### Brain Teasers

# Swindler's Dice

Fun: (2.71)
Difficulty: (2.05)
Puzzle ID: #16401

Submitted By: silver_sword Corrected By: Winner4600

Submitted By: silver_sword Corrected By: Winner4600

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

A swindler once approached an honest man with a die. He handed him the die and told him about the bet. The die had six sides. If the man rolled a ONE, he wins, and gets back twice the amount of his bet. If not, the swindler would keep the bet.

"But...my chances are only one out of six," retorted the man.

"True," grinned the swindler, "But I'll give you three tries to get a one."

The man considered. Three tries, with each try having a 1/6 chance of winning. So his chances of winning are 1/2. Why not give it a try?

Is the bet really fair? If not, what are the chances of the man winning?

"But...my chances are only one out of six," retorted the man.

"True," grinned the swindler, "But I'll give you three tries to get a one."

The man considered. Three tries, with each try having a 1/6 chance of winning. So his chances of winning are 1/2. Why not give it a try?

Is the bet really fair? If not, what are the chances of the man winning?

### Answer

As you have guessed...the bet is not fair. He had calculated the probability wrongly. Probability does not accumulate, like 1/6 x 3.The probability of the man not getting a ONE in three throws is: 5/6 x 5/6 x 5/6, which is 125/216. This is the probability of the swindler winning.

Hence, the remaining fraction, 91/216, is the actual chance of the man winning.

Hide Answer Show Answer

## What Next?

**Solve a Similar Brain Teaser...**

Or, get a random brain teaser.

If you become a registered user you can vote on this brain teaser, keep track of

which ones you have seen, and even make your own.

## Comments

Ah, a probability puzzle! It is indeed a relatively rare breed, so thank you for a good one!

Great brain tease for those who are newbies on this site. What I mean is that it is not a very easy one (you have to know what to count, and the wrong answer which is given might be misleading), but not that hard either. I liked this one!

Yes, a nice way to see the flaw in the man's logic is, what would happen if the man were to throw the dice 6 times or more? As per his logic, the probability of getting a 1 would be greater than 1. Good teaser, by the way.

fyi, plural is DICE and singular is DIE. good teaser!

I like probability stuff

Since the swindler provided the die, who's to say it was a fair die in the first place? Do you think your typical swindler will be satisfied with a mere 125:91 edge?

Reply to dewtell: Indeed! Good point there!

What a way to Die

I am not sure what sparkplug meant by the wrong answer is given. The probability of winning is certainly correct at 91/216. However, the odds are 2:1 for the bet so the return on investment is 182/216 or 84%. The game is not quite fair. Nice puzzle.

The die must be loaded. A 91:125 ratio should only have a return of 125:91 in the same way that a 1:5 ratio should have a 5:1 payoff or that a 3:3 ratio should have a 50:50 payoff. In this case the Swindler is giving a 200:100 payoff. Unless the die is Loaded , the deal is NOT FAIR to the SWINDLER.

Indeed the JOKE is on the Swindler. This works only if the given payoff is 1 to 1 instead of 2 is to 1.

I have to admit, I didn't get this one. But I almost did! I won't give up!

very straightforward teaser, simple.

people with strong prob fundas should have a cakewalk.

people with strong prob fundas should have a cakewalk.

To Jimbo, Scramble and Griphook, the payoff in the teaser is, indeed, 1:1. The man has to place a bet and gets back twice the amount of his bet. In effect, his gain, if he wins, would be equal to the bet placed. The swindler's gain would also be the same amount if he won. So the teaser is correct!

you are right there kris if the payoff is 1:1. twice the amount of the bet in my opinion suggests a 2:1 payoff, but i might have read it wrong.

To put this matter to rest, the probability of winning on a given trial is approximately .57 in favor of the swindler. However, due to the payoff structure, if the swindler kept making bets he should expect a net loss of 29 cents multiplied by the number of trials.

sorry, Eigenvector, but krishnan is correct. Look at it this way: Both the man and the swindler put down 1$, making the 'pot' total 2$. Go back and read the teaser again, and both statements hold true; The man (who put down 1$) will get back twice the money (the 2$ pot) if he wins, and the swindler will get the mans money (plus his own money) if he wins. Either way, the swindler either gains a dollar or loses a dollar; never two.

This teaser is incorrect. Probability is added, not multiplied.

If it is true that the odds of not rolling a one on the die is 5/6 x 5/6 x 5/6, then based on the same logic, would the odds of rolling a one be 1/6 x 1/6 x 1/6?

Dice and odds are what I do for a living. The odds are added as such. 1/6 + 1/6 + 1/6.

What are the odds of getting a rolling a one using 6 rolls? The odds are naturally 1 out of 6 rolls will be a one.

Based on the authors theory and making it 6 rolls instead of 3 rolls, 5/6 x 5/6 x 5/6 x 5/6 x 5/6 x 5/6 = 15625/46656 or 33.48% that wont roll a one. Obviously the authors formula for calculating the odds is incorrect.

Taken one step further. Lets say its a coin flip. Lets use the authors means of calculating the odds but instead of a die, we will use a coin.

Odds it wont come up heads in 3 tosses, 1/2 x 1/2 x 1/2 = 1/6. If you add the chances that it can come up heads, you get the same result. Therefore, incorrect math.

Odds of rolling a one in three rolls IS 50/50. That my friends, is a fact.

If it is true that the odds of not rolling a one on the die is 5/6 x 5/6 x 5/6, then based on the same logic, would the odds of rolling a one be 1/6 x 1/6 x 1/6?

Dice and odds are what I do for a living. The odds are added as such. 1/6 + 1/6 + 1/6.

What are the odds of getting a rolling a one using 6 rolls? The odds are naturally 1 out of 6 rolls will be a one.

Based on the authors theory and making it 6 rolls instead of 3 rolls, 5/6 x 5/6 x 5/6 x 5/6 x 5/6 x 5/6 = 15625/46656 or 33.48% that wont roll a one. Obviously the authors formula for calculating the odds is incorrect.

Taken one step further. Lets say its a coin flip. Lets use the authors means of calculating the odds but instead of a die, we will use a coin.

Odds it wont come up heads in 3 tosses, 1/2 x 1/2 x 1/2 = 1/6. If you add the chances that it can come up heads, you get the same result. Therefore, incorrect math.

Odds of rolling a one in three rolls IS 50/50. That my friends, is a fact.

xpitxbullx: First, I didn't understand how two things coming out the same means one is incorrect. Second, 1/2*1/2*1/2 is 1/8, and 1/2+1/2+1/2=3/2. Neither one is 1/6, and they are not equal. Third, probability is indeed multiplied. If you roll 7 dice, then by your logic the odds of getting a 1 should be 1/6+1/6+1/6+1/6+1/6+1/6+1/6=7/6. The range of probabilities is from 1 to 0. 7/6 is not in that range. Fourth, if the odds of rolling no 1 in rolling 3 dice is (5/6)^3 (and it is), then the odds of rolling a 1 in three dice is 1-((5/6)^3), which is not (1/6)^3. That is all for now.

Forgive me. I was kind of tired when I made that comment and it was a little in error. When you take the probablility of one outcome and then do it again and again, it is added, not multiplied. The way he words it is incorrect leading to an erroneous answer. Instead of caluculating the one not coming up, calculate the odds that it does. Then tell me what answer you get.

Also, range of probability does not lie between 1 and 0. If that were true, and I flipped a coin a thousand times, what is the probablility for the amount of times I can flip heads? That doesn't lie between 1 and 0. That lies at 500. Don't you agree?

At first blush I was seeing things exactly like xpitxbullx, but eventually proved to myself that the chances of winning the bet is truly 91 in 216. If you list all 216 of the three-die combinations (111, 112, 113, etc, up to 666), you will find that there are exactly 91 of the combinations that include at least one 1.

If the game can end prematurely by rolling a 1 on the first or second roll, how many possible outcomes are there that result in a game ending conclusion? I'll tell you: 156 possible game ending combinations (considering that you don't continue rolling if you roll a 1 on the first or second roll). Now, there are 125 losing combinations and 31 winning combinations. You cant roll 1-1-1 because the game ended after the first roll of 1 with no need to try to roll a second or third time (the wager was already won and the game finished). If you look at possible game ending conclusion, there are 156 total game ending possibilities, not 216. 125 combinations that lose, 31 combinations that win. I'M TELLING YOU THIS TO POINT SOMETHING OUT. Just because there are 125 possible losing outcomes and 31 possible winning outcomes (or stick with the 91 winning outcomes if it makes you happy) THE ODDS OF REACHING ONE OF THE 125 LOSING COMBINATIONS IS NOT 125/216 BECAUSE THE GAME CAN END PREMATURELY WITH A ROLL OF 1 ON THE FIRST OR SECOND ROLL. Its not in how many losing combinations there are, but the odds of reaching one of those losing combinations. So to sum it all up...The game has a 1/6 chance of ending with a win on the first roll. If the first roll doesn't roll a 1, then the game has a 1/6 chance of winning on the second roll. If the second roll doesn't roll a 1 then the game has a 1/6 chance again of winning. For this teaser to be a break even chance for the player, he would have to put a dollar in the pot and the swindler put 2 dollars in the pot.

Because the game can end prematurely on the first or second roll by rolling a one, you cannot say there are 216 possibilities. Catch my drift? The math everyone is using is wrong.

(sorry to keep commenting but I just want to be clear). Each conclusion has different odds of being achieved. There is a 1 in 216 possibility of ending the game rolling 6-6-1. There is a 1 in 6 chance of rolling 1-x-x. (X = no need to roll again because the game is over). Each outcome has different probabilities. OK, I'll shut up now.

I suppose it's all in the way you interpret the original post. I read that the swindler is proposing an even money bet. The sucker bets $1 and if he wins, he gets back twice the amount of his bet, or $2. This is a profit of $1 for $1 wagered, for an even money bet. If in fact the swindler is laying 2-1, then it's a bad bet for the swindler, and a good bet for the sucker.

xpitxbullx: There's only a 5 in 6 chance of getting to the second roll. So the odds of getting it on the second roll provided you make it there are 1 in 6, true, but if you say you may not have made it there, the odds are actually 1/6*5/6 (you have a 5 in 6 chance of getting there and a 1 in 6 chance of rolling the right number). Similarly, the odds of getting it on the third roll are 1/6*5/6*5/6. If you add those three true answers up, you get 91/216, which is what the answer says it is. Any questions?

True there is a five in six chance of getting to the second roll. But that second roll has a 1/6 chance of rolling a 1, also. Probability doesn't change from one roll to the next unless you think the die remembers what was rolled last.

It apparent that you wont take my word for it so please consult someone else of mathamatical authority if you want to be enlightened. Perhaps your local COLLEGE math professor. (Even though I learned this in 6th grade first).

If you are going to use multiplication to calculate probability, then you need to use the multiplicative rule and not the rule you just made up.

http://www.netnam.vn/unescocourse/statistics/46.htm

http://www.netnam.vn/unescocourse/statistics/46.htm

Contrary to what you may have thought, xpitxbullx, I did not calculate the odds of rolling a 1 on the second roll. I calculated the odds of rolling a number that is not a 1 on the first roll and then rolling a 1 on the second roll. Same with the third; I calculated the odds of missing on the first two and hitting on the third. If I was to just calculate them separately, then yes, they would be 1/6 each.

That is where you went wrong. You aren't supposed to calculate the odds of the 1 showing up on the second roll or the third. Each roll is its own seperate event.

If you were to calculate them seperately coming up with 1/6 for each roll, then you are on the right track. But you said you don't want to do it that way and that is where you go wrong.

I created a program to simulate Swindlers Dice to prove I was right. After running my program, I was proven wrong. the AVERAGE win/lose ratio of my program was 91 wins and 125 loses per 216 games. My logic was wrong and proven wrong before my very eyes by my own creation over and over. I understood the math used by the author but I didn't agree with it until now. I apologize for my stubborness and bull-headed attitude. This is a great teaser in that it tested my logic and I now agree with the formula. Thanks for putting up with me.

I want to say to Poker, thanks for putting up with me and keeping firm to your arguement. It helped me in the end to see the truth.

The truth will set you free! Kudos for your persistance.

This reminds me of the way infantry fight on Axis and Allies using turns. I had forgotten to apply the probability rule when I did a calculation recently with my friend jiggafet. yeah.. I'm a complete geek. Thanks guys!

yay, i got it right..

it is the same as you roll 3 die at the same time. if one appear at least once then you win. the probability one appear once is 75/216. one appear twice is 15/216. and one appear three times is 1/216. if we add them, they will be 91/216 which is lower than 125/216 (1-91/126 -> one didn't appear)

so it is not a fair game. not really hard but good teaser i really like probability things.

it is the same as you roll 3 die at the same time. if one appear at least once then you win. the probability one appear once is 75/216. one appear twice is 15/216. and one appear three times is 1/216. if we add them, they will be 91/216 which is lower than 125/216 (1-91/126 -> one didn't appear)

so it is not a fair game. not really hard but good teaser i really like probability things.

5/6 x 5/6 x 5/6 is the chance of rolling something other than 1, three times in a row, not the chance of losing the stated game.

According to your teaser, the man can win on roll 1, 2, and 3. In this case, probability DOES add. Statistically, if you were to roll a die six times, you would always get the number you are looking for on one of those rolls (of course, that does not always happen in reality, but that is the nature of probability).

So 1/6 + 1/6 + 1/6 = 3/6 = 1/2 is the probability of rolling a 1 over the course of three rolls.

Remember in general probability ADDS when you are talking about "or" and MULTIPLIES when you are talking about "and."

The probability of flipping a coin twice and getting heads on flip 1 OR 2 is 1/2 + 1/2 = 1

The probability of flipping a coin twice and getting heads on flip 1 AND flip 2 is 1/2 x 1/2 = 1/4

According to your teaser, the man can win on roll 1, 2, and 3. In this case, probability DOES add. Statistically, if you were to roll a die six times, you would always get the number you are looking for on one of those rolls (of course, that does not always happen in reality, but that is the nature of probability).

So 1/6 + 1/6 + 1/6 = 3/6 = 1/2 is the probability of rolling a 1 over the course of three rolls.

Remember in general probability ADDS when you are talking about "or" and MULTIPLIES when you are talking about "and."

The probability of flipping a coin twice and getting heads on flip 1 OR 2 is 1/2 + 1/2 = 1

The probability of flipping a coin twice and getting heads on flip 1 AND flip 2 is 1/2 x 1/2 = 1/4

CORRECTION FOR ABOVE COMMENT.

I jumped the gun, the answer to the teaser is correct. My explanation contradicts itself. The chance of losing three times in a row is, of course, the swindlers chance of winning.

Maybe I'll go back and read some textbooks or something.

I jumped the gun, the answer to the teaser is correct. My explanation contradicts itself. The chance of losing three times in a row is, of course, the swindlers chance of winning.

Maybe I'll go back and read some textbooks or something.

I always thought that probability is always between 0 and 1, ie, 0% and 100%.

PatH wrote that the "The probability of flipping a coin twice and getting heads on flip 1 OR 2 is 1/2 + 1/2 = 1". This means it is a certainty to get a head... Obviously not true.....

If the probability of an event occurring is 100%, it means it is a certainty...

Please inform me if I am wrong...

PatH wrote that the "The probability of flipping a coin twice and getting heads on flip 1 OR 2 is 1/2 + 1/2 = 1". This means it is a certainty to get a head... Obviously not true.....

If the probability of an event occurring is 100%, it means it is a certainty...

Please inform me if I am wrong...

Mar 12, 2007

The fact that it stated that the guy was a swindler gave it away.

If the way you calculate the swindler's possibility is by cubing 5/6, doesn't the cubing of 1/6 make sense? And if so, the probability of the man's winning now would be 1/216. Is this really the right way?

Pojeur, this also threw me at first but you have to realize exactly what event or events you are calculating for. 1/6 is the probability for rolling a one with a single die. if you cube 1/6 you get the probability of rolling three ones (a 1,1,1) which is indeed 1/216. But thats not what we're after. We want the probability of ANY of the three rolls coming up 1 and the easiest way to do that is precisely the way the author shows in the solution: By cubing the probability of NOT getting a one (5/6 cubed =125/216) and subtracting that value from 1 or 216/216 to get 91/216.

Im wondering why no one has adressed the apparent paradox of this teaser. In other words, why is this so when intuitively it appears to be a fair game. Well I will give you an intuitive explanation now:

Suppose we roll all 3 dice at once. If you roll 3 dice at once 1000 times and count the total number of ones that occur you find that it will be around 500, which seems to support the claim that the player would win 50% of the time. But this is misleading. Sometimes you will roll 2 ones and sometimes 3 ones in a single 3 dice roll. The way the game is described, these events result in just a single win for the player with no added bonus for multiple ones rolled (in the game if the player rolls a one on the first try he simply wins and they play another "hand") This is where the swindler gains his advantage. He is offering no benefit or bonus for the possiblility that the player will on occasion roll a one on his first and second rolls; his first and third rolls; his second and third rolls; or his first, second and third rolls, because according to the rules there ARE NO additional rolls after a one (a player win) is achieved.

This game would be fair (exactly even odds for player and house) if the swindler allowed all 3 rolls regardless of whether a one has already shown and then payed 3$ for 2 ones and 5$ for 3 ones rolled. 2 to 1 and 4 to 1 odds respectively on those "hands". There is actually a teaser on this site that contains this variation of the game.

So there you have the intuituive explanation for those who understand the probability calculations but still cant get around the intuitive paradox of the teaser.

Suppose we roll all 3 dice at once. If you roll 3 dice at once 1000 times and count the total number of ones that occur you find that it will be around 500, which seems to support the claim that the player would win 50% of the time. But this is misleading. Sometimes you will roll 2 ones and sometimes 3 ones in a single 3 dice roll. The way the game is described, these events result in just a single win for the player with no added bonus for multiple ones rolled (in the game if the player rolls a one on the first try he simply wins and they play another "hand") This is where the swindler gains his advantage. He is offering no benefit or bonus for the possiblility that the player will on occasion roll a one on his first and second rolls; his first and third rolls; his second and third rolls; or his first, second and third rolls, because according to the rules there ARE NO additional rolls after a one (a player win) is achieved.

This game would be fair (exactly even odds for player and house) if the swindler allowed all 3 rolls regardless of whether a one has already shown and then payed 3$ for 2 ones and 5$ for 3 ones rolled. 2 to 1 and 4 to 1 odds respectively on those "hands". There is actually a teaser on this site that contains this variation of the game.

So there you have the intuituive explanation for those who understand the probability calculations but still cant get around the intuitive paradox of the teaser.

Another important thing to note is that probabilities add only if the events are disjoint. Here, the three events {rolling a 1 on first die, rolling a 1 on 2nd die, rolling a 1 on 3rd die} are NOT disjoint. You can certainly roll a 1 on the first die and the second die. Thus, if you want to really want to calculate P(roll a 1) instead of 1-P(not roll a 1), you will get the following:

Let A, B, C denote event that you roll a 1 on first die, 2nd die, and 3rd die respectively.

P(roll a 1) = P(A or B or C) = P(A) + P(B) + P(C) - P(AB) - P(BC) - P(AC) + P(ABC) = 3 * 1/6 - 3 * (1/6)^2 + (1/6)^3

The above formula is a fundamental one that anyone who has taken a class on probability should know. Some of us learn with the misconception that P(ABC) = P(A) + P(B) + P(C) all the time, but this is clearly not true. It is only true for disjoint events because P(AB), P(BC), P(AC), and P(ABC) are all equal to 0!

Let A, B, C denote event that you roll a 1 on first die, 2nd die, and 3rd die respectively.

P(roll a 1) = P(A or B or C) = P(A) + P(B) + P(C) - P(AB) - P(BC) - P(AC) + P(ABC) = 3 * 1/6 - 3 * (1/6)^2 + (1/6)^3

The above formula is a fundamental one that anyone who has taken a class on probability should know. Some of us learn with the misconception that P(ABC) = P(A) + P(B) + P(C) all the time, but this is clearly not true. It is only true for disjoint events because P(AB), P(BC), P(AC), and P(ABC) are all equal to 0!

Feb 18, 2010

The Honest man wins as soon as he gets a 1 so the chances of him getting a one are (1)+(5_1)+(5_5_1)=31 all the others are irrelevent because (1_1_x) is the same as (1_x_x) as the first time he gets 1 he wins .The question is should we consider the remaining throws as soon as he hits 1 the first time or the second time.

if he dosent consider then thw swindler winning is (5_5_5)=125 and all the permutations are 125+31=156

hence the honest man winning is 31/156 and that of swindler is 125/156

Have tried my best to express my views. Please correct me if i am wrong.

if he dosent consider then thw swindler winning is (5_5_5)=125 and all the permutations are 125+31=156

hence the honest man winning is 31/156 and that of swindler is 125/156

Have tried my best to express my views. Please correct me if i am wrong.

@vishurammohan

31/156 assumes each of the outcomes in the sample space are equally likely, which is not true. For example, the probability he gets the outcome (1) is not the same as the probability he gets the outcome (2, 1). The former is 1/6 while the latter is 5/6 * 1/6 = 5/36.

31/156 assumes each of the outcomes in the sample space are equally likely, which is not true. For example, the probability he gets the outcome (1) is not the same as the probability he gets the outcome (2, 1). The former is 1/6 while the latter is 5/6 * 1/6 = 5/36.

The correct way to do it if we consider the guy stops after a one is rolled is the following:

1/6 + 5/6(1/6) + 5/6(5/6)(1/6).

Note this is the same as 91/216 = 1 - (5/6)^3

1/6 + 5/6(1/6) + 5/6(5/6)(1/6).

Note this is the same as 91/216 = 1 - (5/6)^3

Mar 23, 2011

The answer is incorrect. The expected gains for the gambler (not swindler) is 57/216, iff the swindler does indeed pay back 2 dollars for the initial dollar of the gambler. Also, some people here claiming to work with dice are insanely incorrect in what they are saying about probability, which is disappointing.

## Follow Braingle!