Rachel and the Hustler

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Mark Kudlowski
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Rachel and the Hustler

Post by Mark Kudlowski » Wed Aug 06, 2014 8:08 pm

(I'd posed this as a GCSE question.)

Rachel is in Vegas, and notices a hustler's stall outside one of the casinos.

Hustler: "There are two normal pennies in here, an I'm offering 2-1 on you getting two heads when tossed."
Rachel: "I refuse those odds. They're biased in your favour."
Hustler: "There are three different results, two heads, two tails, or a head and a tail. So 2-1 is fair."
Rachel: "Right then. Can you offer me 2-1 on a head and a tail, then ? "

(Hustler thinks and diverts the question)

Hustler: "All right. There are ten normal pennies in here, and I'll give you the same 2-1 odds for getting 5 heads out of 10. I mean, half of 10 is 5, so I should only be offering you evens."
Rachel: "No. The chance of getting 5 heads out of 10 is even lower than getting 2 heads out of 2."

Show that Rachel was correct on both games.

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Re: Rachel and the Hustler

Post by Sean Fletcher » Thu Aug 07, 2014 12:03 am

Based on the rules of probability every time you toss a coin it's an independent event. You have a 1 in 2 chance of hitting a head or tail each time. So to hit 2 heads it would be 1/2*1/2 equalling 1/4 which is odds of 4-1.
For the 5 heads from ten coins the best chance is 1/2*1/2*1/2*1/2*1/2 equalling 1/32 or 32-1 in terms of odds. However, that relies on hitting 5 heads in the first 5 goes. If that didn't happen the odds would go up in this order: 64-1; 128-1; 256-1; 512-1; 1024-1.

I hope that's right

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Re: Rachel and the Hustler

Post by Fred Mumford » Thu Aug 07, 2014 7:45 am

My understanding is that a 1 in 32 chance would be classed as 31-1 in terms of fair odds.

Do we mean exactly 5 heads or at least 5? For exactly 5 then the probability would be 0.5 to the power of 10, multiplied by the number of different ways this can be achieved eg the heads might be tosses 1 2 3 4 5 or tosses 2 3 6 7 9 etc etc. At school we used Pascal's triangle to arrive at this number. Can't be arsed checking it now but that's the principle anyway.

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Re: Rachel and the Hustler

Post by Fred Mumford » Thu Aug 07, 2014 9:03 am

Change of heart. I now can be arsed.

The relevant number from the triangle is 252, so the probability of exactly 5 heads out of 10 is 24.61%.

I therefore assume this is what the original question was getting at, as the probability of at least 5 is just over 62%.

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Re: Rachel and the Hustler

Post by Mark Kudlowski » Thu Aug 07, 2014 9:17 am

Fred Mumford wrote:Change of heart. I now can be arsed.

The relevant number from the triangle is 252, so the probability of exactly 5 heads out of 10 is 24.61%.

I therefore assume this is what the original question was getting at, as the probability of at least 5 is just over 62%.


Rachel was correct in both cases :

The prob. of 2 heads out of 2 is 1/4, so the true odds would be 3-1 against, not 2-1.
As a head and a tail can be made in 2 ways out of a possible 4, the true odds would be evens,
which is why the hustler changed his mind when Rachel suggested 2-1 for a head and a tail.

As for the case of 5 heads in 10, the Pascal's triangle entry is 252, so the prob. of exactly 5 heads in 10 is 252/1024 or 63/256, which is very slightly less than 1 in 4. The fair odds would be 193 to 63, or about 3.06 to 1.

This is an example of binomial distribution, for all statisticians among you.

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Re: Rachel and the Hustler

Post by Sean Fletcher » Thu Aug 07, 2014 1:30 pm

I was quite wrong with my answer then :( Great question though Mark

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Re: Rachel and the Hustler

Post by sean d » Thu Aug 07, 2014 1:34 pm

This has to go down as the most disappointing thread ever...

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Re: Rachel and the Hustler

Post by Ian Volante » Fri Aug 08, 2014 12:25 pm

sean d wrote:This has to go down as the most disappointing thread ever...
I recommend the recent one about the penis programme then - it easily surpasses this one.
meles meles meles meles meles meles meles meles meles meles meles meles meles meles meles meles

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