I know, 50/50 heads tails right? But help me understand the next step – each coin flip has a 50/50 shot of heads or tails. What I don’t understand is how the likelihood of the next flip doesn’t change. For example if I flipped a coin 10 times and every time it flipped heads, the next flip would be 50/50 tails. Wouldn’t the likelihood of flipping a coin 11 times and having it be heads every time be really low? 0.5^11 = 0.048%?
Here’s the origin of the question. I was at a roulette table and the guy said “it’s been black the last 8 rolls, the next one has to be red.” At first I thought, the next roll will be ~47% black, ~47 red, ~6% 0 or 00 you fucking imbecile. Then I thought to myself, what are the chances that there are no red rolls in 9 rolls, which is well below 1%.
Am I the imbecile?
In: 215
I’m reminded of the Martingale betting system. It basically states you bet one unit (for our sake let’s just say our bet is $1), and then double it when you lose.
In this way if you bet and win, you are up $1. If you bet and lose then double your bet to $2. If you lose again, double it to $4. When you win, In this case $4, you are actually only up your original bet. You won $4, but lost $1 and $2. So net $4.
So if this was your betting system, you could analyze it and determine that if I had $63, which is enough for 6 bets ( 1,2,4,8,16,32) what is your chance of total failure. So 6 lost bets in a row. Well for roulette let’s pretend betting black is 50%. It isn’t as you know because of 0 and 00 but I want easy math.
6 losses in a row would have a probability of 0.5^6 ~ 1.6%. So my odds of success would be about 98.4%, of ending up ahead $1.
That’s the expectation. BUT when I’m at the casino, and I’m that 6th bet, with my $32 down on black, my odds are only 50% for this result not 98%. We are here now. There was a 50% for each spin to come up black, but it hasn’t. But each spin had the same odds.
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