### Brain Teasers

# Card Cash

Probability
Probability puzzles require you to weigh all the possibilities and pick the most likely outcome.

Someone offers you the following deal:

There is a deck of 100 initially blank cards. The dealer is allowed to write ANY positive integer, one per card, leaving none blank. You are then asked to turn over as many cards as you wish. If the last card you turn over is the highest in the deck, you win; otherwise, you lose.

Winning grants you $50, and losing costs you only the $10 you paid to play.

Would you accept this challenge?

There is a deck of 100 initially blank cards. The dealer is allowed to write ANY positive integer, one per card, leaving none blank. You are then asked to turn over as many cards as you wish. If the last card you turn over is the highest in the deck, you win; otherwise, you lose.

Winning grants you $50, and losing costs you only the $10 you paid to play.

Would you accept this challenge?

### Hint

Perhaps thinking in terms of one deck is the wrong approach.### Answer

Yes!A sample strategy:

Divide the deck in half and turn over all lower 50 cards, setting aside the highest number you find. Then turn over the other 50 cards, one by one, until you reach a number that is higher than the card you set aside: this is your chosen "high card."

Now, there is a 50% chance that the highest card is contained in the top 50 cards (it is or it isn't), and a 50% chance that the second-highest card is contained in the lower 50. Combining the probabilities, you have a 25% chance of constructing the above situation (in which you win every time).

This means that you'll lose three out of four games, but for every four games played, you pay $40 while you win one game and $50. Your net profit every four games is $10.

Obviously, you have to have at least $40 to start in order to apply this strategy effectively.

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## Comments

nice teaser. i really loved it. i probably would never guess the answer. o, well, keep up the good work

That was great. I love probability teasers.

Great teaser and nice explanation. The last comment is not correct however. You do not necessarily need $40 to implement this strategy. According to the probability you will win one time in four. You are just as likely to win on the first game as you are on the fourth. On the other hand you might lose 100 games in a row (although that is highly unlikely). Obviously, the bigger the stake you have, the more you will be able to withstand a run of losses and probability dictates that you will win with this strategy in the long term.

Thanks for the kind words. I see your point, Jimbo, and I included the $40 requirement as a way to "feel safe" in taking this bet. For instance, if you started with $10, and your chance to win is one in four, you could easily lose that $10 and never get rolling; however, if you have $40 to start, you are probabalistically guaranteed to get that first win. You might like to live life on the edge, sir, but I like money. =]

Make that "probabilistically."

this is old problem, where usual question is: how can you maximize chance to find highest card (reward...)?

If i remember correctly, best way is to open first 37 cards (not 50), and then after that keep opening until you find next higher. That will give you 9I think) 37% chance to get highest,

With such approach, you could take bet even with 30$ prize.

If i remember correctly, best way is to open first 37 cards (not 50), and then after that keep opening until you find next higher. That will give you 9I think) 37% chance to get highest,

With such approach, you could take bet even with 30$ prize.

2 comments -- 1, having $40 does not make you guaranteed, or even "probabilistically guaranteed" a win. If you are looking for a better than 50% chance of winning (i.e., a less than 50% chance of going broke), you need $30. [Because your losing string would have a probability of (3/4)^3 ~ 42%.] $40 gives you a 68% chance of not going broke before your first win. To have a chance better than 95%, you will need $110. For a chance better than 99%, you will need $170.

Forgot to add my second comment -- 37 is indeed optimal for number of cards to turn over. As the number of cards increases, the optimal number to turn over before the "next highest card" part approaches n/e, where n is the number of cards and e is 2.71828... An excellent discussion of the reasoning behind this can be found at http://www.mathproblems.info/group2.html in problem #26.

I kept it simple because I think the problems that are not overly complicated, but that still make you think, are the best. At the risk of not being entirely correct, the teaser was understandable: no heavy math involved in the explanation. Nevertheless, you do raise several valid points, and based on your knowledge of this category, I'm looking forward to seeing one of your teasers in the future.

I wish you would play this game with me as dealer. I'd write the same integer on every card.

Tree... wouldn't each card then be the highest in the deck, so everyone would be guaranteed to win?

Nice statistical probablity variation. However, the part about the net profit every 4 games needs to be modified to every 5 games.

maybe i misread this teaser, but... if you flip over the first 50 cards... then start with the next 50 cards... you cant"save" the highest card of the first 50 - youve already flipped it and then started flipping others. i dont see how you dont have a 1/100 shot considering any integer is possible and there is no way to tell what the last card will be. maybe its not 1/100, but i still dont see how you get 1/4... an explanation would be appreciated

i said no because it didnt say they had to be DIFFERENT numbers

I agree with Deepsea. I immediately said no way, for they could put the same number on every card, ensuring you would never turn over the highest. Or, if you count ties, then 99 ones and 1 two.

Quick note: "highest" card does not necessarily mean strictly greater than the rest. If all cards are numbered the same, then any card may represent the "highest" card.

In terms of execution, you would keep flipping the cards until a strictly higher one came up; if none did, then the last card in the deck would be your chosen "highest" card.

As was pointed out, if the dealer were especially clever, she could write the same number on every card but the last, and then a lower number on the final card to foil this strategy; however, she does not necessarily know that you will be *using* this strategy. Indeed, if you picked any card but the last, you would win, and that's not good odds for the dealer.

Even assuming that she did know your tactic, I filed this teaser under Probability, not Trick. I applaud your outside-the-box thinking... now stop dodging the actual problem.

In terms of execution, you would keep flipping the cards until a strictly higher one came up; if none did, then the last card in the deck would be your chosen "highest" card.

As was pointed out, if the dealer were especially clever, she could write the same number on every card but the last, and then a lower number on the final card to foil this strategy; however, she does not necessarily know that you will be *using* this strategy. Indeed, if you picked any card but the last, you would win, and that's not good odds for the dealer.

Even assuming that she did know your tactic, I filed this teaser under Probability, not Trick. I applaud your outside-the-box thinking... now stop dodging the actual problem.

Nov 23, 2007

Here's how I solved it ... I turned over all 100 cards, and then turned over the highest one AGAIN. Always the winner!

OK this make no sense to me.

I was among those who instantly said no to the challenge. ALthough 5 2 1 return sounds good the dealer isnt going to play a game where they have a disadvantage.

So the dealer cn write any number on any card. I thought "all the same" too, knowing that would no tbe the highest card in the deck. THey all have the d=same value so it cant be the "highest" or lowest.

Second if you can look at as many cards as u want at once but the last card you turn over has to b the highest, I dont see how looking at 37 or 5o can help.

Assuming the dealer writes the numbers from 100 to 1 each card you look at will be lower, or if they are shuffled the high card could be in the first half or the second half.

I need more help to understand why you would have a 37% chance.

If you have no idea what the high card is you could turn over a card with 12564 on it and still not have the highest card.

I was among those who instantly said no to the challenge. ALthough 5 2 1 return sounds good the dealer isnt going to play a game where they have a disadvantage.

So the dealer cn write any number on any card. I thought "all the same" too, knowing that would no tbe the highest card in the deck. THey all have the d=same value so it cant be the "highest" or lowest.

Second if you can look at as many cards as u want at once but the last card you turn over has to b the highest, I dont see how looking at 37 or 5o can help.

Assuming the dealer writes the numbers from 100 to 1 each card you look at will be lower, or if they are shuffled the high card could be in the first half or the second half.

I need more help to understand why you would have a 37% chance.

If you have no idea what the high card is you could turn over a card with 12564 on it and still not have the highest card.

Your analysis is wrong. It is true you have a 50% chance of your highest card being in the second half. But given your highest card is the the second set, the odds of the first card you flip higher than the highest of the first deck also being the highest of that second deck is not 50%. Through simulation, of 10,000+ tries those odds are about 70.24%. And your odds of winning are half that, so 35.12%.

I'm so disappointed in my nerdiness, but had to do it.

I'm so disappointed in my nerdiness, but had to do it.

The odds are actually somewhat better than the 25% you suggest. (And certainly the previous poster's experiment is useful information - suggesting around 35%.)

The reason for this is you are expecting to win ONLY if (a) the highest card is in the second half AND (b) the 2nd highest is in the first half. That would give you the 25%, I agree. But there is another winning case: (a) highest card in the second half AND (b) 2nd highest card somewhere AFTER the highest card AND (c) 3rd highest card in the first half, or.... you can imagine other such permutations.

All the more reason you should take the bet!!

Thanks belatedly for a great teaser.

The reason for this is you are expecting to win ONLY if (a) the highest card is in the second half AND (b) the 2nd highest is in the first half. That would give you the 25%, I agree. But there is another winning case: (a) highest card in the second half AND (b) 2nd highest card somewhere AFTER the highest card AND (c) 3rd highest card in the first half, or.... you can imagine other such permutations.

All the more reason you should take the bet!!

Thanks belatedly for a great teaser.

Thanks for all the comments. It's good to see that this teaser is still appreciated after all this time! I completely agree that my analysis is by no means rigorous, and certainly better solutions exist.

The root question of the teaser is, "Yes or no?" and no matter how you justify it, whether 25% or 32.5%, or some other percentage, the answer is still most definitely yes. As the Answer portion states, this is only a sample strategy, one of many that can get you to the same answer.

That said, keep the comments coming!

Enjoy the teaser,

TTO

The root question of the teaser is, "Yes or no?" and no matter how you justify it, whether 25% or 32.5%, or some other percentage, the answer is still most definitely yes. As the Answer portion states, this is only a sample strategy, one of many that can get you to the same answer.

That said, keep the comments coming!

Enjoy the teaser,

TTO

Great puzzle!

Random observation: as of this writing the "Fun" rating on this is equal to the first three digits of e (2.71), which figures in the solution.

Random observation: as of this writing the "Fun" rating on this is equal to the first three digits of e (2.71), which figures in the solution.

The answer is ALMOST correct. It has a very small MISTAKE.

If the second highest card is in the first 50 cards that means it occupies one position. So there are 99 remaining positions for the highest card to be. 49 positions in the first pack and 50 to the other. The highest card has probability 50/99 to be in the other pack of cards and not 50/100 as you said.

If the second highest card is in the first 50 cards that means it occupies one position. So there are 99 remaining positions for the highest card to be. 49 positions in the first pack and 50 to the other. The highest card has probability 50/99 to be in the other pack of cards and not 50/100 as you said.

I have just read the solution of 37% and it doesn't make any sence. It has to be wrong. This guy applied his solution for another number of cards as well. He also did this for 3 cards. He says that you should open the first 2 , find the bigger and then open the rest. Guess what!!! He says that the chance of winning is 50%. This is clearly wrong. Because eventually you have to open all three cards and you win only if the highest card is last. that's 1/3

Dec 19, 2010

I would turn over 101 cards. Turn every card face up, find the highest, and turn it again. The dealer didn't say we can't turn a card multiple times.

Dec 19, 2010

Just a small edit; I would flip a card 101 times, not 101 cards.

I don't get it. Doesn't it mean that there is only 1 highest card out of the 100 cards(supposing the dealer wrote different numbers). And since it says that "the last card you turnover should be the highest" then that means that regardless of how you turnover the cards, it depends on whether or not the last one you turn over is the highest. In this case wouldn't the chance be 1/100 only?

Like for exampel if the cards were numbered from 1-100 and the lower half you picked was 1-50. So your first high card is 50 but whatever you pick next makes you lose. Or if you got the top half first so your first high card is 100. But since you have to turn over a card from the other half then whatever you turn over would be the last card you turn over and that would always be lower than 100 so you lose. In this case, isn't the only way that you will win is if you actually turn over the "100" card last?

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