20:1 Odds on a Monotone Flop - Prop Bet from Live Game

[lowpockets]

Back story:
Last night I was forced to play a 10-20 limit game live (shoot me please - I play pretty much exclusively NL). Although I know very little about limit, I forced myself to play (when the $1000NL game didn't go) because it looked like easy money (pshhhh why not).

Ok, so at one point this old guy to my right starts claiming that when he has two suited cards, the flop comes out monotone of a different color about 5 out of 10 times. He was a nice guy, but wouldn't shut up about it and kept asking what the odds of a monotone flop were. I thought about it in my head briefly and figured it was around 5%. I know my math was definitely flawed. It went something like this:

12/51*11/50 ~ .05

I know this number isn't completely accurate (closer to .06), butI finally told the old man that I would give him 20:1 that the flop wouldn't be monotone... who was getting a good deal?

The other twist to the scenario is that the old man, after losing a few times in a row, told me that he would didn't want to make the bet every hand. He said he would tell me when he wanted to make the bet and he only made it when he had two suited cards. How does this change the probability of a monotone flop of any suit when you have two random suited cards?

Thanks for reading this far - any thoughts are appreciated, its been a few years since my last stats class.

[jay_shark]

If you have monotone cards in spades without loss of generality , then the probability the flop comes monotone in spades is :

11C3/50c3=0.008418

monotone in any of the three other suits :
13C3/50c3= 0.0145

Therefore , the probability the flop comes monotone in any of the suits when you hold two suited cards is :

3*13c3/50c3 + 11c3/50c3 = 0.0521

Otherwise , if you don't hold two monotone hold cards , then the probability the flop is monotone is :

2*12c3/50c3 + 2*13c3/50c3 = 0.05163

In the former case , you need to offer him odds of (100-5.21)/5.21=18.19:1 to break even .

For the other case , you need to offer him odds of (100-5.16)/5.16= 18.37:1 which isn't much different .

[jay_shark]

This means that if you give him odds of 20:1 , then he only needs to win at least 1/21 = 0.047 for it to be a good bet for him . This would favor the old man in both cases .