You lead with a tenth card, knowing that the chance your opponent has a 5 card and will play it knowing your need to peg is slim. Odds are more in your favor if you are holding at least one 5.
Your hand consists of four cards and you know which two cards you discarded, which leaves 46 unknown cards. Of the six cards originally in your hand, what are the chances that one of your opponent’s six cards is a 5, considering you have one? The percentage of 5s left in the unknown 46 cards is 3 divided by 46 or 6.5%.
Let’s also say that you have three tenth cards in your hand, which means that there are thirteen tenth cards left out of the 46 unknown cards. What is the possibility that at least one of your opponent’s six cards will be a tenth card? The percentage of tenth cards left in the remaining 46 cards is 13 divided by 46 or 28.3%.
You have four holes that still need to be pegged. This is a great opportunity to get two points quickly so that you only have to worry about two holes and not the full four points.
As the above tenth card example shows, it is less likely that your opponent will have a 5 than a tenth card. Not only that, but if you were your opponent, would you fifteen a tenth card knowing that your opponent probably has a 5 to go along with the tenth card and needs to peg? So you are left with the 5 as the lead.
Categories: Rules and Tips