The Game of Death (Part I)
Five undercover agents from an elite special force unit are caught by a criminal organization. The gangster leader decides to execute them by playing a deadly game. Anyone who can survive from this game will be released without a single wound. These are the details of the game:
- All five agents will be lined up in random order.
- Each agent can see the agents lined up in front of him, but cannot see any of the agents lined up behind him.
- After they line up, the gangster members will spray airbrush paint on each agent's back with blue or red color. No agent can see the color on their back.
- Then one of the gangster members will ask each agent for the color of paint on his back, starting from the most behind one.
- The agent who is asked can only answer one word: "BLUE" or "RED", otherwise he will be killed.
- If the agent answers incorrectly, the gangster will kill him.
- All agents must be silent during the game. If not, they will all be killed.
The agents are given an opportunity to gather for a day before the game is started. After long discussion, one of the agents, the smartest one, has the best strategy to survive from the game. The strategy will yield the greatest number of agents that can be saved. Assume that, no matter what happens, all the agents will follow the strategy and since they are from the elite special force unit, it is reasonable to assume that they are all physically and mentally healthy so no one will ruin the plan. If the expected rate of survival using the best strategy can be expressed as q p , then what is p + q ?
Next step : The Game of Death (Part II)
The answer is 19.
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7 solutions
Actually there is an easy way to know the color on each agent's back without counting or hard thinking. Suppose that the strategy is as following:
If the agent immediately answers when he is asked, it means the color on agent's back in front of him is "RED" and if he answers a bit longer (5 or 6 seconds) it means the color on agent's back in front of him is "BLUE" regardless his answer "RED" or "BLUE". For instance, when the agent in most behind line up is asked, "what color on your back?" and he immediately answers "BLUE", then it means the color on agent's back in front of him is "RED" or vice versa. By using this strategy, the agents do not need to count anything, they don't care whether it's odd or even. They only need to pay attention when agent behind them is asked.
# Q . E . D . #
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This is NOT allowed in this type of problems.
The problem only says "agent in front of him" not agents. So counting number of reds might not be possible.
Feeling stupid.... I worked all of it out, but thought that the add sign was a divide one. So I got 0.9, which I couldn't input. Meh.
Isn't 4+.5=9/2?
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Yeah.....but the probability is 4+0.5 out of 5, which is 9/10......
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It says expected rate of survival, and the expected number of people to survive is 9 / 2 ...
It never said probability.
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@Baby Googa – By expected rate he meant expected rate for each person
I thought that too! I put 11 because that's how many I thought would survive! AAGGG!
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Me too! I almost got a level 5 combinatorics? AAGGG!
(4+0.5)/5=9/10
Can this please be clarified in the wording of the problem? That by rate of survival it is referring to probability?
There is still a possibility that the gangster members could have trick them by making them all have the same color or at least 3 agents are seen with the same color by the last person which also has the same color
You're so clever 😂
Instead of this solution, the person can directly say the color on the back of the agent standing in front of him. Due to this all 4 agents will be 100% safe. And there is 50% probability that the 5th agent is also safe. So 4+0.5/5=9/10...is p+q=19
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Your methodology is wrong in this case. The first agent would say the color in front of him. This would give him a 50 50 chance of having the correct color for himself. The next agent knowing the color on his back could say that color, however, this would leave the third agent with only a 50 percent chance of having the correct color on his back. In other words if the gangsters alternated colors on every agent, and the agents only said the colors on the back in front of them the only the very last agent would be safe.
Actually there is an easy way to know the color on each agent's back without counting or hard thinking. Suppose that the strategy is as following:
If the agent immediately answers when he is asked, it means the color on agent's back in front of him is "RED" and if he answers a bit longer (5 or 6 seconds) it means the color on agent's back in front of him is "BLUE" regardless his answer "RED" or "BLUE". For instance, when the agent in most behind line up is asked, "what color on your back?" and he immediately answers "BLUE", then it means the color on agent's back in front of him is "RED" or vice versa. By using this strategy, the agents do not need to count anything, they don't care whether it's odd or even. They only need to pay attention when agent behind them is asked. Unfortunately, the first agent who is asked can only guess the answer and must sacrifice for the other agents.
Could you post the solution to this problem...thanx in advance...
Please post the solution.
i got 17 as: The first guy will tell the number on the guy in front. His chance of survival is 1/2 and the agent in front is 1. Now the third agent is again 1/2, next is 1 and the last is also 1/2. total probability is 2+3/2/5=7/10
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That was also my first though Siddhant but then I realized there was a way to tell the color of agent in front of you so 4,5 agents can survive. It is a really simple way although the agent can only tell "BLUE" or "RED". Your thinking is on the right track, he must guess but he can 'tell' the color of agent in front of him although his guess completely different with the color in front of him. For example, he can say RED but the agent in front of him know that his color on his back is BLUE.
For Milly and Tanya, please be patient. I'll tell the answer but not now. Just keep following this problem, maybe then you will know the answer.
My only issue is that you never make it clear that the guy in the back can see ALL of the backs in front of him. The question seems to say that you can only see the back of the guy directly in front of him. Thus, 17 should be the correct answer, guys in odd positions tell guy in front of them the correct answer and just hope that they happen to have the same color.
I understand why 19 is correct, but the wording of the question is unclear. Also, might be better to mention that the fraction is reduced integers to avoid people trying to use decimals in the fraction: 4.5+5 = 9.5
The first guy will guess, and the rest can live: First guy dies: 4/5 First guy lives: 5/5
Total 9/10 ->19
Could you tell us how can the rest survive? Because that is the main point of this problem. BTW, you might want to try The Game of Death (Part II) .
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The rest can survive because the strategy is for the "i"th agent to tell the color on the back of "i+1"th person
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How come?? Remember, each agent can only answer with one word, either "BLUE" or "RED". So, if he tells the color on the back of agent in front of him he can get killed.
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@Tunk-Fey Ariawan – answer should be 17
@Tunk-Fey Ariawan – He can get killed, but he has 50% chance of surviving. Hence the answer. I don't know wtf this comment was for.
Strategy: the 1st person will count the no. Of red and blue marks in front of him. Now there are 3 possibilities: (2,2)(1,3)(4,0). He will shout red if the no. Of red's in front of him is even,else if no. Of red's is odd,he will shout blue. Case 1: he shouts red. So either all 4 in front are marked red or 2 red 2 blue. Thenext person will see the no. Of red in front of him. If red's are odd numbered then he is must shout red,else blue. And so n each person should keep track of the count. Case 2: he shouts blue. So 4 in front are marked 1 red 3 blue or 1 blue 3 red. Now similarly as above, each person should keep track of the count and so can predict their color!!!!
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Pawan, there are lots of flaws in your strategy. First of all, how come the agent in most behind one can be so sure the color on agent's back 2, 3, or 4 people in front of him? Of course, he can make sure 100% the color on agent's back in front of him, but the rest?? Even though there are 3 possibilities of number of red colored, but in overall there are 16 ( 2 4 ) different combinations of color. Beside that, with your plan the agent in most behind is really unlucky because he can only say the color on agent's back in front of him. How if he want to say RED but the color on agent's back in front of him BLUE? Your idea is also really complicated considering the huge tense during the game.
There is a simple strategy. It is a really simple plan although the agent can only say "BLUE" or "RED". HINT: the most behind agent must guess but he can 'tell' the color of agent in front of him although his guess completely different with the color in front of him. For example, he can say RED but the agent in front of him know that his color on his back is BLUE.
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Tunk-Fey, I considered that the most behind one is able to see color on back of all in front of him. considering this it is really easy . there would be 2^3 different combinations of color. but the point is we dont have to consider all! main thing is the no. of odd or even color red. in my plan last agent does NOT have to say color on back of agent in front of him !!!and anyways last person will never knw color on his back....he HAS to guess!!!
could not get your hint.... your solution please :-)
answer should be 17
The first agent (in the very back) uses a timing system with which to delay before responding. He waits this amount of seconds, depending on the specific pattern of red and blue backs in front of him (starting with the one directly in front of him, and ending with the one at the very front):
1 seconds: RRRR
2 seconds: BRRR
3 seconds: RBRR
4 seconds: BBRR
5 seconds: RRBR
6 seconds: BRBR
7 seconds: RBBR
and so on, covering all 16 binary possibilities. Every agent knows this system, so once they receive a timing they know their color. Obviously, it's a crapshoot whether or not the first agent guesses his own color right, so it's a 50/50 shot, bumping a 4/5 ratio to a 9/10.
Make the number of 'red's that you see plus the number of 'red's that you hear odd.
The first responder has 50% chance of survival because his/her color is independent of the others.
Every next responder will survive.
PROOF
CASE I If the ith responder (i > 1) hears his/her immediate predecessor say 'blue', this predecessor must have seen+heard an odd number of 'red's. The ith responder thus hears as many 'red's as the predecessor heard and will see the same number of 'red's as his/her predecessor except possibly his/her own. If the ith responder has red, he must make the sum odd again by saying 'red', and will live; otherwise he must say blue to keep the sum odd, and will live also.
CASE II If the ith responder (i > 1) hears his/her immediate predecessor say 'red', this predecessor must have seen+heard an even number of 'red's. The ith responder thus hears one more 'red' and will see the same number of 'red's as his/her predecessor except possibly his/her own. If the ith responder has red, he must make the sum odd again by saying 'red', and will live; otherwise he must say 'blue' to keep the sum odd, and will live also.
MAXIMUM The chance for the first responder is 50%, and with the above strategy all others will live, 100% sure. A higher expected survival rate is not possible.
So: expectation is 5 2 1 + 4 .
To make this easier lets call the guy at the front who can't see anything Agent A and the guy at the back Agent E, with the people in between named in order.
So Agent E can see Agents D, C, B and A, Agent D can see C, B and A and so on.
Agent E must say RED if there is an EVEN number of reds in front of him, or BLUE if there is an ODD number of reds in front of him.
Agent D now knows how many reds and blues there are out of D, C, B and A, and since he can either see an even or odd number of reds in front of him, if it coincides with what Agent E said, then he is blue, however if Agent D sees an odd number of reds, but Agent E saw an even number of reds, Agent D knows he must be a red.
This goes on for each person, however they must remember what each person behind them said, and then realise what colour they are.
This method provides a survival rate of 1/2 for Agent E, as it makes no difference, but a survival rate of 1 for the rest. Average these you get 9/10, and 9 + 10 = 19
How if there are 100 people forced to play this game? Can they still count which one is even and odd?
The last person will count the number of red backs, if it is even, he will go with red, and if it is odd, he will go with blue. So the person in front of him will know if the even or odd has changed, then he is red. If it is not changed, he will say blue. So the third person will know whether is is odd or even. Similarly all can do this way. Thus there is 100% guarantee that 4 will live, but the 5th one has half probability, so we get 4+0.5=9/10. So our answer is 19.
A great problem, Tunk, I got screwed up at first, it was a hard job, but I got it. I have followed you, please keep posting problems like these.