit, you could make it a game of skill. Say the game works as follows:

• You chose (removing it from play) one card from a deck of 12 black and 12 red cards
  
• A predetermined number of cards (N) are turned up at random, revealing their colors
  
• Based on this information, you choose how much to wager (W)
  
• The card you chose is turned up, revealing if you won (black) or lost (red).
  
  • If you lose, you pay the house W
    
  • If you win, the house pays you W*G, where G is the game’s “payoff odds”, determined in advance by the house

Here’s a fun question: for a particular value of N, what should G be for the house to expect to exactly break even against a player who’s a perfect mathematician? Keep in mind that, most of the time, the turning up of N cards won’t gain the player much information on which to base a mathematically correct wager.

 

The answer to this question when N=0 is trivial: G=1. Without any cards turned up, the player has no information on which to base their wager, so the expected value of the game is [math]\frac12WG -\frac12 W = 0[/math].

 

Do there need to be any additional rules for it to be possible to determine the "fair" G for a given N?

[Jay-qu]

A fascinating range of solutions. :P

 

I have considered this carefully. From what I can tell, only one among you has nailed it.

 

Lets see how the discussion plays out before I give you mine.

 

But there's one point I will make that shouldn't give anything away.

 

The chance that your card is black = the chance that one of the other cards is red. Once you know the former, you'll know exactly what advantage or disadvantage, if any, there is in swapping.

 

Ok I'm going to rule out one answer. The probability of your card being black is not 1/4.

 

Simon

In that case I dont think you have defined the question rigorously enough.

 

Would you care to clarify? Is this a swapping question?

[Simon]

Would you care to clarify? Is this a swapping question?

 

Not at all. The question at the end was clear: "What is the probability your card is black?"

 

And I conclude that the answer is 80%. To many, this will seem absurd, considering there are 3 reds and 1 black left unrevealed.

 

It was others who raised the swapping question. Actually, I'll revise my conclusions about that. Essentially they're the same as Qfwfq.

 

Chance your card is black: 80% (or 4/5)

Chance your card is red: 20% (or 1/5)

Chance any one of the other cards is black: 6.66% (or 1/15)

Chance any one of the other cards is red: 93.33% (or 14/15)

 

While some are reeling from that, the following version may be more intriguing. It uses the actual game show and it is a swapping question.

 

Deal or No Deal.

22 boxes hiding money prizes.

11 blue (low). 11 red (high).

You chose one box at random.

Obviously you want it to be red.

The other boxes are revealed one by one

 

The game follows the normal rules of "Deal or No Deal", with the following caveat:

 

With every box you open, the host will always reveal one of the opposite colour. He has been told which boxes have reds and which have blues, but he does not know the values.

 

Thus, at every stage of the game (after the host opens a box) the number of blues and reds unopened is exactly equal.

 

You've just played to the end, refusing all deals.

Twenty boxes have been opened.

There are two remaining.

For the sake of drama, lets say these are 1p and £250,000.

The banker has offered you a swap.

Do you accept it?

 

Does it matter which boxes you or the host opened?

Or are the odds 50/50 regardless?

 

Typical example:

 

Say, in the course of the game you opened 7 blues and 3 reds.

The host complimented each occasion with 7 reds and 3 blues.

 

Should you stick or swap?

Or does it make no difference?

 

Simon

[Jay-qu]

 

Chance your card is black: 80% (or 4/5)

 

why?

[Qfwfq]

Why?

 

Just like in the Monty Hall problem, except that in this case the information that you can partially exploit makes the probability higher (instead of lower :shrug:) for your card.

[CraigD]

why?

I thought I gave an OK explanation in this post, but seem to (as I often do) overrated my writing skills :(. Let me give it another try, without complicating my post with burblings about a follow-up problem.

 

The key to the problem asked by the question “what is the probability that you have picked (and removed from play) a black card” is to calculate the probability of the next event – randomly drawing 9 red cards in 9 tries – for the case of: (A) you picked a black card; and (B) you picked a red card.

 

If you picked a black card, there are now 11 black cards and 12 red cards available. The probability that the first random selection will be red is therefore [math]\frac{12}{11+12}[/math]. After selecting the first (red) card, there are 11 black cards and 11 red cards. The probability that the second selection will be red is [math]\frac{11}{11+11}[/math]. Next probability, [math]\frac{10}{11+10}[/math], and so on.

The probability of a chain of dependent events is the product of the probability of each event. So the probability A of randomly selecting 9 red cards from a pool of 11 black and 12 red cards is [math]\frac{12}{23} \cdot \frac{11}{22}. . . \cdot \frac{4}{15} = \frac{12! (23-9)!}{(12-9)! 23!} \dot= .0002692[/math]

 

If you picked a red card, there are 12 black cards and 11 red cards available. Calculating the probability of randomly B selecting 9 red cards as above gives [math]\frac{11}{23} \cdot \frac{10}{22}. . . \cdot \frac{3}{15} = \frac{11! (23-9)!}{(11-9)! 23!} \dot= .00006730[/math]

 

A and B are mutually exclusive, and it’s a given of the problem that A or B has occurred, so the probability that A has occurred is [math]\frac{A}{A+B}[/math].

Doing the arithmetic (notice that A and B have a common term of [math]\frac{(23-9)!}{23!}[/math], which we can omit), we get:

[math]\frac{12 \cdot 11 \cdot 10 \cdot 9 \cdot 8 \cdot 7 \cdot 6 \cdot 5 \cdot 4}{12 \cdot 11 \cdot 10 \cdot 9 \cdot 8 \cdot 7 \cdot 6 \cdot 5 \cdot 4 + 11 \cdot 10 \cdot 9 \cdot 8 \cdot 7 \cdot 6 \cdot 5 \cdot 4 \cdot 3} = \frac{12}{12+3} = \frac{4}{5} = .8[/math]

[Simon]

I thought I gave an OK explanation

 

You did more than that Craig. Your equations pin down precisely all the conditional probabilities implied in the question.

 

However you came up with the mathematical proof. When you're dealing with counter-intuition, sometimes that isn't enough.

 

I'll rise to the challenge and explain exactly why in English.

 

The key to this is something you alluded to, only you didn't give it enough emphasis. Your own eliminations always change the probability of what your card is. The host's eliminations never do.

 

Once you beleive this, everything else follows.

 

In the 24 card situation, you removed 9 cards, all of them red. There were no other random eliminations. The remaining 15 cards at this time consisted of 3 reds and 12 blacks. This tells you everything you need to know.

 

Chance of your card being red = 3/15 = 1/5 = 20%

Chance of your card being black = 12/15 = 4/5 = 80%

 

Whatever the host does after that, this probability is preserved - so long as there is at least one of each colour remaining.

 

The same principle applies to my second version which uses the actual game. Actually I wish I'd presented this originally, because less people will beleive the answer.

 

Deal or No Deal.

22 boxes hiding money prizes.

11 blue (low). 11 red (high).

You chose one box at random.

Obviously you want it to be red.

The other boxes are revealed one by one

 

The game follows the normal rules of "Deal or No Deal", with the following caveat. With every box you open, the host will always reveal one of the opposite colour. He has been told which boxes have reds and which have blues, but he does not know the values. Thus, at every stage of the game (after the host opens a box) the number of blues and reds unopened is exactly equal. You've just played to the end, refusing all deals. Twenty boxes have been opened. There are two remaining. For the sake of drama, lets say these are 1p and £250,000. The banker has offered you a swap.

 

Do you accept it?

Does it matter which boxes you or the host opened?

Or are the odds 50/50 regardless?

 

Typical example. Say, in the course of the game you opened 7 blues and 3 reds. The host complimented each occasion with 7 reds and 3 blues.

 

Should you stick or swap?

Or does it make no difference?

 

Answer for the above example:

 

Stick. Odds are 2/3 your box contains the red money.

 

 

Simon

[Simon]

Actually I need to re-evaluate the last part of my post above.

 

It seems that the game version with 22 boxes requires knowing not only how many of each colour you revealed, but in what order!

 

Simon