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you are in an orgy

Question

You are in an orgy, involving 4 men and 5 women. Every member of the opposite sex will have sexual relations (e.g. heterosexual contact only). Someone in your group of 9 has an std. This Std is of the nature that on contact of the genitals it is spread immediately meaning that a person who just came in contact with it can now spread it to others.

Assume you are a woman like me, if you only have one form or means of protection from a single encounter (I.e. that encounter is safe).

When should it be used?

What is the probability of you getting an std?

If you are the only one with protection, what percentage of the group would most likely have an std after the orgy?

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16 answers to this question

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The plot for all cases:

Spoiler

 

Round 1: M1W1 (M1W1) M2W2 (    ) M3W3 (    ) M4W4 (    )
Round 2: M1W2 (M1w2) M2W3 (    ) M3W4 (    ) M4W5 (    )
Round 3: M1W3 (M1w3) M2W4 (    ) M3W5 (    ) M4W1 (m4W1)
Round 4: M1W4 (M1w4) M2W5 (    ) M3W1 (m3W1) M4W2 (****)
Round 5: M1W5 (M1w5) M2W1 (m2W1) M3W2 (****) M4W3 (****)

W1: protection in the 1st round (supposing it is M1 who is infected)
W2: cannot escape
W3: cannot escape
W4: last round (was wrong in the previous post)
W5: last round

Round 1: M1W1 (    ) M2W2 (M2W2) M3W3 (    ) M4W4 (    )
Round 2: M1W2 (m1W2) M2W3 (M2w3) M3W4 (    ) M4W5 (    )
Round 3: M1W3 (****) M2W4 (M2w4) M3W5 (    ) M4W1 (    )
Round 4: M1W4 (****) M2W5 (M2w5) M3W1 (    ) M4W2 (m4W2)
Round 5: M1W5 (****) M2W1 (M2w1) M3W2 (m3W2) M4W3 (****)

W1: last round
W2: 1st round (supposing it is M2 who is infected)
W3: cannot escape
W4: cannot escape
W5: cannot escape

Round 1: M1W1 (    ) M2W2 (    ) M3W3 (M3W3) M4W4 (    )
Round 2: M1W2 (    ) M2W3 (m2W3) M3W4 (M3w4) M4W5 (    )
Round 3: M1W3 (m1W3) M2W4 (****) M3W5 (M3w5) M4W1 (    )
Round 4: M1W4 (****) M2W5 (****) M3W1 (M3w1) M4W2 (    )
Round 5: M1W5 (****) M2W1 (****) M3W2 (M3w2) M4W3 (m4W3)

W1: cannot escape
W2: last round
W3: 1st round (supposing it is M3 who is infected)
W4: cannot escape
W5: cannot escape

Round 1: M1W1 (    ) M2W2 (    ) M3W3 (    ) M4W4 (M4W4)
Round 2: M1W2 (    ) M2W3 (    ) M3W4 (m3W4) M4W5 (M4w5)
Round 3: M1W3 (    ) M2W4 (m2W4) M3W5 (****) M4W1 (M4w1)
Round 4: M1W4 (m1W4) M2W5 (****) M3W1 (****) M4W2 (M4w2)
Round 5: M1W5 (****) M2W1 (****) M3W2 (****) M4W3 (M4w3)

W1: cannot escape
W2: cannot escape
W3: last round
W4: cannot escape
W5: cannot escape

Round 1: M1W1 (    ) M2W2 (    ) M3W3 (    ) M4W4 (    )
Round 2: M1W2 (    ) M2W3 (    ) M3W4 (    ) M4W5 (m4W5)
Round 3: M1W3 (    ) M2W4 (    ) M3W5 (m3W5) M4W1 (M4W1)
Round 4: M1W4 (    ) M2W5 (m2W5) M3W1 (****) M4W2 (M4w2)
Round 5: M1W5 (m1W5) M2W1 (****) M3W2 (****) M4W3 (M4w3)

W1: cannot escape
W2: cannot escape
W3: last round
W4: escapes
W5: infected by definition

 

Strategy
 

Spoiler

 

The optimal strategy requires to wear the protection in the last round, quite evident.

M1 infected, escape for: W4, W5    1/9 * 2/5 = 2 / 45
W1 infected, escape for: W4, W5    1/9 * 2/5 = 2 / 45
M2 infected, escape for: W1        1/9 * 1/5 = 1 / 45
W2 infected, escape for: W1        1/9 * 1/5 = 1 / 45
M3 infected, escape for: W2        1/9 * 1/5 = 1 / 45
W3 infected, escape for: W2        1/9 * 1/5 = 1 / 45
M4 infected, escape for: W3        1/9 * 1/5 = 1 / 45
W4 infected, escape for: W3        1/9 * 1/5 = 1 / 45
W5 infected, escape for: W3 + W4   1/9 * 2/5 = 2 / 45

Total                                         12 / 45
=====                                         =======

Men have no chance, just one other woman will escape, if:
- W5 is infected and
- she is W4, giving:

1/9  *  1/4  =  1/36  = 1/12

=> near 8 2/3 members of the group of nine get infected.

e. o. e.

 

I knew it would be hard, but not THAT hard.

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So far, so good?
 

Spoiler

 

Case is M1 or W1 infected.

In brackets:
- empty: none infected on the end of the round
- M1W1:  M1 infects W1 or inversely
- m4W1:  m4 is infected by W1
- ****:  both already infected

Round 1: M1W1 (M1W1) M2W2 (    ) M3W3 (    ) M4W4 (    )
Round 2: M1W2 (M1w2) M2W3 (    ) M3W4 (    ) M4W5 (    )
Round 3: M1W3 (M1w3) M2W4 (    ) M3W5 (    ) M4W1 (m4W1)
Round 4: M1W4 (M1w4) M2W5 (    ) M3W1 (m3W1) M4W2 (****)
Round 5: M1W5 (M1w5) M2W1 (m2W1) M3W2 (****) M4W3 (****)

W1: protection in the 1st round (supposing it is M1 who is infected)
W2: cannot escape
W3: cannot escape
W4: 1st round
W5: last round

At least, did I understand the problem?

 

 

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To be clear the sexual activity will occur in a round robin sense

Edited by BMAD

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Aah. I have been thinking long and hard about this. But no solution comes to hand.
I'm intrigued as to what "a woman like you" is. But I find it refreshing that a lady (such as you) can be so open about their lives.
BMAD, I think I love you
Just FYI, In my avatar, I'm the one one the right. Or left, if you prefer.
xxxx

Edited by fabpig

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The "woman like me" was simply meant to let the reader know that they are to be of that subset and not the male group. Given the different population sizes, it does play into the approach. It is to tell nothing else about my personal habits.

Edited by BMAD

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my guess would be, assuming your having sex with both genders

14/2.718 = 5, so the fifth one.

Edited by phil1882

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To be clear the sexual activity will occur in a round robin sense

What does it mean?

I have never been to an orgy ;)

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after googling it is evident that: Every man and every woman have relations with each other at least once at a party.

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s***! did i choose the wrong disk??!!

I dont understand.

But orgy:

Round 1 everyone hooks up with 1 person, odd person out watches

round 2 everyone hooks up with someone else

Continue in this fashion until all the boys and girls were with each other.

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Since everybody must hook up with every individual of the opposite sex, there needs to be at least 5 "rounds". Since there are 5 girls and 4 guys, each round one girl will not participate. If you are able to schedule the round that you sit out, then you would want to do so on either the 4th or 5th round. Use protection on your last active round (either the 5th or the 4th) to maximize the likelihood of not catching the std.

Using the above strategy, you should only have a 9/16 chance of catching the std.

I would think that anyone who participates in the orgy without protection would have a 100% chance of catching the std.

Edited by BobbyGo

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14 hours ago, harey said:

So far, so good?
 

  Hide contents

 

Case is M1 or W1 infected.

In brackets:
- empty: none infected on the end of the round
- M1W1:  M1 infects W1 or inversely
- m4W1:  m4 is infected by W1
- ****:  both already infected

Round 1: M1W1 (M1W1) M2W2 (    ) M3W3 (    ) M4W4 (    )
Round 2: M1W2 (M1w2) M2W3 (    ) M3W4 (    ) M4W5 (    )
Round 3: M1W3 (M1w3) M2W4 (    ) M3W5 (    ) M4W1 (m4W1)
Round 4: M1W4 (M1w4) M2W5 (    ) M3W1 (m3W1) M4W2 (****)
Round 5: M1W5 (M1w5) M2W1 (m2W1) M3W2 (****) M4W3 (****)

W1: protection in the 1st round (supposing it is M1 who is infected)
W2: cannot escape
W3: cannot escape
W4: 1st round
W5: last round

At least, did I understand the problem?

 

 

so far, so good

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Spoiler

 

Listing all the outcomes (after excluding cases when you take the place of the infected Wn), it comes out that you always get off if you use the protection in the last round - provided you are lucky to be on the right place. (Not very surprising, as the disease spreads, the chances to get infected rise.) That makes 1/5. (A little bit more if you happen to be W4, see below.)

All the men get infected (which was clear from the beginning) and three women certainly.

If W5 (the woman just looking in the first round) is that one that is infected, W4 (the woman preceding her) escapes. That makes 1/8. (To subtract the compound p that you are W4.)

So finally, we will come very close to 8.75 infected members of the group of 9.

 

 

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