The Math Factor Podcast

EX. Gambler’s Ruin!

January 6, 2009 · The Mathcast · Permalink

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After taping this segment, Kyle and I discovered an incredible way to make LOTS of money. We will extend this amazing opportunity to our listeners next week!

 

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3 Comments »

1. indy500 said,

   January 7, 2009 at 7:20 pm

   Listening to the podcast, I think you could make a lot of money on the roulette wheel Kyle mentioned. I think there are just slightly less than 50% black and red spaces on a typical roulette wheel (2 green spaces?) and the person mentioned made $400 off of a $100 bet. I’ll take those odds/payout!

2. indy500 said,

   January 7, 2009 at 9:20 pm

   Here is my submission for what I think the solutions are. Show Spoiler ▼

   For the games, let the number of heads obtained before a tails comes up be ‘n’. Then let the payout for a value n be y(n). And let the probability of getting exactly n tosses be p(n). For these games, p(n) = (1/2)^(n+1). This is because in order to get exactly n heads, you have to have a tail occur on the next toss (gives you the ‘+1′ in the exponent). The expected value is just the sum over all n of y(n)*p(n). For the first game, the payout is equal to the number of heads tossed, or n. So the expected value is the sum over all n of [n * (1/2)^(n+1)]. if we pull out 1/2, then the equation becomes (1/2)*[n * (1/2)^n]. The sum the equation in brackets is a power series and is equal to [(1/2)/{(1-1/2)^2}]. Multiplying by the 1/2, that simplifies to 1, so the first game should cost you $1 to get in. For the second game, the payout is equal to 2^n – 1. So the expected value is the sum over all n of the payout: (2^n – 1)*(1/2)^(n+1). Moing the terms around, the payout can be written as (1/2) * [2^n - 1]/(2^n), or 1/2 * [1 - (1/2^n)]. When you sum this over n to get the expected value, the first term in the brackets is the sum over all n of 1, which is infinity. The second term is a power series which is equal to 2, but 1/2 * [infinity - 2] is still infinity. So, no matter what the entrance fee, you should always play the second game. However, when I look at the second game in excel, for each value n, you have very low odds of getting to the really high payouts. A payout of $31 occurs about 1.5% of the time. A payout of $1023 occurs about 0.05% of the time (10 tosses in a row of heads, followed by a tail). So, if you have to pay a lot to get in, you are going to have to be in the game for a very long time – you need a big bankroll!

3. 4jacks said,

   January 14, 2009 at 3:15 pm

   Hey guys,
   Little Roulette Info, there are 38 spaces, 18 red and 18 black.
   So the odds of Hitting red or black is 47.37%
   The payout for hitting red or black is 100%
   So the man who bet $100 on red won his $100 back and an extra $100

   I don’t gamble and I don’t condone it, but I do agree with the man’s theory.  If you look at the chances of each individual spin, yes it’s 47.37% chance of red or black.

   But if you look at a series of spins, it’s a little different.  For simplification purpose pretend there is no Green 0 and 00 and only red and black.

   So for a series of four spins to all be black is a 6.25% chance
   A series of five spins to all be black is a 3.125% chance
   A series of six spins to all be black is 1.5625% chance
   and so on. 

   So what the man was doing was waiting and watching for the 3.125% chance to happen, becuase it’s only a matter of time before it does. 

   When he saw that the 3.125% chance did happen, he bet money that the 1.5625% chance WOULD NOT happen.

   Had he LOST his money and the 1.5625% chance DID happen, then he would probably double up his bet on black and bet $200.  Becuase now he has there is a 0.78125% chance that any 7 spins will come out all red.

   It’s true that in theory if the wheel spins infinite number of times, eventually you will have really long streaks of nothing but one color.  But it’s still a safer bet than walking in blind.  

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