Joined: Feb 7, 2012
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March 6th, 2020 at 2:22:37 PM permalink
I just had 2 very unlikely events happen to me on back to back nights. I would like to know the odds of each occurring if someone would like to do the math for me.

Night 1) I spent a few hours betting on some races at the live racetrack and on my way back to my truck while passing the craps table, I saw a bunch of guys I knew playing, so I decided to play a while while saying hello and not having to go to the ATM. I had $11 cash in pocket and bought in for a whopping $11. I placed a $5 bet on the Don't Pass Line, a point of 6 rolls, and then I lay $6 in odds. I am immediately with my $11 all in. I luckily end up winning the bet and now have a whopping $21. I ended up playing for about 45 minutes with that $11 and turning it into $90 when I cashed out. I felt like that was an extremely lucky 45 minutes and wondered what the odds were of actually turning $11 into $90 before busting playing don't pass at craps. Is that something you could even calculate?

Night 2) I show up to actually play craps, so I bring a $300 bankroll. 4 times in a row, the point is either a 10 or 4. All 4 times I lay $30 in odds with my $5 flat bet and then hedge the hard way for $5. So I net $15 if a 7 comes, and break even if the point comes the hard way. The only way I get beat is 2 combinations out of 36. The point coming easy! Only 1 in 18 combinations of the dice can beat me. I lost all 4 bets, in a row, by the 4 OR 10 coming EASY! I considered this just as unlucky as the luck was the night before!

I'd consider my math skills above average for an average American, but below average for the quality of minds in this forum. I have no idea how to calculate scenario 1. I tried to calculate the odds of scenario 2 and came up with 0.16%. There are 2 in 36 combinations that could beat me. 8 in 36 combinations that I win or break even. Meaning I should only lose, on average, 1 time out of 5 rolls or 20%. I then figured the odds of a 20% event happening 4 times in a row is 20% to the 4th power or 0.16% Do I do this correct?

Any help with scenario 1?
Joined: Jun 1, 2014
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March 6th, 2020 at 4:01:39 PM permalink
Q1. About 1 in 8, ignoring house edge.
Applies to any low edge wager session where the only possibilities are success or total loss.
Take care out there. Spare a thought for the newly poor who were happy in their world just a few days ago, but whose whole way of life just collapsed..
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March 6th, 2020 at 10:31:21 PM permalink
You have a 949/1925 = p probability of winning a Donít Pass bet. (1 - p) / p = 976/949 = r

Betting $5 and starting with a bankroll of $10 (2 units), the chance of turning that into $90 (18 units) is:

(1 - r^2) / (1 - r^18) = 8.78%
Last edited by: Ace2 on Mar 7, 2020
Itís all about making that GTA

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