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Subject: Fun math challenge (if you like math): rss

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Mike K
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I enjoy watching "Let's Make a Deal" during the summer. Here's a game I just saw for the second time (in reruns).

8 envelopes. 4 have 'X', 4 have 'O'. Object: pick 3-of-a-kind. Win a car if you do so.

So you pick your three envelope. Wayne Brady shows that one is an 'X'. Then he shows that another one has an 'X'. Then he offers you stuff to call the whole thing off. Over the course of the game, he shows you every other envelope but one: you see three 'O's and a 3rd 'X'. Finally, the moment of truth: do you go for the car, or do you take the sure thing?

So here's the question: What are the chances that you win the car?

HINT: it's not 50%
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I'll make a drive-by guess, and perhaps change my answer later after I've thought about it some more.


You have a 25% chance of winning.
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Spoiler (click to reveal)
Choosing three of a kind is really unlikely. There are only 8 ways of selecting three of a kind, out of 56 total selections. So, 1 in 7, or about 14.3%
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Assuming he knows which envelopes are which, and is picking them on purpose to lead you on, the probability of winning isn't affected by seeing all those Os and Xs. (After all, you have to draw at least a pair of one or the other.)

There are 56 ways to choose 3 out of 8 envelopes, and 8 ways of drawing three of a kind (4 each in Xs and Os). So based on the assumptions above, I'm going with a win probability of 1 in 7.


Edit: D'oh! ninja
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Not that I'm certain but here's my logic:

Spoiler (click to reveal)

1/3

You will always get 2 of a kind. The chance of picking a matching mark on your 3rd envelope will be 2 out of the 6 remaining envelopes: 1/3
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Mike K
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CrankyPants wrote:
Not that I'm certain but here's my logic:

Spoiler (click to reveal)

1/3

You will always get 2 of a kind. The chance of picking a matching mark on your 3rd envelope will be 2 out of the 6 remaining envelopes: 1/3

Spoiler (click to reveal)
Sorry, but you assume that your first two picks are the pair. The '1/7' answer IS correct.
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The answers have been making what is, I would guess never having seen tje programme, the right assumption about the problem. This makes it closely related to another well-known problem. But even more so than that one, it's absolutely critical to know the rules of the host's knowledge and behaviour in order to get the right answer. If you don't know that then you could instead go for most pessimistic answer, and that's what's been assumed.

Edit: There's another factor. The answers given have assumed knowledge by the host, which means the host knows if you've won. That allows tje possibility of the bribes being affected by that. Good news and bad news there. Bad news is more variables to understand (presumably by watching earlier shows and analysing results). Good news is more information to guide your decision. Unless the bribes are independent of your win/loss status.
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Coyotek4 wrote:
I enjoy watching "Let's Make a Deal" during the summer. Here's a game I just saw for the second time (in reruns).

8 envelopes. 4 have 'X', 4 have 'O'. Object: pick 3-of-a-kind. Win a car if you do so.

So you pick your three envelope. Wayne Brady shows that one is an 'X'. Then he shows that another one has an 'X'. Then he offers you stuff to call the whole thing off. Over the course of the game, he shows you every other envelope but one: you see three 'O's and a 3rd 'X'. Finally, the moment of truth: do you go for the car, or do you take the sure thing?

So here's the question: What are the chances that you win the car?

HINT: it's not 50%


I'm confused. What do you mean that he 'shows' them to you?

After picking three envelopes you've got two 'x' envelopes in hand, right?

When does the next envelope get revealed that you picked, and what is it?

And the other envelopes he 'shows', are those permanently placed so it simply becomes a memory game at that point? Or are they permanently revealed?

I first understood this to mean that all but two envelopes were left to be revealed. You'd picked three of each, right?

In that case, with two envelopes remaining, you should know that one is an 'x' and one is an 'o', and it would be a 50-50 proposition, right?

What am I not understanding? Do the envelopes get re-hidden? Do they get shifted around?
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diehard4life wrote:
Coyotek4 wrote:
I enjoy watching "Let's Make a Deal" during the summer. Here's a game I just saw for the second time (in reruns).

8 envelopes. 4 have 'X', 4 have 'O'. Object: pick 3-of-a-kind. Win a car if you do so.

So you pick your three envelope. Wayne Brady shows that one is an 'X'. Then he shows that another one has an 'X'. Then he offers you stuff to call the whole thing off. Over the course of the game, he shows you every other envelope but one: you see three 'O's and a 3rd 'X'. Finally, the moment of truth: do you go for the car, or do you take the sure thing?

So here's the question: What are the chances that you win the car?

HINT: it's not 50%


I'm confused. What do you mean that he 'shows' them to you?

After picking three envelopes you've got two 'x' envelopes in hand, right?

When does the next envelope get revealed that you picked, and what is it?

And the other envelopes he 'shows', are those permanently placed so it simply becomes a memory game at that point? Or are they permanently revealed?

I first understood this to mean that all but two envelopes were left to be revealed. You'd picked three of each, right?

In that case, with two envelopes remaining, you should know that one is an 'x' and one is an 'o', and it would be a 50-50 proposition, right?

What am I not understanding? Do the envelopes get re-hidden? Do they get shifted around?

You pick three envelopes. He shows all but one of yours and all but one of the other five. He rigs it so that the two of yours are a pair (in this example, Xs), and that the two still unknown envelopes are an X and an O. So we now have:

(Yours) X X ?
(Remaining) O O O X ?

The important thing is that if he chose those particular six envelopes to reveal on purpose, the probability that you win does not change compared to seeing none of the envelopes revealed. If you drew two Os and one X, he would reveal your two Os, plus three Xs and one more O from the other five envelopes.

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Brian Bankler
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Spoilers, if I can do math.
Spoiler (click to reveal)

Let's assume that "X" is dominant. You picked two or three "X"s. (If you picked O's, symmetry applies. I'm assuming the host knows (or is told via Mic) what is in each envelope, so that no information is available from the reveals)

Call the envelopes X1, X2, X3, X4, 01, 02, 03, 04)
Ways you can pick 3 Xs == 4 ( X123, 124, 134, 234)
Ways you can pick 2 Xs == 6 (12, 13, 14, 23, 24)
Ways you can pick 1 0 == 4 (obvious)

So the Total combinations of XX0 are 6 * 4 = 24.

So there are 24 "Lose" and 4 "Win" cases, so 6:1. So you are a winner one time in seven.


Like most things in probability (that I understand), if you can correctly lay out all possible combinations the answer isn't far behind.

Of course, this all depends on my assumptions about what the host knows.
His revealing all (but one) of the other envelopes doesn't change your odds of winning. Of having won, I should say. Information (and the assumptions about how it was revealed) makes things tricky.

I speak as an engineer ("Someone who doesn't actually know math or care much, but just tries to get the right answer.") So I've probably used math terms incorrectly.
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Slow Motion Walter
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Spoiler (click to reveal)
After you pick two X's, what remains are two X's and four O's. So it seems like there should be a 2/6 chance that you win.


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Even though I got the math wrong—♪ lit major ♫—I got it right enough to understand you don't want to go for the car. The moment you win the car you're into the IRS for at least 10 large, with the following possible outcomes:

- win half a car & a big-ass tax bill

- win a whole car & problems with the government

- win a whole car that you can only sell for half a car & an epic tax-time headache

The obvious answer, sans math, is to declare that you banged the host's mother, light something on fire and make your escape in the confusion.
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Sean Ahern
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Spoiler (click to reveal)
My reasoning is that given what we know, there's four possible picks:

XXX
XX0
X0X
0XX

Everything else is irrelevant, so I'm going with Leezar's drive by guess of 25%.
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The 2/6 seems logical, but that would only be the case if the two you were shown were picked randomly, and they weren't.

Imagine there are 999,999 X's and one O, and the way you win is if you manage to pick all 999,999 X's. Pretty slim chance, right? Now you are shown that 999,998 of those you picked were indeed X's. There's one X and one O left, but the chance you correctly picked all 999,999 X's doesn't suddenly become 1 in 2.
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Mark Delano
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Spoiler (click to reveal)
There are two possible answers here, depending on how the host picked your revealed "X"s. If he picked the first two envelopes that you picked to reveal, you have already corrrectly picked two Xs at that point, which is normally only a 3/14 chance. However since you already were lucky enough for that to happen that probability doesn't enter into the chance of the last envelope being an "X". At the point when you picked the last envelope there were 4 "O"s and 2 "X"s, which gives you a 1/3 chance of picking another "X".

If the host revealed any two selected envelopes that were "X"s then the probabilities are different. Then your pre-reveal chance of getting at least two "X"s is 50% (XXX, XXO, XOX, OXX) (1/14, 2/14, 2/14, 2/14 respectively). Unsurprisingly the odds of at least two "O"s is also 50% (OOO, OOX, OXO, XOO) (same probs as above), which means with the game in this form the host can always reveal two that are the same. The chances of winning by drawing all 3 "X"s at the start of the game was only 1/14. Now with 2 "X"s revealed I know that it has to be one of the 4 double "X" results above, of which there is only a 1/7 chance that it's the triple "X".

In either case what he does with the other unselected envelopes doesn't matter unless he reveals 4 unpicked "O"s or the last two unpicked "X"s.
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Sean Ahern
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Spoiler (click to reveal)
Thinking more on it: possible ways of ordering XXXX0000: 8!/4!4! = 70. How many of those 70 start with XXX:

"XXX"X0000, so how many ways to arrange X0000? 5!/4! = 5 ways to win.

But we can eliminate some of those losing combinations, all the ones that start with "00X", "0X0", "X00", and "000", so thats

"XXX00" rearranged times 3, so 5!/3!2! * 3 or 30.
"XXXX0" rearranged, so 5!/4! or 5.

So we know that 35 combinations are not possible, so the odds are now 5/35 or 1/7.

I'd like to change my answer to 1/7!
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diehard4life wrote:
Coyotek4 wrote:
I enjoy watching "Let's Make a Deal" during the summer. Here's a game I just saw for the second time (in reruns).

8 envelopes. 4 have 'X', 4 have 'O'. Object: pick 3-of-a-kind. Win a car if you do so.

So you pick your three envelope. Wayne Brady shows that one is an 'X'. Then he shows that another one has an 'X'. Then he offers you stuff to call the whole thing off. Over the course of the game, he shows you every other envelope but one: you see three 'O's and a 3rd 'X'. Finally, the moment of truth: do you go for the car, or do you take the sure thing?

So here's the question: What are the chances that you win the car?

HINT: it's not 50%


I'm confused. What do you mean that he 'shows' them to you?

After picking three envelopes you've got two 'x' envelopes in hand, right?

When does the next envelope get revealed that you picked, and what is it?

And the other envelopes he 'shows', are those permanently placed so it simply becomes a memory game at that point? Or are they permanently revealed?

I first understood this to mean that all but two envelopes were left to be revealed. You'd picked three of each, right?

In that case, with two envelopes remaining, you should know that one is an 'x' and one is an 'o', and it would be a 50-50 proposition, right?

What am I not understanding? Do the envelopes get re-hidden? Do they get shifted around?


What you're missing is this: The Monty Hall Problem. Although it seems, intuitively like there is a 50-50 chance, that is not, in fact, the case.
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I have not got the time to work this out right now, but you need to be using Bayes Theorem,

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Can I switch to the other envelope? Then I've got much higher chances of winning.
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purplewurple wrote:
Spoiler (click to reveal)
The 2/6 seems logical, but that would only be the case if the two you were shown were picked randomly, and they weren't.

Imagine there are 999,999 X's and one O, and the way you win is if you manage to pick all 999,999 X's. Pretty slim chance, right? Now you are shown that 999,998 of those you picked were indeed X's. There's one X and one O left, but the chance you correctly picked all 999,999 X's doesn't suddenly become 1 in 2.



This makes sense, but why do you state that the two revealed envelopes were not picked randomly? It reads to me like they were.
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Brian Bankler
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jaa17 wrote:
I have not got the time to work this out right now, but you need to be using Bayes Theorem,



Only if you assume new information is being added to the system by the host's revelations. If he already knew everything, the the fact that he could reveal all the other envelopes (but one) doesn't change anything.
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Bankler wrote:
jaa17 wrote:
I have not got the time to work this out right now, but you need to be using Bayes Theorem,



Only if you assume new information is being added to the system by the host's revelations. If he already knew everything, the the fact that he could reveal all the other envelopes (but one) doesn't change anything.


I don't think that's true. The Monty Hall problem showed that Bayes works in a case like this.

Spoiler (click to reveal)
The probability of 3 x's given you have at least 2 x's is:
P(3 x's|2+ x's) =
P(2+ x's|3 x's) * P(3 x's) / P(2+ x's) =
1 * (4/8 * 3/7 * 2/6) / (4/8 * 3/7 * 2/6 + 3 * (4/8 * 3/7 * 4/6)) =
1/7

There are three ways of getting 2 x's that's why the second part of the denominator is multiplied by 3.
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Brian Bankler
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greyareabeyond wrote:
Bankler wrote:
jaa17 wrote:
I have not got the time to work this out right now, but you need to be using Bayes Theorem,



Only if you assume new information is being added to the system by the host's revelations. If he already knew everything, the the fact that he could reveal all the other envelopes (but one) doesn't change anything.


I don't think that's true. The Monty Hall problem showed that Bayes works in a case like this.

You don't NEED to use bayes theorem to solve the monty hall problem, which is what the quote said. But sure, you can use it.
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Spoiler (click to reveal)
It is really pretty simple:

There are 8!/(3!5!) = 56 ways to draw three envelopes from 8.
There are 4 ways to draw 3 X's.
There are 4 ways to draw 3 O's.
This means that there is an 8/56 = 1/7 chance that you originally drew three of the same envelopes. Opening any envelopes does not change that chance (it might change the knowledge, but the host is of course careful not to provide any extra knowledge).
So 1/7 is the correct answer.
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