The HIV Paradox: A Tale of Two Males

[Voltron87]

[ QUOTE ]
So I can [censored] random women without a condom, including hookers, and I won't get the HIV? SWEET!!!

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high five, dude. the cops wont hassle us either.

"did you see who stole that car?"
"uh, sort of, I wasn't able to see his face but I could tell he was puerto rican"

problem solved.

[shant]

I think you meant strippers, not prostitutes.

[Talk2BigSteve]

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Crotch Crickets

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That phrase is disgusting. Congratulations.

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Welcome to Word Heard From Rednecks 101.

Steve

[BigBaitsim (milo)]

150 times a year. Not a married guy, clearly.

[CallMeIshmael]

[ QUOTE ]
I think you meant strippers, not prostitutes.

[/ QUOTE ]

HAHAHAH


FWIW, this idea started along the lines of:

Dear Bison,

What are the chances Jason_t is bring some friends home from Vegas?



Then, I remembered that I read about this 'paradox' once, and decided to go this way.


Its interesting, because, for the most part, people mention the safeness of the partner far more than the frequency of the act when discussing the chances of HIV contraction.

[CallMeIshmael]

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interesting choice of names...

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Really? How so? [img]/images/graemlins/tongue.gif[/img]

[gumpzilla]

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Its interesting, because, for the most part, people mention the safeness of the partner far more than the frequency of the act when discussing the chances of HIV contraction.

[/ QUOTE ]

For a sufficiently safe partner, the frequency of the act will become irrelevant.

[CallMeIshmael]

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For a sufficiently safe partner, the frequency of the act will become irrelevant.

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Define 'sufficiently safe'

[1800GAMBLER]

I checked the math, seems right.

0.6% chance of HIV shocks me, a lot. 1/167 americans have HIV?

0.002% chance of getting HIV with an infected partner shocks me too, i thought this would be way higher.

[mmbt0ne]

</font><blockquote><font class="small">En réponse à:</font><hr />

Jordan: He is in an 'Either/Or' situation. If Sarah doesnt have HIV, then he will never contract HIV over that year. IF she is, however, the probability of him contracting HIV =

1 - (499/500)^180 = 0.3026

Since, Jordan estimates the probability of Sarah having HIV is 0.3%, his final chance of contracting is:

= 0.0907%

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Ok, fine.

</font><blockquote><font class="small">En réponse à:</font><hr />
Jason: His equation looks like:

1 - P(Getting HIV)

= 1 -

(1 - (1.2/100*1/500))^20*(1-(4/100)*(1/500))^6

= 0.0959%

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BUZZ!!!!

Sorry, but now you're doing it differently. Now you have the percentage she is HIV+ included with the base of the exponential.

OK, I'M CHANGING THIS.

Correct form for Jason:

Bar Skanks
1-(499/500)^20 = 0.0392
0.0392*0.012 = 0.0004710

Whores
1-(499/500)^6 = 0.0119
0.0119*0.04 = 0.0004776

TOTAL
0.0004710+0.0004776 = 0.0949%


Don't be so mean to Jason, he's not that much more likely at all. Of course, Jordan's situation could be resolved very easily, and he would know automatically how safe it is to bang away.